Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy

Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy

Mastering Physics Solutions

Chapter 8 Potential Energy And Conservation Of Energy Q.1CQ
Is it possible for the kinetic energy of an object to be negative? Is it possible for the gravitational potential energy of an object to be negative? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy1cqs

Chapter 8 Potential Energy And Conservation Of Energy Q.1P
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Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy1ps

Chapter 8 Potential Energy And Conservation Of Energy Q.2CQ
An avalanche occurs when a mass of snow slides down a steep mountain slope. Discuss the energy conversions responsible for water vapor rising to form clouds, falling as snow on a mountain, and then sliding down a slope as an avalanche.
Solution:
As water vapor rises, there is an increase in the gravitational potential energy of the system. Part of this potential energy is released as snow and falls onto the mountain. If an avalanche occurs, the snow on the mountain accelerates down the slope, converting more gravitational potential energy into kinetic energy.

Chapter 8 Potential Energy And Conservation Of Energy Q.2P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.3CQ
If the stretch of a spring is doubled, the force it exerts is also doubled. By what factor does the spring’s potential energy increase?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy3cqs1

Chapter 8 Potential Energy And Conservation Of Energy Q.3P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.4CQ
When a mass is placed on top of a vertical spring, the spring compresses and the mass moves downward. Analyze this system in terms of its mechanical energy.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy4cqs

The initial mechanical energy of the system is the gravitational potential energy of the mass-Earth system. As the mass moves downward, the gravitational potential energy of the system decreases.
At the same time, the potential energy of the spring increases because it is compressed. Initially, the decrease in gravitational potential energy is greater than the increase in the spring’s potential energy, which means that the mass gains kinetic energy. Eventually, the increase in the spring’s energy equals the decrease in the gravitational energy, and the mass comes to rest.

Chapter 8 Potential Energy And Conservation Of Energy Q.4P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.5CQ
If a spring is stretched so far that it is permanently deformed, its force is no longer conservative. Why?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy5cqs

We know that the external force must be equal to the restoring force, and its direction is opposite to the direction of the restoring force. If the external force is greater than the restoring force, then the spring gets permanently deformed. In this situation, the work that was done to stretch the spring is not fully recovered. Some of the work is converted into the energy of the deformation. For this reason, the spring force is not conservative during deformation.

Chapter 8 Potential Energy And Conservation Of Energy Q.5P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.6CQ
An object is thrown upward to a person on a roof. At what point is the object’s kinetic energy at maximum? At what point is the potential energy of the system at maximum? At what locations do these energies have their minimum values?
Solution:
When the object is first thrown upward, its speed and its kinetic energy are at a maximum. Its potential energy is zero at that moment.
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy6cqs
(i) The potential energy of the system is at a maximum at the highest point of the object’s flight and is at a minimum at the starting point of its journey (when it has just been released).
(ii) The kinetic energy of the system is at maximum when the object has just been thrown up and is at a minimum when it reaches its highest point of flight.

Chapter 8 Potential Energy And Conservation Of Energy Q.6P
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy6p
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.7CQ
It is a law of nature that the total energy of the universe is conserved. What do politicians mean, then, when they urge “energy conservation”?
Solution:
When the term “energy conservation” is used in everyday language, it doesn’t refer to the total amount of energy in the universe. Instead it refers to using energy wisely, especially when a particular source of energy like oil or natural gas is finite and non-renewable.

Chapter 8 Potential Energy And Conservation Of Energy Q.7P
Predict/Explain Ball 1 is thrown to the ground with an initial downward speed; ball 2 is dropped to the ground from rest. Assuming the balls have the same mass and are released from the same height, is the change in gravitational potential energy of ball 1 greater than, less than, or equal to the change in gravitational potential energy of ball 2? (b) Choose the best explanation from among the following:
I. Ball 1 has the greater total energy, and therefore more energy can go into gravitational potential energy.
II. The gravitational potential energy depends only on the mass of the ball and the drop height.
III. All of the initial energy of ball 2 is gravitational potential energy.
Solution:
(a) The change in gravitational potential energy of the ball 1 is equal to the change in gravitational potential energy of the ball 2.
(b) This is because the change in gravitational potential energy depends only on the mass of the ball and the height from which the ball is dropped. Therefore option II is the best explanation.

Chapter 8 Potential Energy And Conservation Of Energy Q.8CQ
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy8cq
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy8cqs
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Chapter 8 Potential Energy And Conservation Of Energy Q.8P
A mass is attached to the bottom of a vertical spring. This causes the spring to stretch and the mass to move downward. (a) Does the potential energy of the spring increase, decrease, or stay the same during this process? Explain. (b) Does the gravitational potential energy of the Earth-mass system increase, decrease, or stay the same during this process? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy8ps
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Chapter 8 Potential Energy And Conservation Of Energy Q.9CQ
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy9cq
Solution:
We have a photograph which shows an earthmover digging the earth and lifting it and loading it on the truck. In this situation, there are both conservative and non-conservative works involved.
Initially, the engine of the earthmover does positive, non-conservative work as it digs out and lifts a load of rocks. At the same time, gravity does negative, conservative work on the rocks as the gravitational potential energy of the system increases. As the rock is transported to the truck, the earthmover does positive, non-conservative work. When the rocks are released, gravity does positive, conservative work as the gravitational potential energy of the system is converted into kinetic energy. The kinetic energy of the rocks is converted into sound and heat when they are loaded into the truck.
As the truck is loaded, the spring is compressed. The potential energy is stored in the form of the spring’s potential energy, and the work done by the truck is conservative.

Chapter 8 Potential Energy And Conservation Of Energy Q.9P
As an Acapulco cliff diver drops to the water from a height of 46 m, his gravitational potential energy decreases by 25,000 J. What is the diver’s weight in newtons?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy9ps

Chapter 8 Potential Energy And Conservation Of Energy Q.10CQ
A toy frog consists of a suction cup and a spring. When the suction cup is pressed against a smooth surface, the frog is held down. When the suction cup lets go, the frog leaps into the air. Discuss the behavior of the frog in terms of energy conversions.
Solution:
The toy frog consists of a suction cup and a spring. When the suction cup is pressed, the energy of the toy is stored in the form of the spring as potential energy. When the suction cup lets go, then all its potential energy is converted into kinetic energy. As a result, the frog leaps into the air. The total energy of the system is conserved.

Chapter 8 Potential Energy And Conservation Of Energy Q.10P
Find the gravitational potential energy of an 88-kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as the location for y = 0.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy10ps

Chapter 8 Potential Energy And Conservation Of Energy Q.11CQ
If the force on an object is zero, does that mean the potential energy of the system is zero? If the potential energy of a system is zero, is the force zero?
Solution:
No
Zero force implies a zero rate of change in the potential energy. However, the value of the potential energy can be anything at all. Similarly, if the potential energy is zero, it does not mean that the force is zero. Again, what matters is the rate of change of the potential energy.

Chapter 8 Potential Energy And Conservation Of Energy Q.11P
Jeopardy! Contestante on the game show Jeopardy! depress spring-loaded buttons to “buzz in” and provide the question corresponding to the revealed answer. The force constant on these buttons is about 130 N/m. Estimate the amount of energy it takes—at a minimum—to buzz in.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.12CQ
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy12cq
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.12
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.13CQ
When a ball is thrown upward, it spends the same amount of time on the way up as on the way down—as long as air resistance can be ignored. If air resistance is taken into account, is the time on the way down the same as, greater than, or less than the time on the way up? Explain.
Solution:
If the air resistance is taken into account, then the total mechanical energy of the system decreases. The distance covered by the ball is the same on the way down as it is on the way up, and so the amount of time will be determined by the average speed of the ball on the two portions of its trip. Note that air resistance does negative, non-conservative work continuously on the ball as it moves. Therefore, its total mechanical energy is less on the way down than it is on the way up, which means that its speed at any given elevation is less on the way down. It follows that more time is required for the downward portion of the trip.

Chapter 8 Potential Energy And Conservation Of Energy Q.13P
A vertical spring stores 0.962 J in spring potential energy when a 3.5-kg mass is suspended from it. (a) By what multiplicative factor does the spring potential energy change if the mass attached to the spring is doubled? (b) Verify your answer to part (a) by calculating the spring potential energy when a 7.0-kg mass is attached to the spring.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.14P
Pushing on the pump of a soap dispenser compresses a small spring. When the spring is compressed 0.50 cm, its potential energy is 0.0025 J. (a) What is the force constant of the spring? (b) What compression is required for the spring potential energy to equal 0.0084 J?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy14ps

Chapter 8 Potential Energy And Conservation Of Energy Q.15P
A force of 4.1 N is required to stretch a certain spring by 1.4 cm. (a) How far must this spring be stretched for its potential energy to be 0.020 J? (b) How much stretch is required for the spring potential energy to be 0.080 J?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.16P
The work required to stretch a certain spring from an elongation of 4.00 cm to an elongation of 5.00 cm is 30.5 J. (a) Is the work required to increase the elongation of the spring from 5.00 an to 6.00 cm greater than, less than, or equal to 30.5 J? Explain. (b) Verify your answer to part (a) by calculating the required work.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy16ps

Chapter 8 Potential Energy And Conservation Of Energy Q.17P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.18P
Predict/Explain You throw a ball upward and let it fall to the ground. Your friend drops an identical ball straight down to the ground from the same height. Is the change in kinetic energy of your ball greater than, less than, or equal to the change in kinetic energy of your friend’s ball? (b) Choose the best explanation from among the following:
I. Your friend’s ball converts all its initial energy into kinetic energy.
II. Your ball is in the air longer, which results in a greater change in kinetic energy.
III. The change in gravitational potential energy is the same for each ball, which means the change in kinetic energy must be the same also.
Solution:
(a) The Change in kinetic energy of your ball is equal to the change in kinetic energy of your friend’s ball.
(b) As both of you are at the same height, so when the balls come to the ground both balls lose same amount of potential energy. Therefore change in potential energy of the two balls is same. Then from conservation of energy the change in kinetic energy is also same.
So, option III is the best explanation.

Chapter 8 Potential Energy And Conservation Of Energy Q.19P
Suppose the situation described in Conceptual Checkpoint 8-2 is repeated on the fictional planet Epsilon, where the acceleration due to gravity is less than it is on the Earth. (a) Would the height of a hill on Epsilon that causes a reduction in speed from 4 m/s to 0 be greater than, less than, or equal to the height of the corresponding hill on Earth? Explain. (b) Consider the hill on Epsilon discussed in part (a). If the initial speed at the bottom of the hill is 5 m/s, will the final speed at the top of the hill be greater than, less than, or equal to 3 m/s? Explain.
Solution:
Solution:
(a) As we are considering on a planet Epsilon where the acceleration due to gravity is less than that on the earth and as the gravitational potential energy is mgh so larger height would be needed to gain the same level of gravitational potential energy. Thus, a higher hill would be needed.
(b) The initial speed at the bottom of the Epsilon is 5m/s which are the same on the earth. Since the values of the initial kinetic energy & final potential energies are the same on the Epsilon.
So therefore the final speed at the top of the hill on Epsilon is the same which is 3m/s

Chapter 8 Potential Energy And Conservation Of Energy Q.20P
Predict/Explain When a ball of mass m is dropped from. rest from a height h, its kinetic energy just before landing is K. Now, suppose a second ball of mass Am is dropped from rest from a height h/4. (a) Just before ball 2 lands, is its kinetic energy 4K, 2K, K, K/2, or K/4? (b) Choose the best explanation from among the following:
I. The two balls have the same initial energy.
II. The more massive ball will have the greater kinetic energy.
III. The reduced drop height results in a reduced kinetic energy.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy20ps
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Chapter 8 Potential Energy And Conservation Of Energy Q.21P
Predict/Explain When a ball of mass m is dropped from rest from a height h, its speed just before landing is v. Now, suppose a second ball of mass 4m is dropped from rest from a height h/4. (a) Just before ball 2 lands, is its speed 4v, 2v, v, v/2, or v/4? (b) Choose the best explanation from among the following:
I. The factors of 4 cancel; therefore, the landing speed is the same.
II. The two balls land with the same kinetic energy; therefore, the ball of mass 4m has the speed v/2.
III. Reducing the height by a factor of 4 reduces the speed by a factor of 4.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.22P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.23P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.24P
At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.31 m, what is her speed at the bottom of the slide?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.25P
In the previous problem, find the swimmer’s speed at the bottom of the slide if she starts with an initial speed of 0.840 m/s.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy25ps

Chapter 8 Potential Energy And Conservation Of Energy Q.26P
A player passes a 0.600-kg basket ball downcourt for a fast break. The ball leaves the player’s hands with a speed of 8.30 m/s and slows down to 7.10 m/s at its highest point. (a) Ignoring air resistance, how high above the release point is the ball when it is at its maximum height? (b) How would doubling the ball’s mass affect the result in part (a)? Explain.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.27P
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Solution:
Here the all the three balls are at same height when they reached to the dashed line. And all the balls started from same height. Therefore the gain in potential energy of the three balls is same. Therefore the loss in kinetic energy of each ball is same. Also all the three balls have same initial speed. So all the three balls started with same speed and lost equal amount of kinetic energy. Therefore all the three balls have same speed at the dashes line. So, option (C) is correct.
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Chapter 8 Potential Energy And Conservation Of Energy Q.28P
In a tennis match, a player wins a point by hitting the ball sharply to the ground on the opponent’s side of the net. (a) If the ball bounces upward from the ground with a speed of 16 m/s, and is caught by a fan in the stands with a speed of 12 m/s, how high above the court is the fan? Ignore air resistance. (b) Explain why it is not necessary to know the mass of the tennis ball.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.29P
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Solution:
Here the all the three balls are at same height when they reached to the dashed line. And all the balls started from same height. Therefore the gain in potential energy of the three balls is same. Therefore the loss in kinetic energy of each ball is same. Also all the three balls have same initial speed. So all the three balls started with same speed and lost equal amount of kinetic energy. Therefore all the three balls have same speed at the dashes line. So, option (C) is correct.

Chapter 8 Potential Energy And Conservation Of Energy Q.30P
A 2.9-kg block slides with a speed of 1.6 m/s on a frictionless horizontal surface until it encounters a spring. (a) If the block compresses the spring 4.8 cm before coming to rest, what is the force constant of the spring? (b) What initial speed should the block have to compress the spring by 1.2 cm?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy30ps
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Chapter 8 Potential Energy And Conservation Of Energy Q.31P
A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.32P
A 1.40-kg block slides with a speed of 0.950 m/s on a frictionless horizontal surface until it encounters a spring with a force constant of 734 N/m. The block comes to rest after compressing the spring 4.15 cm. Find the spring potential energy, u, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for compressions of (a) 0 cm, (b) 1.00 cm, (c) 2.00 cm, (d) 3.00 cm, and (e) 4.00 cm.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.33P
A 5.76-kg rock is dropped and allowed to fall freely. Find the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for (a) the first 2.00 m of fall and (b) the second 2.00 m of fall.
Solution:
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If the mass of the bob increases, then the first part in the above equation decreases hence the speed of bob at point A increases. Therefore, the answer to part (b) increases as the mass increases.

Chapter 8 Potential Energy And Conservation Of Energy Q.34P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.35P
In the previous problem, (a) what is the bob’s kinetic energy at point B? (b) At some point the bob will come to rest momentarily. Without doing an additional calculation, determine the change in the system’s gravitational potential energy between point B and the point where the bob comes to rest. (c) Find the maximum angle the string makes with the vertical as the bob swings back and forth. Ignore air resistance.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.36P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.37P
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Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.38P
You coast up a hill on your bicycle with decreasing speed. Your friend pedals up the hill with constant speed. (a) Ignoring friction, does the mechanical energy of the you-bike-Earth system increase, decrease, or stay the same? Explain. (b) Does the mechanical energy of the friend-bike-Earth system increase, decrease, or stay the same? Explain.
Solution:
(A) When a person coasts up a hill on his bicycle with decreasing speed, no non-conservative work is done on his bicycle. Therefore, his mechanical energy is conserved even though his speed decreases.
(B)
To maintain a constant speed, his friend will have to do positive non-conservative work while going uphill. Thus, his friend’s mechanical energy will increase.

Chapter 8 Potential Energy And Conservation Of Energy Q.39P
Predict/Explain On reentry, the space shuttle’s protective heat tiles become extremely hot. (a) Is the mechanical energy of the shuttle-Earth system when the shuttle lands greater than, less than, or the same as when it is in orbit? (b) Choose the best explanation from among the following:
I. Dropping out of orbit increases the mechanical energy of the shuttle.
II. Gravity is a conservative force.
III. A portion of the mechanical energy has been converted to heat energy.
Solution:
The mechanical energy E of a system is the sum of potential energy U and kinetic energy K.
Non conservative forces might decrease the mechanical energy by converting it to heat energy, or increase it by converting muscular work to kinetic or potential energy.
(a)
The space shuttle’s protective heat tiles become extremely hot, means that some amount of mechanical energy is converted to heat energy due to some non-conservative forces, and total mechanical energy of the system will decrease.
Thus, the mechanical energy of shuttle-Earth system when the shuttle lands is that of the mechanical energy when the shuttle is in orbit.
(b)
The space shuttle’s protective heat tiles become extremely hot, means that some amount of mechanical energy has been converted to heat energy.
Thus, the correct option is .

Chapter 8 Potential Energy And Conservation Of Energy Q.40P
Catching a wave, a 77-kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much nonconservative work was done on the surfer?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.41P
At a playground, a 19-kg child plays on a slide that drops through a height of 2.3 m. The child starts at rest at the top of the slide. On the way down, the slide does a nonconservative work of —361 J on the child. What is the child’s speed at the bottom of the slide?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.42P
Starting at rest at the edge of a swimming pool, a 72.0-kg athlete swims along the surface of the water and reaches a speed of 1.20 m/s by doing the work Wnc1 = + 161 J. Find the nonconservative work, Wnc2, done by the water on the athlete.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.43P
A 17,000-kg airplane lands with a speed of 82 m/s on a stationary aircraft carrier deck that is 115 m long. Find the work done by nonconservative forces in stopping the plane.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy43ps

Chapter 8 Potential Energy And Conservation Of Energy Q.44P
The driver of a 1300-kg car moving at 17 m/s brakes quickly to 11 m/s when he spots a local garage sale. (a) Find the change in the car’s kinetic energy. (b) Explain where the “missing” kinetic energy has gone.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.45P
You ride your bicycle down a hill, maintaining a constant speed the entire time. (a) As you ride, does the gravitational potential energy of the you-bike-Earth system increase, decrease, or stay the same? Explain. (b) Does the kinetic energy of you and your bike increase, decrease, or stay the same? Explain. (c) Does the mechanical energy of the you-bike-Earth system increase, decrease, or stay the same? Explain.
Solution:
Solution:
(a) The gravitational potential energy of the earth-bike-rider system should decrease.
Since the position of the biker could eventually drop to ground level height, the potential energy of the system could go down to zero.
(b) Since the speed is constant throughout the entire time. The kinetic energy of you & your bike remains the same
(c) The mechanical energy of the system decrease. Since the total mechanical energy is equal to the sum of potential energy and kinetic energy. As the potential energy decreases & kinetic energy remains the same. So therefore the mechanical energy of the system decreases.

Chapter 8 Potential Energy And Conservation Of Energy Q.46P
Suppose the system in Example starts with m2 moving downward with a speed of 1.3 m/s. What speed do the masses have just before m2 lands?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.47P
A 42.0-kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.75 m higher than the surface of the water, and the ramp is inclined at an angle of 35.0° above the horizontal. If the seal reaches the water with a speed of 4.40 m/s, what are (a) the work done by kinetic friction and (b) the coefficient of kinetic friction between the seal and the ramp?
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.48P
A 1.9-kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 4.6 N is exerted on it by water resistance. Calculate the nonconservative work, Wnc, done by water resistance on the rock, the gravitational potential energy of the system, u, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, when the depth of the rock below the water’s surface is (a) 0 m, (b) 0.50 m, and (c) 1.0 m. Let y = 0 be at the bottom of the pond.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy48ps
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Chapter 8 Potential Energy And Conservation Of Energy Q.49P
A 1250-kg car drives up a hill that is 16.2 m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car’s engine. The work done by friction is −3.11 × 105 J; the work done by the engine is + 6.44 × 105 J. Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy49ps

Chapter 8 Potential Energy And Conservation Of Energy Q.50P
An 81.0-kg in-line skater does + 3420 J of nonconservative work by pushing against the ground with his skates. In addition, friction does −715 J of nonconservative work on the skater. The skater’s initial and final speeds are 2.50 m/s and 1.22 m/s, respectively. (a) Has the skater gone uphill, downhill, or remained at the same level? Explain. (b) Calculate the change in height of the skater.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy50ps

Chapter 8 Potential Energy And Conservation Of Energy Q.51P
In Example, suppose the two masses start from rest and are moving with a speed of 2.05 m/s just before m2 hits the floor. (a) If the coefficient of kinetic friction is µk = 0.350, what is the distance of travel, d, for the masses? (b) How much conservative work was done on this system? (c) How much non-conservative work was done on this system? (d) Verify the three work relations given in Equations 8-10.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.52P
A 15,800-kg truck is moving at 12.0 m/s when it starts down a 6.00° incline in the Canadian Rockies. At the start of the descent the driver notices that the altitude is 1630 m. When she reaches an altitude of 1440 m, her speed is 29.0 m/s. Find the change in (a) the gravitational potential energy of the system and (b) the truck’s kinetic energy. (c) Is the total mechanical energy of the system conserved? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy52ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy52ps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.53P
A 1.80-kg block slides on a rough horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is µk = 0.560, what is the force constant of the spring?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy53ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy53ps1
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy53ps2

Chapter 8 Potential Energy And Conservation Of Energy Q.54P
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy54p
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy54ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy54ps1
The above figure shows that, the object is at rest at initial point A. As the object moves from A to B, some of its potential energy is converted into kinetic energy, and the speed of the object increases at B. So the kinetic energy of the object increases when potential energy decreases.
Now the object moves from point B to point C. In this process, some of its kinetic energy is again converted into potential energy, thus, the speed of the object decreases as it moves from B to C. The object’s speed again increases as it moves from point C to point D, so its potential energy decreases while its kinetic energy increases.
Finally from point D to point E, the kinetic energy of the object decreases as curve rises. Thus, the speed of the object decreases. Therefore, at point E, the speed of the object momentarily becomes zero.

Chapter 8 Potential Energy And Conservation Of Energy Q.55P
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy55p
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.56P
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy56p
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy56ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy56ps1

Chapter 8 Potential Energy And Conservation Of Energy Q.57P
A 23-kg child swings back and forth on a swing suspended by 2.5-m-long ropes. Plot the gravitational potential energy of this system as a function of the angle the ropes make with the vertical, assuming the potential energy is zero when the ropes are vertical. Consider angles up to 90° on either side of the vertical.
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.58P
Find the turning-point angles in the previous problem if the child has a speed of 0.89 m/s when the ropes are vertical. Indicate the turning points on a plot of the system’s potential energy.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy58ps
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Chapter 8 Potential Energy And Conservation Of Energy Q.59P
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy59p
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy59ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy59ps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.60P
A block of mass m = 0.95 kg is connected to a spring of force constant k = 775 N/m on a smooth, horizontal surface. (a) Plot the potential energy of the spring from x = −5.00 cm to x = 5.00 cm. (b) Determine the turning points of the block if its speed at x = 0 is 1.3 m/s.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy60ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy60ps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.61P
A ball of mass m = 0.75 kg is thrown straight upward with an initial speed of 8.9 m/s. (a) Plot the gravitational potential energy of the block from its launch height, y = 0, to the height y = 5.0 m. Let u = 0 correspond to y = 0. (b) Determine the turning point (maximum height) of this mass.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy61ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy61ps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.62P
Two blocks, the of mass m, are connected on a frictionless horizontal table by a spring of force constant k and equilibrium length L. Find the maximum and minimum separation between the two blocks in terms of their maximum speed, vmax,relative to the table. (The two blocks always move in opposite directions as they oscillate back and forth about a fixed position.)
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy62ps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy62ps1

Chapter 8 Potential Energy And Conservation Of Energy Q.63GP
You and a friend both solve a problem involving a skier going down a slope. When comparing solutions, you notice that your choice for the y = 0 level is different than the y = 0 level chosen by your friend. Will your answers agree or disagree on the following quantities: (a) the skier’s potential energy; (b) the skier’s change in potential energy; (c) the skier’s kinetic energy?
Solution:
Answer:
Your answers will disagree on (a), but agree on (b) and (c)
The gravitational potential energy depends upon the reference level but not the change in potential energy. The work done by gravity must be the same in the two solutions so change in potential energy and change in kinetic energy should be same.

Chapter 8 Potential Energy And Conservation Of Energy Q.64GP
A particle moves under the influence of a conservative force. At point A the particle has a kinetic energy of 12 J; at point 13 the particle is momentarily at rest, and the potential energy of the system is 25 J; at point C the potential energy of the system is 5 J. (a) What is the potential energy of the system when the particle is at point A? (b) What is the kinetic energy of the particle at point C?
Solution:
(a) In the given data the particle is at rest at point B having potential energy 25J. Therefore the total energy of the particle is 25J.
Now at point A the particle has kinetic energy of 12J.
Therefore the potential energy of the system at point A is 25J-12J=13J
(b) Now at point C the particle has potential energy of 5J.
Therefore the kinetic energy of the system at point C is 25J-5J=20J

Chapter 8 Potential Energy And Conservation Of Energy Q.65GP
A leaf falls to the ground with constant speed. Is ui + Ki for this system greater than, less than, or the same as uf + Kf for this system? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy65ps

Chapter 8 Potential Energy And Conservation Of Energy Q.66GP
Consider the two-block system shown in Example. (a) As block 2 descends through the distance d, does its mechanical energy increase, decrease, or stay the same? Explain. (b) Is the nonconservative work done on block 2 by the tension in the rope positive, negative, or zero? Explain.
Solution:
Solution:
(a) Since there is no conservative work with this system (the friction from mass 1) the mechanical energy of the system will change. Specifically, the mechanical energy of the system will decrease due to energy being converted into heat by the friction forces.
(b) The tension in the rope deals with the weight of the block that’s dropping down. The non conservative force comes from the friction of the block on the table. The tension in the rope simply deals with mass and gravitational acceleration thus, the tension in the rope is involved in potential/kinetic energy relations, and the amount of non conservative work it does is zero.

Chapter 8 Potential Energy And Conservation Of Energy Q.67GP
Taking a leap of faith, a bungee jumper steps off a plat-form and falls until the cord brings her to rest. Suppose you analyze this system by choosing y = 0 at the platform level, and your friend chooses y = 0 at ground level. (a) Is the jumper’s initial potential energy in your calculation greater than, less than, or equal to the same quantity in your friend’s calculation? Explain. (b) Is the change in the jumper’s potential energy in your calculation greater than, less than, or equal to the same quantity in your friend’s calculation? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy67gps

Chapter 8 Potential Energy And Conservation Of Energy Q.68GP
A sled slides without friction down a small, ice-covered hill. If the sled starts from rest at the top of the hill, its speed at the bottom is 7.50 m/s. (a) On a second run, the sled starts with a speed of 1.50 m/s at the top. When it reaches the bottom of the hill, is its speed 9.00 m/s, more than 9.00 m/s, or less than 9.00 m/s? Explain. (b) Find the speed of the sled at the bottom of the hill after the second run.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy68gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy68gps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.69GP
In the previous problem, what is the height of the hill?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy69gps

Chapter 8 Potential Energy And Conservation Of Energy Q.70GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy70gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy70gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy70gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.71GP
Running Shoes The soles of a popular make of running shoe have a force constant of 2.0 × 105 N/m. Treat the soles as ideal springs for the following questions. (a) If a 62-kg person stands in a pair of these shoes, with her weight distributed equally on both feet, how much does she compress the soles? (b) How much energy is stored in the soles of her shoes when she’s standing?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy71gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy71gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.72GP
Nasal Strips The force required to flex a nasal strip and apply it to the nose is 0.25 N; the energy stored in the strip when flexed is 0.0022 J. Assume the strip to be an ideal spring for the following calculations. Find (a) the distance through which the strip is flexed and (b) the force constant of the strip.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy72gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy72gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.73GP
A pendulum bob with a mass of 0.13 kg is attached to a string with a length of 0.95 m. We choose the potential energy to be zero when the string makes an angle of 90° with the vertical. (a) Find the potential energy of this system when the string makes an angle of 45° with the vertical. (b) Is the magnitude of the change in potential energy from an angle of 90° to 45° greater than, less than, or the same as the magnitude of the change from 45° to 0°? Explain. (c) Calculate the potential energy of the system when the string is vertical.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy73gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy73gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.74GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy74gps
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy74gps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.75GP
An 1865-kg airplane starts at rest on an airport runway at sea level. (a) What is the change in mechanical energy of the airplane if it climbs to a cruising altitude of 2420 m and maintains a constant speed of 96.5 m/s? (b) What cruising speed would the plane need at this altitude if its increase in kinetic energy is to be equal to its increase in potential energy?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy75gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy75gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.76GP
At the local playground a child on a swing has a speed of 2.02 m/s when the swing is at its lowest-point. (a) To what maximum vertical height does the child rise, assuming he sits still and “coasts”? Ignore air resistance. (b) How do your results change if the initial speed of the child is halved?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy76gps

Chapter 8 Potential Energy And Conservation Of Energy Q.77GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy77gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy77gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.78GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy78gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy78gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.79GP
A person is to be released from rest on a swing pulled away from the vertical by an angle of 20.0°. The two frayed ropes of the swing are 2.75 m long, and will break if the tension in either of them exceeds 355 N. (a) What is the maximum weight the person can have and not break the ropes? (b) If the person is released at an angle greater than 20.0°, does the maximum weight increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy79gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.80GP
A car is coasting without friction toward a hill of height h and radius of curvature r. (a) What initial speed, v0, will result in the car’s wheels just losing contact with the roadway as the car crests the hill? (b) What happens if the initial speed of the car is greater than the value found in part (a)?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy80gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy80gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.81GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy81gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy81gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.82GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy82gp
Solution:
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Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy82gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.83GP
An 8.70-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.62. Use energy conservation to find the distance the block slides before coming to rest.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy83gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.84GP
Repeat the previous problem for the case of an 8.70-kg block sliding down the ramp, with an initial speed of 1.56 m/s.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy84gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy84gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.85GP
Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 37° with the vertical. If Jeff starts at restand has a mass of 78 kg, what is the tension in the vine at the lowest point of the swing?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy85gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy85gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.86GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy86gp
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.87GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy87gp
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.88GP
Compressing the Ground A running track at Harvard University uses a surface with a force constant of 2.5 × 105 N/m This surface is compressed slightly every time a runner’s foot lands on it. The force exerted by the foot, according to the Saucony shoe company, has a magnitude of 2700 N for a typical runner. Treating the track’s surface as an ideal spring, find (a) the amount of compression caused by a foot hitting the track and (b) the energy stored briefly in the track every time a foot lands.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy88gps

Chapter 8 Potential Energy And Conservation Of Energy Q.89GP
A Flea’s Jump The resilin in the upper leg (coxa) of a flea has a force constant of about 26 N/m, and when the Flea cocks its jumping legs, the resilin in the leg is stretched by approximately 0.10 mm. Given that the flea has a mass of 0.50 mg, and that two legs are used in a jump, estimate the maximum height a flea can attain by using the energy stored in the resilin. (Assume the resilin to be an ideal spring.)
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy89gps

Chapter 8 Potential Energy And Conservation Of Energy Q.90GP
A trapeze artist of mass m swings on a rope of length L. Initially, the trapeze artist is at rest and the rope makes an angle θ with the vertical. (a) Find the tension in the rope when it is vertical. (b) Explain why your result for part (a) depends on L in the way it does.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy90gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy90gps1
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Chapter 8 Potential Energy And Conservation Of Energy Q.91GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy91gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy91gps

Chapter 8 Potential Energy And Conservation Of Energy Q.92GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy92gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy92gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.93GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy93gp
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.94GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy94gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy94gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.95GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy95gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy95gps
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Chapter 8 Potential Energy And Conservation Of Energy Q.96GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy96gp
Solution:
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Chapter 8 Potential Energy And Conservation Of Energy Q.97GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy97gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy97gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy97gps1

Chapter 8 Potential Energy And Conservation Of Energy Q.98GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy98gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy98gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy98gps1
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy98gps2

Chapter 8 Potential Energy And Conservation Of Energy Q.99GP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy99gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy99gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy99gps1
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy99gps2

Chapter 8 Potential Energy And Conservation Of Energy Q.100PP
The Flight of the Dragonflies
Of all the animals you’re likely to see on a summer’s day, the most ancient is the dragonfly. In fact, the fossil record for dragonflies extends back over 250 million years, more than twice as long as for birds. Ancient dragonflies could be as large as a hawk, and were surely buzzing around the heads of both T. Rex and Triceratops.
Dragonflies belong to the order Odonata (“toothed jaws”) and the suborder Anisoptera (“different wings”), a reference to the fact that their hindwings are wider front-to-back than their forewings. (Damselflies, in contrast, have forewings and hind-wings that are the same.) Although ancient in their lineage, dragonflies are the fastest flying and most acrobatic of all insects; some of their maneuvers subject them to accelerations as great as 20g.
The properties of dragonfly wings, and how they account for such speed and mobility, have been of great interest to biologists. Figure shows an experimental setup designed to measure the force constant of Plexiglas models of wings, which are used in wind tunnel tests. A downward force is applied to the model wing at the tip (1 for hindwing, 2 for forewing) or at two-thirds the distance to the tip (3 for hindwing, 4 for forewing). As the force is varied in magnitude, the resulting deflection of the wing is measured. The results are shown in Figure. Notice that significant differences are seen between the hindwings and forewings, as one might expect from their different shapes.
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy100gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy100gps

Chapter 8 Potential Energy And Conservation Of Energy Q.101PP
The Flight of the Dragonflies
Of all the animals you’re likely to see on a summer’s day, the most ancient is the dragonfly. In fact, the fossil record for dragonflies extends back over 250 million years, more than twice as long as for birds. Ancient dragonflies could be as large as a hawk, and were surely buzzing around the heads of both T. Rex and Triceratops.
Dragonflies belong to the order Odonata (“toothed jaws”) and the suborder Anisoptera (“different wings”), a reference to the fact that their hindwings are wider front-to-back than their forewings. (Damselflies, in contrast, have forewings and hind-wings that are the same.) Although ancient in their lineage, dragonflies are the fastest flying and most acrobatic of all insects; some of their maneuvers subject them to accelerations as great as 20g.
The properties of dragonfly wings, and how they account for such speed and mobility, have been of great interest to biologists. Figure shows an experimental setup designed to measure the force constant of Plexiglas models of wings, which are used in wind tunnel tests. A downward force is applied to the model wing at the tip (1 for hindwing, 2 for forewing) or at two-thirds the distance to the tip (3 for hindwing, 4 for forewing). As the force is varied in magnitude, the resulting deflection of the wing is measured. The results are shown in Figure. Notice that significant differences are seen between the hindwings and forewings, as one might expect from their different shapes.
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy101gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy101gps

Chapter 8 Potential Energy And Conservation Of Energy Q.102PP
The Flight of the Dragonflies
Of all the animals you’re likely to see on a summer’s day, the most ancient is the dragonfly. In fact, the fossil record for dragonflies extends back over 250 million years, more than twice as long as for birds. Ancient dragonflies could be as large as a hawk, and were surely buzzing around the heads of both T. Rex and Triceratops.
Dragonflies belong to the order Odonata (“toothed jaws”) and the suborder Anisoptera (“different wings”), a reference to the fact that their hindwings are wider front-to-back than their forewings. (Damselflies, in contrast, have forewings and hind-wings that are the same.) Although ancient in their lineage, dragonflies are the fastest flying and most acrobatic of all insects; some of their maneuvers subject them to accelerations as great as 20g.
The properties of dragonfly wings, and how they account for such speed and mobility, have been of great interest to biologists. Figure shows an experimental setup designed to measure the force constant of Plexiglas models of wings, which are used in wind tunnel tests. A downward force is applied to the model wing at the tip (1 for hindwing, 2 for forewing) or at two-thirds the distance to the tip (3 for hindwing, 4 for forewing). As the force is varied in magnitude, the resulting deflection of the wing is measured. The results are shown in Figure. Notice that significant differences are seen between the hindwings and forewings, as one might expect from their different shapes.
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy102gp
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy102gp1

Solution:
From the graph it is clear that by the application of same force the deflection of wing is more for hindwing than the forewing.
Therefore force constant of forewing is greater than the force constant of hindwing.
So forewing is stiffer than the hindwing.
Therefore option B. is correct.

Chapter 8 Potential Energy And Conservation Of Energy Q.103PP
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy103gp
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy103gps

Chapter 8 Potential Energy And Conservation Of Energy Q.104IP
Referring to Example Consider a spring with a force constant of 955 N/m. (a) Suppose the mass of the block is 1.70 kg, but its initial speed can be varied. What initial speed is required to give a maximum spring compression of 4.00 cm? (b) Suppose the initial speed of the block is 1.09 m/s, but its mass can be varied. What mass is required to give a maximum spring compression of 4.00 cm?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy104gps

Chapter 8 Potential Energy And Conservation Of Energy Q.105IP
Referring to Example Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force constant of the spring is 955 N/m. (a) What is the speed of the block when the spring expands to a compression of only 2.50 cm? (b) What is the speed of the block after it leaves the spring?
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy105gps
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy105gps1
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy105gps2

Chapter 8 Potential Energy And Conservation Of Energy Q.106IP
Referring to Example Suppose we would like the landing speed of block 2 to be increased to 1.50 m/s. (a) Should the coefficient of kinetic friction between block 1 and the table-top be increased or decreased? (b) Find the required coefficient of kinetic friction for a landing speed of 1.50 m/s. Note that m1 = 2.40 kg, m2 = 1.80 kg, and d = 0.500 m.
Solution:
Mastering Physics Solutions Chapter 8 Potential Energy And Conservation Of Energy106gps

Mastering Physics Solutions Chapter 7 Work And Kinetic Energy

Mastering Physics Solutions Chapter 7 Work And Kinetic Energy

Mastering Physics Solutions

Chapter 7 Work And Kinetic Energy Q.1CQ
Is it possible to do work on an object that remains at rest?
Solution:
No.
We know that work is said to be done only when a body moves a certain distance in the direction of an applied force. In other words, if no external force is applied or the body fails to move in the direction of an applied force, the work done is said to be zero.
Consider a block of mass m (kg) moving a distance d (m) by application of force F (N) in the direction of the motion. Then, work done
W = F·d = force distance
If d = 0 [i.e., the body is at rest]
W = F 0
= 0
Therefore, it is not possible to do work on an object that is at rest.

Chapter 7 Work And Kinetic Energy Q.1P
The International Space Station orbits the Earth in an approximately circular orbit at a height of h = 375 km above the Earth’s surface. In one complete orbit, is the work done by the Earth on the space station positive, negative, or zero? Explain.
Solution:
The work done by Earth on the space station is zero. This is because during the movement of the satellite, the force acting on it is always perpendicular to the direction of motion. If the force does not have non-zero components along the direction of motion, the work done is zero.

Chapter 7 Work And Kinetic Energy Q.2CQ
A friend makes the statement, “Only the total force acting on an object can do work.” Is this statement true or false? If it is true, state why; if it is false, give a counterexample.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy2cqs

Chapter 7 Work And Kinetic Energy Q.2P
A pendulum bob swings from point I to point II along the circular arc indicated in Figure. (a) Is the work done on the bob by gravity positive, negative, or zero? Explain. (b) Is the work done on the bob by the string positive, negative, or zero? Explain.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy2p
Solution:
(a) When the pendulum bob swings from point I to point II along the circular arc, its displacement is downward. This is in the direction of force of gravity. Therefore the work done by the gravity is positive when the pendulum bob swings from point I to pint II along the circular arc.
(b) When the pendulum bob swings from point I to point II along the circular arc, the tension in the string is always perpendicular to its direction of motion. Therefore the work done by the string is zero when the pendulum bob swings from point I to pint II along the circular arc.

Chapter 7 Work And Kinetic Energy Q.3CQ
A friend makes the statement, “A force that is always perpendicular to the velocity of a particle does no work on the particle.” Is this statement true or false? If it is hue, state why; if it is false, give a counterexample.
Solution:
True. A force that is always perpendicular to displacement does not have a non-zero component along the direction of motion. As a result, work will not be done on the particle.
Work done W=(F cosθ)d
W is positive if F has a component in the direction of motion.
W is positive if the angle between the force F and displacement d is -90< <90.

Chapter 7 Work And Kinetic Energy Q.3P
A pendulum bob swings from point II to point III along the circular arc indicated in Figure. (a) Is the work done on the bob by gravity positive, negative, or zero? Explain. (b) Is the work done on the bob by the string positive, negative, or zero? Explain.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy3p

Solution:
(a) When the pendulum bob swings from point II to point III along the circular arc, its displacement is upward. This is opposite to the direction of force of gravity. Therefore the work done by the gravity is negative when the pendulum bob swings from point II to pint III along the circular arc.
(b) When the pendulum bob swings from point II to point III along the circular arc, the tension in the string is always perpendicular to its direction of motion. Therefore the work done by the string is zero when the pendulum bob swings from point II to pint III along the circular arc.

Chapter 7 Work And Kinetic Energy Q.4CQ
The net work done on a certain object is zero. What can you say about its speed?
Solution:
The work done is equal to the change in kinetic energy. So if the net work done on an object is zero, its change in kinetic energy is also zero. Thus, the speed of the object will remain the same.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy4cqps

Chapter 7 Work And Kinetic Energy Q.4P
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy4ps

Chapter 7 Work And Kinetic Energy Q.5CQ
To get out of bed in the morning, do you have to do work? Explain.
Solution:
Yes, we have to do work against the force of gravity to get out of bed in the morning because the constant force of gravity acts downward on our body when we are sleeping in bed. To get up we have to apply force upward; thus, we are doing work against the force of gravity.

Chapter 7 Work And Kinetic Energy Q.5P
Children in a tree house lift a small dog in a basket 4.70 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy5ps

Chapter 7 Work And Kinetic Energy Q.6CQ
Give an example of a frictional force doing negative work.
Solution:
Frictional forces do negative work whenever they act in a direction that opposes the motion.
For example, friction does negative work when you push a box across the floor, or when you hit the brakes while driving.

Chapter 7 Work And Kinetic Energy Q.6P
Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 3.2-kg pumpkin to a height of 1.2 in, then carry it 50.0 m (on level ground) to the check-out stand. (a) Calculate the work you do on the pumpkin as you lift it from the ground. (b) How much work do you do on the pumpkin as you carry it from the field?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy6ps

Chapter 7 Work And Kinetic Energy Q.7CQ
Give an example of a frictional force doing positive work.
Solution:
The force of friction always opposes relative motion, but if the body moves along the direction of applied force, for ◊ = 0º, the work done is positive.
So when man walks on the ground, the frictional force between his shoes and the ground does positive work whenever he begins to walk, as ◊ = 0º.

Chapter 7 Work And Kinetic Energy Q.7P
The coefficient of kinetic friction between a suitcase and the floor is 0.272. If the suitcase has a mass of 71.5 kg, how far can it be pushed across the level floor with 642 J of work?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy7ps

Chapter 7 Work And Kinetic Energy Q.8CQ
A ski boat moves with constant velocity. Is the net force acting on the boat doing work? Explain.
Solution:
Since the velocity of the boat is constant, there will not be any change in the kinetic energy of the boat.
Thus, the change in kinetic energy = 0 joules ————- (1)
Work done = change in kinetic energy —————- (2)
From equations (1) and (2),
work done = 0 joules ——————– (3)
Work done = net force displacement ————— (4)
Since the boat is moving at a constant speed, then from equations (3) and (4),
net force = 0 newtons
Thus, the net force is not doing any work on the boat.

Chapter 7 Work And Kinetic Energy Q.8P
You pick up a 3.4-kg can of paint from the ground and lift it to a height of 1.8 m. (a) How much work do you do on the can of paint? (b) You hold the can stationary for half a minute, waiting for a friend on a ladder to take it. How much work do you do during this time? (c) Your friend decides against the paint, so you lower it back to the ground. How much work do you do on the can as you lower it?
Solution:
Mass of the can m = 3.4 kg
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy8ps

Chapter 7 Work And Kinetic Energy Q.9CQ
A package rests on the floor of an elevator that is rising with constant speed. The elevator exerts an upward normal force on the package, and hence does positive work on it. Why doesn’t the kinetic energy of the package increase?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy9cqs

Chapter 7 Work And Kinetic Energy Q.9P
A tow rope, parallel to the water, pulls a water skier directly behind the boat with constant velocity for a distance of 65 m before the skier falls. The tension in the rope is 120 N. (a) Is the work done on the skier by the rope positive, negative, or zero? Explain. (b) Calculate the work done by the rope on the skier.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy9ps

Chapter 7 Work And Kinetic Energy Q.10CQ
An object moves with constant velocity. Is it safe to conclude that no force acts on the object? Why, or why not?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy10cqs

Chapter 7 Work And Kinetic Energy Q.10P
In the situation described in the previous problem, (a) is the work done on the boat by the rope positive, negative, or zero? Explain. (b) Calculate the work done by the rope on the boat.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy10ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy10ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy10ps2

Chapter 7 Work And Kinetic Energy Q.11CQ
Engine 1 does twice the work of engine 2. Is it correct to conclude that engine 1 produces twice as much power as engine 2? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy11cqs
The power produced is defined as the rate of the work done. Here it is mentioned that engine 1 does twice the amount of work as engine 2 but their individual time of doing that work is not specified.
Hence, it is incorrect to conclude that engine 1 produces twice the amount of power as produced by engine 2.

Chapter 7 Work And Kinetic Energy Q.11P
A child pulls a friend in a little red wagon with constant speed. If the child pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 25° above the horizontal, how much work does the child do on the wagon?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy11ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy11ps1

Chapter 7 Work And Kinetic Energy Q.12CQ
Engine 1 produces twice the power of engine 2. Is it correct to conclude that engine 1 does twice as much work as engine 2? Explain.
Solution:
No.
Power depends both on the amount of work done by the engine, and the amount of time during which the work is performed.

Chapter 7 Work And Kinetic Energy Q.12P
A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 43.5° above the horizontal. If the tension in the rope is 115 N, how much work is done on the crate to move it 8.0 m?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy12ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy12ps1

Chapter 7 Work And Kinetic Energy Q.13P
To clean a floor, a janitor pushes on a mop handle with a force of 50.0 N. (a) If the mop handle is at an angle of 55° above the horizontal, how much work is required to push the mop 0.50 m? (b) If the angle the mop handle makes with the horizontal is increased to 65°, does the work done by the janitor increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy13ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy13ps1

Chapter 7 Work And Kinetic Energy Q.14P
A small plane tows a glider at constant speed and altitude. If the plane does 2.00 × 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy14ps

Chapter 7 Work And Kinetic Energy Q.15P
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy15p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy15ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy15ps1

Chapter 7 Work And Kinetic Energy Q.16P
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy16p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy16ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy16ps1

Chapter 7 Work And Kinetic Energy Q.17P
Water skiers often ride to one side of the center line of a boat, as shown in Figure. In this case, the ski boat is traveling at 15 m/s and the tension in the rope is 75 N. If the boat does 3500 J of work on the skier in 50.0 m, what is the angle θ between the tow rope and the center line of the boat?
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy17p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy17ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy17ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy17ps2

Chapter 7 Work And Kinetic Energy Q.18P
A pitcher throws a ball at 90 mi /h and the catcher stops it in her glove. (a) Is the work done on the ball by the pitcher positive, negative, or zero? Explain. (b) Is the work done on the ball by the catcher positive, negative, or zero? Explain.
Solution:
a) The pitcher does positive work on the ball because the direction of force is along the direction of displacement of the ball.
b) The catcher does negative work on the ball by exerting a force in the direction opposite to the motion of the ball, in order to stop the ball.

Chapter 7 Work And Kinetic Energy Q.19P
How much work is needed for a 73-kg runner to accelerate from rest to 7.7 m/s?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy19ps

Chapter 7 Work And Kinetic Energy Q.20P
Skylab’s Reentry When Skylab reentered the Earth’s atmosphere on July 11, 1979, it broke into a myriad of pieces. One of the largest fragments was a I770-kg lead-lined film vault, and it landed with an estimated speed of 120 m/s. What was the kinetic energy of the film vault when it landed?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy20ps

Chapter 7 Work And Kinetic Energy Q.21P
A 9.50-g bullet has a speed of 1.30 km/s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy21ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy21ps1

Chapter 7 Work And Kinetic Energy Q.22P
The work W0 accelerates a car from 0 to 50 km/h. (a) Is the work required to accelerate the car from 50 km/h to 150 km/h equal to 2W0, 3W0, 8W0, or 91w0? (b) Choose the best explanation from among the following:
I. The work to accelerate the car depends on the speed squared.
II. The final speed is three times the speed that was produced by the work W0.
III. The increase in speed from 50 km/h to 150 km/h is twice the increase in speed from 0 to 50 km/h.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy22ps

Chapter 7 Work And Kinetic Energy Q.23P
Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2, and jogger D has a mass 4m and a speed v/2. Rank the joggers in order of increasing kinetic energy. Indicate ties where appropriate.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy23ps
>Mastering Physics Solutions Chapter 7 Work And Kinetic Energy23ps1

Chapter 7 Work And Kinetic Energy Q.24P
A 0.14-kg pinecone falls 16 m to the ground, where it lands with a speed of 13 m/s. (a) With what speed would the pinecone have landed if there had been no air resistance? (b) Did air resistance do positive work, negative work, or zero work on the pinecone? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy24ps

Chapter 7 Work And Kinetic Energy Q.25P
In the previous problem, (a) how much work was done on the pinecone by air resistance? (b) What was the average force of air resistance exerted on the pinecone?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy25ps
The negative sign indicates that the force acts in the upward direction.

Chapter 7 Work And Kinetic Energy Q.26P
At t = 1.0 s, a 0.40-kg object is falling with a speed of 6.0 m/s. At t = 2.0 s, it has a kinetic energy of 25 J. (a) What is the kinetic energy of the object at t = 1.0 s? (b) What is the speed of the object at t = 2.0 s? (c) How much work was done on the object between t = 1.0 s and t = 2.0 s?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy26ps

Chapter 7 Work And Kinetic Energy Q.27P
After hitting a long fly ball that goes over the right fielder’s head and lands in the outfield, the batter decides to keep going past second base and try for third base. The 62.0-kg player begins sliding 3.40 m from the base with a speed of 4.35 m/s. If the player comes to rest at third base, (a) how much work was done on the player by friction? (b) What was the coefficient of kinetic friction between the player and the ground?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy27ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy27ps1

Chapter 7 Work And Kinetic Energy Q.28P
A 1100-kg car coasts on a horizontal road with a speed of 19 m/s. After crossing an unpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy28ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy28ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy28ps2

Chapter 7 Work And Kinetic Energy Q.29P
(a) In the previous problem, the car’s speed decreased by 7.0 m/s as it coasted across a sandy section of road 32 m long. If the sandy portion of the road had been only 16 m long, would the car’s speed have decreased by 3.5 m/s, more than 3.5 m/s, or less than 3.5 m/s? Explain. (b) Calculate the change in speed in this case.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy29ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy29ps1

Chapter 7 Work And Kinetic Energy Q.30P
A 65-kg bicyclist rides his 8.8-kg bicycle with a speed of 14 m/s. (a) How much work must be done by the brakes to bring the bike and rider to a stop? (b) How far does the bicycle travel if it takes 4.0 s to come to rest? (c) What is the magnitude of the braking force?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy30ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy30ps1

Chapter 7 Work And Kinetic Energy Q.31P
A block of mass m and speed v collides with a spring, compressing it a distance △x. What is the compression of the spring if the force constant of the spring is increased by a factor of four?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy31ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy31ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy31ps2

Chapter 7 Work And Kinetic Energy Q.32P
A spring with a force constant of 3.5 × 104 N/m is initially at its equilibrium length. (a) How much work must you do to stretch the spring 0.050 m? (b) How much work must you do to compress it 0.050 m?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy32ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy32ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy32ps2

Chapter 7 Work And Kinetic Energy Q.33P
A 1.2-kg block is held against a spring of force constant 1.0 × 104 N/m, compressing it a distance of 0.15 m. How fast is the block moving after it is released and the spring pushes it away?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy33ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy33ps1

Chapter 7 Work And Kinetic Energy Q.34P
Initially sliding with a speed of 2.2 m/s, a 1.8-kg block collides with a spring and compresses it 0.31 m before coming to rest. What is the force constant of the spring?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy34ps

Chapter 7 Work And Kinetic Energy Q.35P
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy35p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy35ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy35ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy35ps2

Chapter 7 Work And Kinetic Energy Q.36P
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy36p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy36ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy36ps1

Chapter 7 Work And Kinetic Energy Q.37P
CE A block of mass m and speed v collides with a spring, compressing it a distance △x. What is the compression of the spring if the mass of the block is halved and its speed is doubled?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy37ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy37ps1

Chapter 7 Work And Kinetic Energy Q.38P
To compress spring 1 by 0.20 m takes 150 J of work. Stretching spring 2 by 0.30 m requires 210 J of work. Which spring is stiffer?
Solution:
38ps
Work done to compress spring 1 by x = 0.2 m is W = 150 J We know that
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy38ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy38ps1

Chapter 7 Work And Kinetic Energy Q.39P
It takes 180 J of work to compress a certain spring 0.15 m. (a) What is the force constant of this spring? (b) To compress the spring an additional 0.15 m, does it take 180 J, more than 180 J, or less than 180 J? Verify your answer with a calculation.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy39ps

Chapter 7 Work And Kinetic Energy Q.40P
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy40p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy40ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy40ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy40ps2
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy40ps3

Chapter 7 Work And Kinetic Energy Q.41P
A block is acted on by a force that varies as (2.0 × 104 N/m)x for 0 ≤ x ≤ 0.21 m, and then remains constant at 4200 N for larger x. How much work does the force do on the block in moving it (a) from x = 0 to x = 0.30 m, or (b) from x = 0.10 m to x = 0.40 m?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy41ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy41ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy41ps2

Chapter 7 Work And Kinetic Energy Q.42P
CE Force F1 does 5 J of work in 10 seconds, force F2 does 3 J of work in 5 seconds, force F3 does 6 J of work in 18 seconds, and force F4 does 25 J of work in 125 seconds. Rank these forces in order of increasing power they produce. Indicate ties where appropriate.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy42ps

Chapter 7 Work And Kinetic Energy Q.43P
Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February 3, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of the step was 0.20 m, and the mass of the runner was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy43ps

Chapter 7 Work And Kinetic Energy Q.44P
How many joules of energy are in a kilowatt-hour?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy44ps

Chapter 7 Work And Kinetic Energy Q.45P
Calculate the power output of a 1.4-g fly as it walks straight tip a windowpane at 2.3 cm/s.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy45ps

Chapter 7 Work And Kinetic Energy Q.46P
An ice cube is placed in a microwave oven. Suppose the oven delivers 105 W of power to the ice cube and that it takes 32,200 J to melt it. How long docs it take for the ice cube to melt?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy46ps

Chapter 7 Work And Kinetic Energy Q.47P
You raise a bucket of water from the bottom of a deep well. If your power output is 108 W, and the mass of the bucket and the water in it is 5.00 kg, with what speed can you raise the bucket? Ignore the weight of the rope.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy47ps

Chapter 7 Work And Kinetic Energy Q.48P
In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water the second from below deck up a height of 2.00 m and over the side. What is the minimum horse power motor that can be used to save the ship?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy48ps

Chapter 7 Work And Kinetic Energy Q.49P
A kayaker paddles with a power output of 50.0 W to maintain a steady speed of 1.50 m/s. (a) Calculate the resistive force exerted by the water on the kayak. (b) If the kayaker doubles her power output, and the resistive force due to the water remains the same, by what factor does the kayaker’s speed change?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy49ps

Chapter 7 Work And Kinetic Energy Q.50P
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy50ps

Chapter 7 Work And Kinetic Energy Q.51P
A grandfather clock is powered by the descent of a 4.35-kg weight. (a) If the weight descends through a distance of 0.760 m in 3.25 days, how much power docs it deliver to the clock? (b) To increase the power delivered to the clock, should the time it takes for the mass to descend be increased or decreased? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy51ps

Chapter 7 Work And Kinetic Energy Q.52P
The Power You Produce Estimate the power you produce in running up a flight of stairs. Give your answer in horsepower.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy52ps

Chapter 7 Work And Kinetic Energy Q.53P
A certain car can accelerate from rest to the speed v in T seconds. If the power output of the car remains constant, (a) how long does it take for the car to accelerate from v to 2v? (b) How fast is the car moving at 2T seconds after starting?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy53ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy53ps1

Chapter 7 Work And Kinetic Energy Q.54GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy54p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy54ps
>Mastering Physics Solutions Chapter 7 Work And Kinetic Energy54ps1
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Chapter 7 Work And Kinetic Energy Q.55GP
A youngster rides on a skateboard with a speed of 2 m/s. After a force acts on the youngster, her speed is 3 m/s. Was the work done by the force positive, negative, or zero? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy55gps

Chapter 7 Work And Kinetic Energy Q.56GP
Predict/Explain A car is accelerated by a constant force, F. The distance required to accelerate the car from rest to the speed v is △x. (a) Is the distance required to accelerate the car from the speed v to the speed 2v equal to △x, 2△x, 3△x, or 4△x? (b) Choose the best explanation from among the following:
I. The final speed is twice the initial speed.
II. The increase in speed is the same in each case.
III. Work is force times distance, and work depends on the speed squared.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy56gps

Chapter 7 Work And Kinetic Energy Q.57GP
Car 1 has four times the mass of car 2, but they both have the same kinetic energy. If the speed of car 2 is v, is the speed of car 1 equal to v/4, v/2, 2v, or 4v? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy57gps

Chapter 7 Work And Kinetic Energy Q.58GP
Muscle Cells Biological muscle cells can be thought of as nanomotors that use the chemical energy of ATP to produce mechanical work. Measurements show that the active proteins within a muscle cell (such as myosin and actin) can produce a force of about 7.5 pN and displacements of 8.0 run. How much work is done by such proteins?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy58gps

Chapter 7 Work And Kinetic Energy Q.59GP
When you take a bite out of an apple, you do about 19 J of work. Estimate (a) the force and (b) the power produced by your jaw muscles during the bite.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy59gps

Chapter 7 Work And Kinetic Energy Q.60GP
A Mountain bar has a mass of 0.045 kg and a calorie rating of 210 Cal. What speed would this candy bar have if its kinetic energy were equal to its metabolic energy? [See the note following Problem.]
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy60gps

Chapter 7 Work And Kinetic Energy Q.61GP
A small motor runs a lift that raises a load of bricks weighing 836 N to a height of 10.7 m in 23.2 s. Assuming that the bricks are lifted with constant speed, what is the minimum power the motor must produce?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy61gps

Chapter 7 Work And Kinetic Energy Q.62GP
You push a 67-kg box across a floor where the coefficient of kinetic friction is µk = 0.55. The force you exert is horizontal. (a) How much power is needed to push the box at a speed of 0.50 m/s? (b) How much work do you do if you push the box for 35 s?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy62gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy62gps1

Chapter 7 Work And Kinetic Energy Q.63GP
The Beating Heart The average power output of the human heart is 1.33 watts. (a) How much energy does the heart produce in a day? (b) Compare the energy found in part (a) with the energy required to walk up a flight of stairs. Estimate the height a person could attain on a set of stairs using nothing more than the daily energy produced by the heart.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy63gps

Chapter 7 Work And Kinetic Energy Q.64GP
The Atmos Clock The Atmos clock (the so-called perpetual motion clock) gets its name from the fact that it runs off pressure variations in the atmosphere, which drive a bellows containing a mixture of gas and liquid ethyl chloride. Because the power to drive these clocks is so limited, they must be very efficient. In fact, a single 60.0-W lightbulb could power 240 million Atmos clocks simultaneously. Find the amount of energy, in joules, required to run an Atmos clock for one day.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy64gps

Chapter 7 Work And Kinetic Energy Q.65GP
The work W0 is required to accelerate a car from rest to the speed v0. How much work is required to accelerate the car (a) from rest to the speed v0/2 and (b) from v0/2 to v0?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy65gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy65gps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy65gps2
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy65gps3

Chapter 7 Work And Kinetic Energy Q.66GP
A work W0 is required to stretch a certain spring 2 cm from its equilibrium position. (a) How much work is required to stretch the spring 1 cm from equilibrium? (b) Suppose the spring is already stretched 2 cm from equilibrium. How much additional work is required to stretch it to 3 cm from equilibrium?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy66gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy66gps1

Chapter 7 Work And Kinetic Energy Q.67GP
After a tornado, a 0.55-g straw was found embedded 2.3 cm into the trunk of a tree. If the average force exerted on the straw by the tree was 65 N, what was the speed of the straw when it hit the tree?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy67gps

Chapter 7 Work And Kinetic Energy Q.68GP
You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. What is the kinetic energy of the glove when it is at the height h/2?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy68gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy68gps1

Chapter 7 Work And Kinetic Energy Q.69GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy69gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy69gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy69gps1

Chapter 7 Work And Kinetic Energy Q.70GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy70gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy70gps
(C) If the pulling force remains the e, then the work done remains the e. If the increased mass results in a larger friction force and an increase in the pulling force, then the work done increases.

Chapter 7 Work And Kinetic Energy Q.71GP
A 0.19-kg apple falls from a branch 3.5 m above the ground. (a) Does the power delivered to the apple by gravity increase, de crease, or stay the same during the time the apple falls to the ground? Explain. Find the power delivered by gravity to the apple when the apple is (b) 2.5 m and (c) 1.5 m above the ground.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy71ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy71ps1

Chapter 7 Work And Kinetic Energy Q.72GP
A juggling ball of mass m is thrown straight upward from an initial height h with an initial speed v0. How much work has gravity done on the ball (a) when it reaches its greatest height, h max, and (b) when it reaches ground level? (c) Find an expression for the kinetic energy of the ball as it lands.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy72ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy72ps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy72ps2

Chapter 7 Work And Kinetic Energy Q.73GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy73p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy73ps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy73ps1

Chapter 7 Work And Kinetic Energy Q.74GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy74p
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy74ps

Chapter 7 Work And Kinetic Energy Q.75GP
The motor of a ski boat produces a power of 36,600 W to maintain a constant speed of 14.0 m/s. To pull a water skier at the same constant speed, the motor must produce a power of 37,800 W. What is the tension in the rope pulling the skier?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy75ps

Chapter 7 Work And Kinetic Energy Q.76GP
Cookie Power To make a batch of cookies, you mix half a bag of chocolate chips into a bowl of cookie dough, exerting a 21-N force on the stirring spoon. Assume that your force is always in the direction of motion of the spoon. (a) What power is needed to move the spoon at a speed of 0.23 m/s? (b) How much work do you do if you stir the mixture for 1.5 min?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy76gps

Chapter 7 Work And Kinetic Energy Q.77GP
A pitcher accelerates a 0.14-kg hardball from rest to 42.5 m/s in 0.060 s. (a) How much work does the pitcher do on the ball? (b) What is the pitcher’s power output during the pitch? (c) Suppose the ball reaches 42.5 m/s in less than 0.060 s. Is the power produced by the pitcher in this case more than, less than, or the same as the power found in part (b)? Explain.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy77gps

Chapter 7 Work And Kinetic Energy Q.78GP
Catapult Launcher A catapult launcher on an aircraft carrier accelerates a jet from rest to 72 m/s. The work done by the catapult during the launch is 7.6 × 107 J. (a) What is the mass of the jet? (b) If the jet is in contact with the catapult for 2.0 s, what is the power output of the catapult?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy78gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy78gps1

Chapter 7 Work And Kinetic Energy Q.79GP
Brain Power The human brain consumes about 22 W of power under normal conditions, though more power may be required during exams. (a) How long can one Snickers bar (see the note following Problem) power the normally functioning brain? (b) At what rate must you lift a 3.6-kg container of milk (one gallon) if the power output of your arm is to be 22 W? (c) How long does it take to lift the milk container through a distance of 1.0 m at this rate?
Human-Powered Flight Human-powered aircraft re quire a pilot to pedal, as in a bicycle, and produce a sustained power output of about 0.30 hp. The Gossamer Albatross flew across the English Channel on June 12, 1979, in 2 h 49 min. (a) How much energy did the pilot expend during the flight? (b) How many Snickers candy bars (280 Cal per bar) would the pilot have to consume to be “fueled up” for the flight? [Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as defined in physics. In addition, the conversion factor between calories and joules is as follows: 1 Cal = 1000 cal = 1 kcal = 4186 J.]
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy79gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy79gps1

Chapter 7 Work And Kinetic Energy Q.80GP
A 1300-kg car delivers a constant 49 hp to the drive wheels. We assume the car is traveling on a level road and that all fractional forces may be ignored. (a) What is the acceleration of this car when its speed is 14 m/s? (b) If the speed of the car is doubled, does its acceleration increase, decrease, or stay the same? Explain. (c) Calculate the car’s acceleration when its speed is 28 m/s.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy80gps

Chapter 7 Work And Kinetic Energy Q.81GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy81gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy81gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy81gps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy81gps2

Chapter 7 Work And Kinetic Energy Q.82GP
Powering a Pigeon A pigeon in flight experiences a force of air resistance given approximately by F = bv2, where v is the flight speed and b is a constant. (a) What are the units of the constant b? (b) What is the largest possible speed of the pigeon if its maximum power output is P? (c) By what factor does the largest possible speed increase if the maximum power is doubled?
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy82gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy82gps1

Chapter 7 Work And Kinetic Energy Q.83GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy83gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy83gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy83gps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy83gps2

Chapter 7 Work And Kinetic Energy Q.84GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy84gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy84gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy84gps1

Chapter 7 Work And Kinetic Energy Q.85GP
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy85gp
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy85gps
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy85gps1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy85gps2

Chapter 7 Work And Kinetic Energy Q.86PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy86gp
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy86gp1
Solution:
From the given figure it is clear that the range of flight speeds for power out put of 9.8W is 7.7m/s to 15m/s
So, option B. is correct.

Chapter 7 Work And Kinetic Energy Q.87PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy87gp
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy87gp1
Solution:
From the figure it is clear that the approximate range of flight speeds would be possible if Microraptor gui could produce 20W of power is 2.5m/s to 25m/s.
This is because below 2.5m/s and above 25m/s, the power out put is greater than 20W.
Therefore option C. is correct.

Chapter 7 Work And Kinetic Energy Q.88PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy88gp
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy88gp1
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy88gps

Chapter 7 Work And Kinetic Energy Q.89PP
Microraptor gui: The Biplane Dinosaur
The evolution of flight is a subject of intense interest in paleontology. Some subscribe to the “cursorial” (or ground-up) hypothesis, in which flight began with ground-dwelling animals running and jumping after prey. Others favor the “arboreal” (or trees-down) hypothesis, in which tree-dwelling animals, like modern-day flying squirrels, developed flight as an extension of gliding from tree to tree.
A recently discovered fossil from the Cretaceous period in China supports the arboreal hypothesis and adds a new element—it suggests that feathers on both the wings and the lower legs and feet allowed this dinosaur, Microraptor gui, to glide much like a biplane, as shown in Figure. Re searchers have produced a detailed computer simulation of Microraptor, and with its help have obtained the power-versus-speed plot presented in Figure. This curve shows how much power is required for flight at speeds between 0 and 30 m/s. Notice that the power increases at high speeds, as expected, but is also high for low speeds, where the dinosaur is almost hovering. A minimum of 8.1 W is needed for flight at 10 m/s. The lower horizontal line shows the estimated 9.8-W power output of Microraptor, indicating the small range of speeds for which flight would be possible. The upper horizontal line shows the wider range of flight speeds that would be available if Microraptor were able to produce 20 W of power.
Also of interest are the two dashed, straight lines labeled 1 and 2. These lines represent constant ratios of power to speed; that is, a constant value for P/v. Referring to Equation 7-13, we see that P/v = Fv/v = F, so the lines 1 and 2 correspond to lines of constant force. Line 2 is interesting in that it has the smallest slope that still touches the power-versus-speed curve.
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy89gp
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy89gp1
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy89gps

Chapter 7 Work And Kinetic Energy Q.90IP
Referring to Figure Suppose the block has a mass of 1.4 kg and an initial speed of 0.62 m/s. (a) What force constant must the spring have if the maximum compression is to be 2.4 cm? (b) If the spring has the force constant found in part (a), find the maximum compression if the mass of the block is doubled and its initial speed is halved.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy90ips
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy90ips1
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy90ips2

Chapter 7 Work And Kinetic Energy Q.91IP
Referring to Figure In the situation shown in Figure (d), a spring with a force constant of 750 N/m is compressed by 4.1 cm. (a) If the speed of the block in Figure (f) is 0.88 m/s, what is its mass? (b) If the mass of the block is doubled, is the final speed greater than, less than, or equal to 0.44 m/s? (c) Find the final speed for the case described in part (b).
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy91ips

Chapter 7 Work And Kinetic Energy Q.92IP
Referring to Example Suppose the car has a mass of 1400 kg and delivers 48 hp to the wheels. (a) How long does it take for the car to increase its speed from 15 m/s to 25 m/s? (b) Would the time required to increase the speed from 5.0 m/s to 15 m/s be greater than, less than, or equal to the time found in part (a)? (c) Determine the time required to accelerate from 5.0 m/s to 15 m/s.
Solution:
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy92ips
Mastering Physics Solutions Chapter 7 Work And Kinetic Energy92ips1

Mastering Physics Solutions Chapter 6 Applications Of Newton’s Laws

Mastering Physics Solutions Chapter 6 Applications Of Newton’s Laws

Mastering Physics Solutions

Chapter 6 Applications Of Newton’s Laws Q.1CQ
A clothesline always sags a little, even if nothing hangs from it. Explain.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws1cqs

Chapter 6 Applications Of Newton’s Laws Q.1P
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws1p
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws1ps

Chapter 6 Applications Of Newton’s Laws Q.2CQ
In the Jurassic Park sequel, The Lost World, a man tries to keep a large vehicle from going over a cliff by connecting a cable from his Jeep to the vehicle. The man then puts the Jeep in gear and spins the rear wheels. Do you expect that spinning the tires will increase the force exerted by the Jeep on the vehicle? Why or why not?
Solution:
No
The man puts the jeep in gear and spins the rear wheels, but spinning will not provide the friction needed to rise above. Spinning the wheels actually decrease the force exerted by the jeep because the force exerted by the spinning wheels is kinetic friction, and the coefficient of kinetic friction is generally less than the coefficient of static friction.

Chapter 6 Applications Of Newton’s Laws Q.2P
Predict/Explain Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among the following:
I. Locking up the brakes gives the greatest possible braking force.
II. The same tires on the same road result in the same force of friction.
III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
Solution:
(a) The stopping distance of driver 1 is greater than the stopping distance of driver 2.
(b) For driver 2 the force stopping the car is the static friction force. And for driver 1 the force stopping the car is the kinetic friction force. But the static friction force is greater than the kinetic friction force. Therefore the driver 1 travels greater than the driver 2. So option III is the best explanation.

Chapter 6 Applications Of Newton’s Laws Q.3CQ
When a traffic accident is investigated, it is common for the length of the skid marks to be measured. How could this information be used to estimate the initial speed of the vehicle that left the skid marks?
Solution:
Braking distance depends on the initial speed and the kinetic friction. If the braking distance and the kinetic friction are known, then the initial speeds of the car can be found.

Chapter 6 Applications Of Newton’s Laws Q.3P
Abaseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of kinetic friction between the player and the ground is 0.46, how far does the player slide before coming to rest?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws3ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws3ps1

Chapter 6 Applications Of Newton’s Laws Q.4CQ
In a car with rear-wheel drive, the maximum acceleration is often less than the maximum deceleration. Why?
Solution:
The maximum acceleration is determined by the normal force exerted on the drive wheels. If the engine of the car is in the front and the drive wheels are in the rear, the normal force is less than it would be with front-wheel drive. During braking, however, all four wheels participate – including the wheels that sit under the engine.

Chapter 6 Applications Of Newton’s Laws Q.4P
A child goes down a playground slide with an acceleration of 1.26 m/s2. Find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 33.0° below the horizontal.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws4ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws4ps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws4ps2

Chapter 6 Applications Of Newton’s Laws Q.5CQ
A train typically requires a much greater distance to come to rest, for a given initial speed, than does a car. Why?
Solution:
The frictional force is responsible for moving the object when the brakes or the driving force are applied. In the case of the train, the frictional force between the rails and the wheels are comparatively low because both are smooth surfaces.
In the case of the car, the road and the rubber tires are both irregular surfaces, so the frictional force is comparatively greater. Therefore, the car stops at a shorter distance than the train.

Chapter 6 Applications Of Newton’s Laws Q.5P
Hopping into your Porsche, you floor it and accelerate at 12 m/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws5ps

Chapter 6 Applications Of Newton’s Laws Q.6CQ
Give some everyday examples of situations in which friction is beneficial.
Solution:
Frictional force is beneficial in the following cases.
(1) Without friction, we cannot walk
(2) Without friction, a car cannot run on the road
(3) Without friction, we cannot hammer the nails inside the walls

Chapter 6 Applications Of Newton’s Laws Q.6P
When you push a 1.80-kg book resting on a tabletop, it takes 2.25 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws6ps

Chapter 6 Applications Of Newton’s Laws Q.7CQ
At the local farm, you buy a flat of strawberries and place them on the backseat of the car. On the way home, you begin to brake as you approach a stop sign. At first the strawberries stay put, but as you brake a bit harder, they begin to slide off the seat. Explain.
Solution:
As you brake harder, your car has a greater acceleration. The greater the acceleration of the car, the greater the force required to give the flat strawberries the same acceleration. When the required force exceeds the maximum force of static friction, the strawberries begin to slide.

Chapter 6 Applications Of Newton’s Laws Q.7P
In Problem, what is the frictional force exerted on the book when you push on it with a force of 0.75 N? When you push a 1.80-kg book resting on a tabletop, it takes 2.25 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws7ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws7ps1

Chapter 6 Applications Of Newton’s Laws Q.8CQ
It is possible to spin a bucket of water in a vertical circle and have none of the water spill when the bucket is upside down. How would you explain this to members of your family?
Solution:
When we rotate the bucket vertically, a centripetal force comes to counter the weight of the water that falls down. A sufficiently high speed of rotation gives a sufficient force opposite to the mouth of the bucket, so the water does not fall at all.

Chapter 6 Applications Of Newton’s Laws Q.8P
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws8p
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws8ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws8ps1

Chapter 6 Applications Of Newton’s Laws Q.9CQ
Water sprays off a rapidly turning bicycle wheel. Why?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws9cqs

Chapter 6 Applications Of Newton’s Laws Q.9P
A tie of uniform width is laid out on a table, with a fraction of its length hanging over the edge. Initially, the tie is at rest. (a) If the fraction hanging from the table is increased, the tie eventually slides to the ground. Explain. (b) What is the coefficient of static friction between the tie and the table if the tie begins to slide when one-fourth of its length hangs over the edge?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws9ps

Chapter 6 Applications Of Newton’s Laws Q.10CQ
Can an object be in equilibrium if it is moving? Explain.
Solution:
Answer: Yes
Explanation:
If a body moving with constant velocity, it acted upon by zero net force. The body is said to be in equilibrium, if net force acting on it is zero. Therefore, object can be in equilibrium if it is moving with a constant velocity.

Chapter 6 Applications Of Newton’s Laws Q.10P
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws10p
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws10ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws10ps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws10ps2

Chapter 6 Applications Of Newton’s Laws Q.11CQ
In a dramatic circus act, a motorcyclist drives his bike around the inside of a vertical circle. How is this possible, considering that the motorcycle is upside down at the top of the circle?
Solution:
The motorcyclist drives his bike around the inside vertical circle at a very high speed. Because of this high speed, sufficient centripetal force appears away from the center and is much higher than the weight of the motorcyclist and his bike when it is upside down.

Chapter 6 Applications Of Newton’s Laws Q.11P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.12CQ
The gravitational attraction of the Earth is only slightly less at the altitude of an orbiting spacecraft than it is on the Earth’s surface. Why is it, then, that astronauts feel weightless?
Solution:
Answer:
In this case, two astronauts are in the constant-free fall motion as they are in orbiting.
For constant free-fall motion, the net gravitational force of attraction acting on the astronauts is zero. Hence, astronauts feel weightless. This is just resembles to the case, if elevator drops downward in free fall motion, you will feel weightless inside the elevator.

Chapter 6 Applications Of Newton’s Laws Q.12P
A 48-kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 26°, the crate begins to slide downward. (a) What is the coefficient of static friction between the crate and the ramp? (b) At what angle does the crate begin to slide if its mass is doubled?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.13CQ
A popular carnival ride has passengers stand with their backs against the inside wall of a cylinder. As the cylinder begins to spin, the passengers feel as if they are being pushed against the wall. Explain.
Solution:
During the carnival ride, when the cylinder begins to spin, its centripetal force is exerted on the passengers. This force, which is radially inward, is supplied by the wall of the cylinder.

Chapter 6 Applications Of Newton’s Laws Q.13P
A 97-kg sprinter wishes to accelerate from rest to a speed of 13 m/s in a distance of 22 m. (a) What coefficient of static friction is required between the sprinter’s shoes and the track? (b) Explain the strategy used to find the answer to part (a).
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.14CQ
Referring to Question, after the cylinder reaches operating speed, the floor is lowered away, leaving the passengers “stuck” to the wall. Explain.
(Answers to odd-numbered Conceptual Questions can be found in the back of the book.) A popular carnival ride has passengers stand with their backs against the inside wall of a cylinder. As the cylinder begins to spin, the passengers feel as if they are being pushed against the wall. Explain
Solution:
Reaching the operating speed, the centripetal force acting on the man is sufficient to counter his weight, which is responsible for the fall. Thus, even without the base, the man sticks to the wall.

Chapter 6 Applications Of Newton’s Laws Q.14P
Coffee To Go A person places a cup of coffee on the roof of her car while she dashes back into the house for a forgotten item. When she returns to the car, she hops in and takes off with the coffee cup still on the roof. (a) If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, what is the maximum acceleration the car can have without Causing the cup to slide? Ignore the effects of air resistance. (b) What is the smallest amount of time in which the person can accelerate the car from rest to 15 m/s and still keep the coffee cup on the roof?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws14ps

Chapter 6 Applications Of Newton’s Laws Q.15CQ
Your car is stuck on an icy side street. Some students on their way to class see your predicament and help out by sitting on the trunk of your car to increase its traction. Why does this help?
Solution:
Students sitting on the trunk of the car increase the normal force between the tires and the road. The force of friction is directly proportional to the normal, so this increases the frictional force enough so that the car moves.

Chapter 6 Applications Of Newton’s Laws Q.15P
Force Times Distance I At the local hockey rink, a puck with a mass of 0.12 kg is given an initial speed of v = 5.3 m/s. (a) If the coefficient of kinetic friction between the ice and the puck is 0.11, what distance d does the puck slide before coming to rest? (b) If the mass of the puck is doubled, does the frictional force F exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping distance of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that . (The significance of this result will be discussed in Chapter 7, where we will see that is the kinetic energy of an object.)
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.16CQ
The parking brake on a car causes the rear wheels to lock up. What would be the likely consequence of applying the parking brake in a car that is in rapid motion? (Note: Do not try this at home.)
Solution:
If the parking brake is applied while the car is in motion, and the rear wheels begin to skid across the pavement. This means that the friction acting on the rear wheels is kinetic friction. This kinetic friction is smaller than the static friction experienced by the front wheels. As a result, the rear wheels overtake the front wheels causing the car to spin around, and the rear wheels begin to move first.

Chapter 6 Applications Of Newton’s Laws Q.16P
Force Times Time At the local hockey rink, a puck with a mass of 0.12 kg is given an initial speed of v0 = 6.7 m/s. (a) If the coefficient of kinetic friction between the ice and the puck is 0.13, how much time f does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force F exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that Ft = mv0. (The significance of this result will be discussed in Chapter 9, where we will see that mv is the momentum of an object.)
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.17CQ
The foot of your average gecko is covered with billions of tiny hair tips—called spatulae—that are made of keratin, the protein found in human hair. A subtle shift of the electron distribution in both the spatulae and the wall to which a gecko clings produces an adhesive force by means of the van der Waals interaction between molecules. Suppose a gecko uses its spatulae to cling to a vertical windowpane. If you were to describe this situation in terms of a coefficient of static friction, µs, what value would you assign to µs? Is this a sensible way to model the gecko’s feat? Explain.
Solution:
The normal force exerted on a gecko by the vertical wall is zero. For the gecko to stay in place, its static friction must exert an upward force equal to the gecko’s weight. In order for this to happen, the normal force should be zero, and the coefficient of static friction should be infinite.

Chapter 6 Applications Of Newton’s Laws Q.17P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.18CQ
Discuss the physics involved in the spin cycle of a washing machine. In particular, how is circular motion related to the removal of water from the clothes?
Solution:
As the basket within a washing machine rotates, the clothes collect on the rim of the basket because of the centripetal force acting on the rotation. The basket exerts an inward force on the clothes, causing them to follow a circular path. The water contained in the clothes, however, is able to pass through the holes of the basket where it can be drained from the machine.

Chapter 6 Applications Of Newton’s Laws Q.18P
The coefficient of kinetic friction between the tires of your car and the roadway is µ. (a) If your initial speed is v and you lock your tires during braking, how far do you skid? Give your answer in terms of v, µ, and m, the mass of your car. (b) If you double your speed, what happens to the stopping distance? (c) What is the stopping distance for a truck with twice the mass of your car, assuming the same initial speed and coefficient of kinetic friction?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.19CQ
The gas pedal and the brake pedal are capable of causing a car to accelerate. Can the steering wheel also produce an acceleration? Explain.
Solution:
Yes, the steering wheel can accelerate a car by changing its direction of motion.

Chapter 6 Applications Of Newton’s Laws Q.19P
A certain spring has a force constant k. (a) If this spring is cut in half, does the resulting half spring have a force constant that is greater than, less than, or equal to k? (b) If two of the original full-length springs are connected end to end, does the resulting double spring have a force constant that is greater than, less than, or equal to k?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.20CQ
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws20cqp
Solution:
People on the outer rim of a rotating space station must experience a force directed toward the center of the station in order to follow a circular path. This force is applied by the floor of the station, which is really its outermost wall. Because people feel the upward force acting on them from the floor, just as they would on Earth, the sensation is like an artificial gravity.

Chapter 6 Applications Of Newton’s Laws Q.20P
Pulling up on a rope, you lift a 4.35-kg bucket of water from a well with an acceleration of 1.78 m/s2. What is the tension in the rope?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws20ps

Chapter 6 Applications Of Newton’s Laws Q.21CQ
When rounding a corner on a bicycle or a motorcycle, the driver leans inward, toward the center of the circle. Why?
Solution:
When a bicycle rider leans inward on a turn, the force applied to the bicycle wheels by the ground is both upward and inward. It is this inward force that produces the rider’s centripetal acceleration.

Chapter 6 Applications Of Newton’s Laws Q.21P
When a 9.09-kg mass is placed on top of a vertical spring, the spring compresses 4.18 cm. Find the force constant of the spring.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws21ps

Chapter 6 Applications Of Newton’s Laws Q.22CQ
In Robin Hood: Prince of Thieves, starring Kevin Costner, Robin swings between trees on a vine that is on fire. At the lowest point of his swing, the vine bums through and Robin begins to fall. The next shot, from high up in the trees, shows Robin falling straight downward. Would you rate the physics of this scene “Good,” “Bad,” or “Ugly”? Explain.
PROBLEMS AND CONCEPTUAL EXERCISES
Note: Answers to odd-numbered Problems and Conceptual Exercises can be found in the back of the book. IP denotes an integrated problem, with both conceptual and numerical parts; BIO identifies problems of biological or medical interest; CE indicates a conceptual exercise, Predict/Explain problems ask for two responses: (a) your prediction of a physical outcome, and (b) the best explanation among three provided. On all problems, red bullets (,,) are used to indicate the level of difficulty.
SECTION 6-1 FRICTIONAL FORCES
Solution:
The physics of this scene is somewhere between bad and ugly. When the rope burns through, the robin is moving horizontally. This horizontal motion should continue as the robin falls, leading to a parabolic trajectory rather than the straight downward drop shown in the movie.

Chapter 6 Applications Of Newton’s Laws Q.22P
A 110-kg box is loaded into the trunk of a car. If the height of the car’s bumper decreases by 13 cm, what is the force constant of its rear suspension?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws22ps

Chapter 6 Applications Of Newton’s Laws Q.23P
A 50.0-kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 15.0° above the horizontal, Find the tension in the ropes.
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.24P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.25P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.26P
The equilibrium length of a certain spring with a force constant of k = 250 N/m is 0.18 m. (a) What is the magnitude of the force that is required to hold this spring at twice its equilibrium length? (b) Is the magnitude of the force required to keep the spring compressed to half its equilibrium length greater than, less than, or equal to the force found in part (a)? Explain.
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.27P
Illinois Jones is being pulled from a snake pit with a rope that breaks if the tension in it exceeds 755 N. (a) If Illinois Jones has a mass of 70.0 kg and the snake pit is 3.40 m deep, what is the minimum tune that is required to pull our intrepid explorer from the pit? (b) Explain why the rope breaks if Jones is polled from the pit in less time than that calculated in part (a).
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.28P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.29P
Your friend’s 13.6-g graduation tassel hangs from his rearview mirror. (a) When he acceleration stoplight, the tassel deflects backward toward the car. Explain. (b) If the tassel hangs at an angle of 6.4 the vertical, what is the acceleration of the car?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.30P
In Problem 29, (a) find the tension in the siring holding the tassel. (b) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
Problem 29
Your friend’s 13.6-g graduation tassel hangs on a string from his rearview mirror. (a) When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car. Explain. (b) If the tassel hangs at an angle of 6.44° relative to the vertical, what is the acceleration of the car?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.31P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.32P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.33P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.34P
Pulling the string on a bow back with a force of 28.7 lb, an archer prepares to shoot an arrow. If the archer pulls in the cen­ter of the string, and the angle between the two halves is 138°, what is the tension in the string?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.35P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.36P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.37P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.38P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.39P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.40P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.41P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.42P
In Example (Connected Blocks), suppose m1 and m2 are both increased by a factor of 2. (a) Does the acceleration of the blocks increase, decrease, or stay the same? (b) Does the tension in the string increase, decrease, or stay the same?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.43P
CE Predict/Explain Suppose m1 and m2 in Example (Atwood’s Machine) are both increased by 1 kg. Does the acceleration of the blocks increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. The net force acting on the blocks is the same, but the total mass that must be accelerated is greater.
II. The difference in the masses is the same, and this is what determines the net force on the system.
III. The force exerted on each block is greater, leading to an increased acceleration.
Solution:
(a) The acceleration decreases when both the masses are increased by 1kg.
(b) This is because of the net force acting on the blocks is same, but the total mass that must be accelerated is greater. Therefore option I is the best explanation.

Chapter 6 Applications Of Newton’s Laws Q.44P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.45P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.46P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.47P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.48P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.49P
A 7.7-N force pulls horizontally on a 1.6-kg block that slides on a smooth horizontal surface. This block is connected by a horizontal string to a second block of mass m2 = 0.83 kg on the same surface, (a) What is the acceleration of the blocks? (b) What is the tension in the string? (c) If the mass of block 1 is increased, does the tension in the string increase, decrease, or stay the same?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.50P
Buckets and a Pulley Two buckets of sand hang from opposite ends of a rope that passes over an ideal pulley. One bucket is full and weighs 120 N; the other bucket is only partly filled and weighs 63 N. (a) Initially, you hold onto the lighter bucket to keep it from moving. What is the tension in the rope? (b) You release the lighter bucket and the heavier one descends. What is the tension in the rope now? (c) Eventually the heavier bucket lands and the two buckets come to rest. What is the tension in the rope now?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.51P
Suppose you stand on a bathroom scale and get a reading of 700 N. In principle, would the scale read more, less, or the same if the Earth did not rotate?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.52P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.53P
A car is driven with constant speed around a circular track. Answer the of the following questions with “Yes” or “No.” (a) Is the car’s velocity constant? (b) Is its speed constant? (c) Is the magnitude of its acceleration constant? (d) Is the direction of its acceleration constant?
Solution:
(a) No.
As the direction of the car is changing time to time, therefore the velocity of the car is not constant.
(c) Yes
The magnitude of acceleration is constant, as it is proportional to the square of the speed of the car which is constant.
(d) No, the direction of acceleration is not constant. It changes from point to point.

Chapter 6 Applications Of Newton’s Laws Q.54P
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws54p
Solution:
Consider a puck attached to a string undergoes circular motion on an air table. If the string breaks at a point on the circumference of the circle,
The motion of the puck directed along a tangential line drawn to the circle at that mentioned point.
From given diagram the motion of the puck will be path B from diagram clearly know that path B is the tangential to the circle at given point.

Chapter 6 Applications Of Newton’s Laws Q.55P
When you take your 1300-kg car out for a spin, you go around a corner of radius 59 m with a speed of 16 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn’t skid, what is the force exerted on it by static friction?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws55ps

Chapter 6 Applications Of Newton’s Laws Q.56P
Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 52,000 times the acceleration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.5 cm.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws56ps

Chapter 6 Applications Of Newton’s Laws Q.57P
A Human Centrifuge To test the effects of high acceleration on the human body, the National Aeronautics and Space Administration (NASA) has constructed a large centrifuge at the Manned Spacecraft Center in Houston. In this device, astronauts are placed in a capsule that moves in a circular path with a radius of 15 m. If the astronauts in this centrifuge experience a centripetal acceleration 9.0 times that of gravity, what is the linear speed of the capsule?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws57ps

Chapter 6 Applications Of Newton’s Laws Q.58P
A car goes around a curve on a road that is banked at an angle of 33.5°. Even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is 22.7 m/s. What is the radius of the curve?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.59P
Jill of the Jungle swings on a vine 6.9 m long. What is the tension in the vine if Jill, whose mass is 63 kg, is moving at 2.4 m/s when the vine is vertical?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws59ps

Chapter 6 Applications Of Newton’s Laws Q.60P
In Problem, (a) how does the tension in the vine change if Jill’s speed is doubled? Explain. (b) How does the tension change if her mass is doubled instead? Explain.
Jill of the Jungle swings on a vine 6.9 m long. What is the tension in the vine if Jill, whose mass is 63 kg, is moving at 2.4 m/s when the vine is vertical?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.61P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.62P
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Solution:

Chapter 6 Applications Of Newton’s Laws Q.63P
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Solution:
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Chapter 6 Applications Of Newton’s Laws Q.64P
You swing a 4.6-kg bucket of water in a vertical circle of radius 1.3 m. (a) What speed must the bucket have if it is to complete the circle without spilling any water? (b) How does your answer depend on the mass of the bucket?
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.65GP
If you weigh yourself on a bathroom scale at the equator, is the reading you get greater than, less than, or equal to the reading you get if you weigh yourself at the North Pole?
Solution:
Your weight at North Pole is greater than your weight at the equator.
At equator you are moving in a circular path because of the rotation of earth. Therefore part of the force of gravity acting on you is providing your centripetal acceleration. And the rest shows up as a reduced weight on the scale. But at North Pole you are not moving in a circular path. Therefore the force of gravity shows up as weight on the scale. Therefore the reading you get on the bath room scale for your weight at equator is less than the reading you get on the bath room scale for your weight at North Pole.

Chapter 6 Applications Of Newton’s Laws Q.66GP
An object moves on a flat surface with an acceleration of constant magnitude. If the acceleration is always perpendicular to the object’s direction of motion, (a) is the shape of the object’s path circular, linear, or parabolic? (b) During its motion, does the object’s velocity change in direction but not magnitude, change in magnitude but not direction, or change in both magnitude and direction? (c) Does its speed increase, decrease, or stay the same?
Solution:
(a) Here the magnitude of acceleration of the object is constant and the acceleration is always perpendicular to the object’s direction of motion. Therefore the shape of the path is circular.
(b) As the acceleration is perpendicular to the direction of motion, therefore the direction of the velocity changes but not its magnitude.
(c) As there is no acceleration in the direction of the motion of the object, therefore the speed of the object stay the same.

Chapter 6 Applications Of Newton’s Laws Q.67GP
BIO Maneuvering a Jet Humans lose consciousness if exposed to prolonged accelerations of more than about 7g. This is of concern to jet fighter pilots, who may experience centripetal accelerations of this magnitude when making high-speed turns. Suppose we would like to decrease the centripetal acceleration of a jet. Rank the following changes in flight path in order of how effective they are in decreasing the centripetal acceleration, starting with the least effective: A, decrease the turning radius by a factor of two; B, decrease the speed by a factor of three; or C; increase the turning radius by a factor of four.
Solution:
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Chapter 6 Applications Of Newton’s Laws Q.68GP
Gravitropism As plants grow, they tend to align their stems and roots along the direction of the gravitational field. This tendency, which is related to differential concentrations of plant hormones known as auxins, is referred to as gravitropism. As an illustration of gravitropism, experiments show that seedlings placed in pots on the rim of a rotating turntable do not grow in the vertical direction. Do you expect their stems to tilt inward—toward the axis of rotation—or outward—away from the axis of rotation?
Solution:
The direction of growth of a plant is opposite to the effective direction of gravitational force at that place. If a growing plant is kept on the rim of a rotating table, it tilts inwards.
Consider the example of a simple pendulum kept on the rim of the turntable. When the table is at rest, the pendulum hangs vertically downwards while the plant grows vertically upwards. When the table is rotating, the pendulum tilts outward, and plant tilts in the opposite direction, i.e., inward.

Chapter 6 Applications Of Newton’s Laws Q.69GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws69p
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws69ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws69ps1

Chapter 6 Applications Of Newton’s Laws Q.70GP
Find the centripetal acceleration at the top of a test tube in a centrifuge, given that the top is 4.2 cm from the axis of rotation and that its linear speed is 77 m/s.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws70ps

Chapter 6 Applications Of Newton’s Laws Q.71GP
Find the coefficient of kinetic friction between a 3.85-kg block and the horizontal surface on which it rests if an 850-N/m spring must be stretched by 6.20 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws71ps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws71ps1

Chapter 6 Applications Of Newton’s Laws Q.72GP
A child goes down a playground slide that is inclined at an angle of 26.5° below the horizontal. Find the acceleration of the child given that the coefficient of kinetic friction between the child and the slide is 0.315.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws72gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws72gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws72gps2

Chapter 6 Applications Of Newton’s Laws Q.73GP
When a block is placed on top of a vertical spring, the spring compresses 3.15 cm. Find the mass of the block, given that the force constant of the spring is 1750 N/m.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws73gps

Chapter 6 Applications Of Newton’s Laws Q.74GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws74gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws74gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws74gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws74gps2

Chapter 6 Applications Of Newton’s Laws Q.75GP
A force of 9.4 N pulls horizontally on a 1.1-kg block that slides on a rough, horizontal surface. This block is connected by a horizontal string to a second block of mass m2 = 1.92 kg on the same surface. The coefficient of kinetic friction is µk = 0.24 for both blocks. (a) What is the acceleration of the blocks? (b) What is the tension in the string?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws75gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws75gps1

Chapter 6 Applications Of Newton’s Laws Q.76GP
You swing a 3.25-kg bucket of water in a vertical circle of radius 0.950 m. At the top of the circle the speed of the bucket is 3.23 m/s; at the bottom of the circle its speed is 6.91 m/s. Find the tension in the rope tied to the bucket at (a) the top and (b) the bottom of the circle.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws76gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws76gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws76gps2

Chapter 6 Applications Of Newton’s Laws Q.77GP
A 14-g coin slides upward on a surface that is inclined at an angle of 18° above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.23; the coefficient of static friction is 0.35. Find the magnitude and direction of the force of friction (a) when the coin is sliding and (b) after it comes to rest.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws77gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws77gps1

Chapter 6 Applications Of Newton’s Laws Q.78GP
In Problem, the angle of the incline is increased to 25°. Find the magnitude and direction of the force of friction when the coin is (a) sliding upward initially and (b) sliding back downward later.
A 14-g coin slides upward on a surface that is inclined at an angle of 18° above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.23; the coefficient of static friction is 0.35. Find the magnitude and direction of the force of friction (a) when the coin is sliding and (b) after it comes to rest.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws78gps

Chapter 6 Applications Of Newton’s Laws Q.79GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws79gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws79gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws79gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws79gps2

Chapter 6 Applications Of Newton’s Laws Q.80GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws80gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws80gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws80gps1

Chapter 6 Applications Of Newton’s Laws Q.81GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps3
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps4
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws81gps5

Chapter 6 Applications Of Newton’s Laws Q.82GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws82gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws82gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws82gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws82gps2

Chapter 6 Applications Of Newton’s Laws Q.83GP
A picture hangs on the wall suspended by two strings, as shown in Figure. The tension in string 2 is 1.7 N. (a) Is the tension in string 1 greater than, less than, or equal to 1.7 N? Explain. (b) Verify your answer to part (a) by calculating the tension in string 1. (c) What is the mass of the picture?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws83gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws83gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws83gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws83gps3

Chapter 6 Applications Of Newton’s Laws Q.84GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws84gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws84gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws84gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws84gps2

Chapter 6 Applications Of Newton’s Laws Q.85GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws85gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws85gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws85gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws85gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws85gps3

Chapter 6 Applications Of Newton’s Laws Q.86GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws86gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws86gps

Chapter 6 Applications Of Newton’s Laws Q.87GP
Find the coefficient of kinetic friction between a 4.7-kg block and the horizontal surface on which it rests if an 89-N/m spring must be stretched by 2.2 cm to pull the block with constant speed. Assume the spring pulls in a direction 13° above the horizontal.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws87gp
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws87gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws87gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws87gps2

Chapter 6 Applications Of Newton’s Laws Q.88GP
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws88gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws88gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws88gps2

Chapter 6 Applications Of Newton’s Laws Q.89GP
In a daring rescue by helicopter, two men with a combined mass of 172 kg are lifted to safety. (a) If the helicopter lifts the men straight up with constant acceleration, is the tension in the rescue cable greater than, less than, or equal to the combined weight of the men? Explain. (b) Determine the tension in the cable if the men are lifted with a constant acceleration of 1.10 m/s2.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws89gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws89gps1

Chapter 6 Applications Of Newton’s Laws Q.90GP
At the airport, you pull a 18-kg suitcase across the floor with a strap that is at an angle of 45° above the horizontal. Find (a) the normal force and (b) the tension in the strap, given that the suitcase moves with constant speed and that the coefficient of kinetic friction between the suitcase and the floor is 0.38.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws90gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws90gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws90gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws90gps3

Chapter 6 Applications Of Newton’s Laws Q.91GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws91gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws91gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws91gps1

Chapter 6 Applications Of Newton’s Laws Q.92GP
A 0.16-g spider hangs from the middle of the first thread of its future web. The thread makes an angle of 7.2° with the horizontal on both sides of the spider. (a) What is the tension in the thread? (b) If the angle made by the thread had been less than 7.2°, would its tension have been greater than, less than, or the same as in part (a)? Explain.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws92gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws92gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws92gps2

Chapter 6 Applications Of Newton’s Laws Q.93GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws93gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws93gps

Chapter 6 Applications Of Newton’s Laws Q.94GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws94gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws94gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws94gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws94gps2

Chapter 6 Applications Of Newton’s Laws Q.95GP
A pair of fuzzy dice hangs from a string attached to your rearview mirror. As you turn a corner with a radius of 98 m and a constant speed of 27 mi/h, what angle will the dice make with the vertical? Why is it unnecessary to give the mass of the dice?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws95gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws95gps1

Chapter 6 Applications Of Newton’s Laws Q.96GP
Find the tension in the of the two ropes supporting a hammock if one is at an angle of 18° above the horizontal and the other is at an angle of 35° above the horizontal. The person sleeping in the hammock (unconcerned about tensions and ropes) has a mass of 68 kg.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws96gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws96gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws96gps2

Chapter 6 Applications Of Newton’s Laws Q.97GP
As your plane circles an airport, it moves in a horizontal circle of radius 2300 in with a speed of 390 km/h. If the lift of the airplane’s wings is perpendicular to the wings, at what angle should the plane be banked so that it doesn’t tend to slip sideways?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws97gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws97gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws97gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws97gps3

Chapter 6 Applications Of Newton’s Laws Q.98GP
A block with a mass of 3.1 kg is placed at rest on a surface inclined at an angle of 45° above the horizontal. The coefficient of static friction between the block and the surface is 0.50, and a force of magnitude F pushes upward on the block, parallel to the inclined surface. (a) The block will remain at rest only if F is greater than a minimum value, Fmin, and less than a maximum value, Fmax. Explain the reasons for this behavior. (b) Calculate Fmin. (c) Calculate Fmax.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws98gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws98gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws98gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws98gps3
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws98gps4

Chapter 6 Applications Of Newton’s Laws Q.99GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws99gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws99gps

Chapter 6 Applications Of Newton’s Laws Q.100GP
A child sits on a rotating merry-go-round, 2.3 m from its center. If the speed of the child is 2.2 m/s, what is the minimum coefficient of static friction between the child and the merry-go-round that will prevent the child from slipping?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws100gps

Chapter 6 Applications Of Newton’s Laws Q.101GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws101gp
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws101gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws101gps1
Solution:

Chapter 6 Applications Of Newton’s Laws Q.102GP
A wood block of mass m rests on a larger wood block of mass M that rests on a wooden table. The coefficients of static and kinetic friction between all surfaces are µs and µk, respectively. What is the minimum horizontal force, F, applied to the lower block that will cause it to slide out from under the upper block?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws102gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws102gps1

Chapter 6 Applications Of Newton’s Laws Q.103GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws103gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws103gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws103gps1

Chapter 6 Applications Of Newton’s Laws Q.104GP
The coefficient of static friction between a rope and the table on which it rests is µs. Find the fraction of the rope that can hang over the edge of the table before it begins to slip.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws104gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws104gps1

Chapter 6 Applications Of Newton’s Laws Q.105GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws105gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws105gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws105gps1

Chapter 6 Applications Of Newton’s Laws Q.106GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws106gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws106gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws106gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws106gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws106gps3

Chapter 6 Applications Of Newton’s Laws Q.107GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws107gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws107gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws107gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws107gps2

Chapter 6 Applications Of Newton’s Laws Q.108GP
A Conveyor Belt A box is placed on a conveyor belt that moves with a constant speed of 1.25 m/s. The coefficient of kinetic friction between the box and the belt is 0.780. (a) How much time does it take for the box to stop sliding relative to the belt? (b) How far docs the box move in this time?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws108gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws108gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws108gps2

Chapter 6 Applications Of Newton’s Laws Q.109GP
You push a box along the floor against a constant force of friction. When you push with a horizontal force of 75 N, the acceleration of the box is 0.50 m/s2; when you increase the force to 81 N, the acceleration is 0.75 m/s2. Find (a) the mass of the box and (b) the coefficient of kinetic friction between the box and the floor.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws109gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws109gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws109gps2

Chapter 6 Applications Of Newton’s Laws Q.110GP
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps1
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps2
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps3
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps4
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps5
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws110gps6

Chapter 6 Applications Of Newton’s Laws Q.111PP
On the straight-line segment I in Figure we see that increasing the applied mass from 26 g to 44 g results in a reduction of the end-to-end distance from 21 mm to 14 mm. What is the force constant in N/m on segment I?
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws111gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws111gps

Chapter 6 Applications Of Newton’s Laws Q.112PP
Is the force constant on segment II greater than, less than, or equal to the force constant on segment I?
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws112gp
Solution:
As the slope of the curve in the segment II is less than the slope of the curve in the segment I, therefore the force constant on segment II is less than the force constant on segment I.

Chapter 6 Applications Of Newton’s Laws Q.113PP
Which of the following is the best estimate for the force constant on segment II?
A. 0.83 N/m
B. 1.3 N/m
C. 2.5 N/m
D. 25 N/m
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws113gp
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws113gps

Chapter 6 Applications Of Newton’s Laws Q.114PP
Rank the straight segments I, II, and III in order of increasing “stiffness” of the nasal strip.
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws114gp
Solution:
The stiffness is more for the segment having more force constant. Here the force constant is more for the segment III and less for the segment II. Therefore the rank of the segments in the increasing order of the stiffness is III, I and then II.

Chapter 6 Applications Of Newton’s Laws Q.115IP
Referring to Example Suppose the coefficients of static and kinetic friction between the crate and the truck bed are 0.415 and 0.382, respectively. (a) Does the crate begin to slide at a tilt angle that is greater than, less than, or equal to 23.2°? (b) Verify your answer to part (a) by determining the angle at which the crate begins to slide. (c) Find the length of time it takes for the crate to slide a distance of 2.75 m when the tilt angle has the value found in part (b).
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws115gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws115gps1

Chapter 6 Applications Of Newton’s Laws Q.116IP
Referring to The crate begins to slide when the tilt angle is 17.5°. When the crate reaches the bottom of the flatbed, after sliding a distance of 2.75 m, its speed is 3.11 m/s. Find (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the crate and the flatbed.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws116gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws116gps1

Chapter 6 Applications Of Newton’s Laws Q.117IP
Referring to Example Suppose that the mass on the frictionless tabletop has the value m1 = 2.45 kg. (a) Find the value of m2 that gives an acceleration of 2.85 m/s2. (b) What is the corresponding tension; T, in the string? (c) Calculate the ratio T/m2g and show that it is less than 1, as expected.
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws117gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws117gps1

Chapter 6 Applications Of Newton’s Laws Q.118IP
Referring to Example At what speed will the force of static friction exerted on the car by the road be equal to half the weight of the car? The mass of the car is m = 1200 kg, the radius of the corner is r = 45 m, and the coefficient of static friction between the tires and the road is µs = 0.82. (b) Suppose that the mass of the car is now doubled, and that it moves with a speed that again makes the force of static friction equal to half the car’s weight. Is this new speed greater than, less than, or equal to the speed in part (a)?
Solution:
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws118gps
Mastering Physics Solutions Chapter 6 Applications Of Newton's Laws118gps1

Mastering Physics Solutions Chapter 5 Newton’s Laws Of Motion

Mastering Physics Solutions Chapter 5 Newton’s Laws Of Motion

Mastering Physics Solutions

Chapter 5 Newton’s Laws Of Motion Q.1CQ
Driving down the road, you hit the brakes suddenly. As a result, your body moves toward the front of the car. Explain, using Newton’s laws.
Solution:
When the brakes are applied, the car slows down. The body, however, keeps moving at the same speed. Thus, because of inertia, the body moves toward the front of the car.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion1cqs

Chapter 5 Newton’s Laws Of Motion Q.1P
CE An object of mass m is initially at rest. After a force of magnitude F acts on it for a time T, the object has a speed v. Suppose the mass of the object is doubled, and the magnitude of the force acting on it is quadrupled. In terms of T, how long does it take for the object to accelerate from rest to a speed v now?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion1ps

Chapter 5 Newton’s Laws Of Motion Q.2CQ
You’ve probably seen pictures of someone pulling a tablecloth out from under glasses, plates, and silverware set out for a formal dinner. Perhaps you’ve even tried it yourself. Using Newton’s laws of motion, explain how this stunt works.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion2cqs

Chapter 5 Newton’s Laws Of Motion Q.2P
On a planet far, far away, an astronaut picks up a rock. The rock has a mass of 5.00 kg, and on this particular planet its weight is 40.0 N. Tf the astronaut exerts an upward force of 46.2 N on the rock, what is its acceleration?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion2ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion2ps1

Chapter 5 Newton’s Laws Of Motion Q.3CQ
As you read this, you arc most likely sitting quietly in a chair. Can you conclude, therefore, that you are at rest? Explain.
Solution:
You are not at absolute rest, you are at rest with respect to the other objects in your surroundings. However when you are viewed from other planets, you are not at rest relative to the vantage point of the other planets.

Chapter 5 Newton’s Laws Of Motion Q.3P
In a grocery store, you push a 12.3-kg shopping cart with a force of 10.1 N. If the cart starts at rest, how far does it move in 2.50 s?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion3ps

Chapter 5 Newton’s Laws Of Motion Q.4CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion4cq
Solution:
When a dog shakes its body, the water remains at rest because of inertia. The water drops fall away from the body when the position of the dog’s body is changed rapidly.

Chapter 5 Newton’s Laws Of Motion Q.4P
You are pulling your little sister on her sled across an icy (fric-tionless) surface. When you exert a constant horizontal force of 120 N, the sled has an acceleration of 2.5 m/s2. If the sled has a mass of 7.4 kg, what is the mass of your little sister?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion4ps

Chapter 5 Newton’s Laws Of Motion Q.5CQ
A young girl slides down a rope. As she slides faster and faster she tightens her grip, increasing the force exerted on her by the rope. What happens when this force is equal in magnitude to her weight? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion5cqs

Chapter 5 Newton’s Laws Of Motion Q.5P
· A 0.53-kg billiard ball initially at rest is given a speed of 12 m/s during a time interval of 4.0 ms. What average force acted on the ball during this time?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion5ps

Chapter 5 Newton’s Laws Of Motion Q.6CQ
A drag-racing car accelerates forward because of the force exerted on it by the road. Why, then, does it need an engine? Explain.
Solution:
The drag racer needs an engine to turn the wheels, which in turn makes them push against the ground. When the wheels push against the ground, the ground is able to exert a reaction force on the car to move it.

Chapter 5 Newton’s Laws Of Motion Q.6P
A 92-kg water skier floating in a lake is pulled from rest to a speed of 12 m/s in a distance of 25 m. What is the net force exerted on the skier, assuming his acceleration is constant?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion6ps

Chapter 5 Newton’s Laws Of Motion Q.7CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion7cq
Solution:
(A) The upper string is exposed to two forces. One is the downward weight attached and the other is the force applied. So, in this case, the upper string will break.
(B) Because of the inertia of the block, the lower string will break.

Chapter 5 Newton’s Laws Of Motion Q.7P
CE Predict/Explain You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball 2, which is made of wood. The upward force due to air resistance is the same for both balls, (a) Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball 2? (b) Choose the best explanation from among the following:
I. The acceleration of gravity is the same for all objects, regardless of mass.
II. The more massive ball is harder to accelerate.
III. Air resistance has less effect on the more massive ball.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion7ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion7ps1

Chapter 5 Newton’s Laws Of Motion Q.8CQ
An astronaut on a space walk discovers that his jet pack no longer works, leaving him stranded 50 m from the spacecraft. If the jet pack is removable, explain how the astronaut can still use it to return to the ship.
Solution:
The astronaut should push the jetpack away from him, in the opposite direction from the spaceship. As a result, the reaction force exerted on him by the pack will accelerate him toward the ship.

Chapter 5 Newton’s Laws Of Motion Q.8P
IP A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.750 m, what force does the ground exert on her? (b) If the parachutist comes to rest over a shorter distance, is the force exerted by the ground greater than, less than, or the same as in part (a)? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion8ps

Chapter 5 Newton’s Laws Of Motion Q.9CQ
Two untethered astronauts on a space walk decide to take a break and play catch with a baseball. Describe what happens as the game of catch progresses.
Solution:
Each time the astronauts catch or throw, an equal and opposite force acts on them. This causes the astronauts to move farther from each other, with increasing speed.

Chapter 5 Newton’s Laws Of Motion Q.9P
IP In baseball, a pitcher can accelerate a 0.15-kg ball from rest to 98 mi/h in a distance of 1.7 m. (a) What is the average force exerted on the ball during the pitch? (b) If the mass of the ball is increased, is the force required of the pitcher increased, decreased, or unchanged? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion9ps

Chapter 5 Newton’s Laws Of Motion Q.10CQ
What are the action-reaction forces when a baseball bat hits a fast ball? What is the effect of each force?
Solution:
Because of action-reaction forces, the ball will be returned with the same force that the ball applied to the bat.
The force exerted on the bat by the ball is action force, and the force exerted on the ball by the bat is reaction force.
The force exerted on the ball changes its direction.

Chapter 5 Newton’s Laws Of Motion Q.10P
A major-league catcher gloves a 92-mi/h pitch and brings it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion10ps

Chapter 5 Newton’s Laws Of Motion Q.11CQ
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion11cq
Solution:
Mr. Ed’s reasoning is incorrect because he is adding two action-reaction forces that act on different objects.
Wilbur should point out that the net force exerted on the cart is simply the force exerted on it by Mr. Ed, and so the cart will accelerate.
The equal and opposite reaction force acts on Mr. Ed, and does not cancel the force acting on the cart.

Chapter 5 Newton’s Laws Of Motion Q.11P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion11p
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion11ps

Chapter 5 Newton’s Laws Of Motion Q.12CQ
A whole brick has more mass than half a brick, thus the whole brick is harder to accelerate. Why doesn’t a whole brick fall more slowly than half a brick? Explain.
Solution:
The whole brick also experiences twice the gravitational force.
As a result, these two effects (more inertia, more force) cancel each other out exactly. The free-fall acceleration is independent of mass.

Chapter 5 Newton’s Laws Of Motion Q.12P
Stopping a 747 A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50 × 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 × 105 N, (a) what is its speed 7.50 s later? (b) How far has it traveled in this time?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion12ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion12ps1

Chapter 5 Newton’s Laws Of Motion Q.13CQ
The force exerted by gravity on a whole brick is greater than the force exerted by gravity on half a brick. Why, then, doesn’t a whole brick fall faster than half a brick? Explain.
Solution:
Acceleration of the whole brick is the same as that of half the brick. This is because acceleration is directly proportional to force and inversely proportional to mass.

Chapter 5 Newton’s Laws Of Motion Q.13P
IP A drag racer crosses the finish line doing 202 mi/h and promptly deploys her drag chute (the small parachute used for braking), (a) What force must the drag chute exert on the 891-kg car to slow it to 45.0 mi/h in a distance of 185 m? (b) Describe the strategy you used to solve part (a).
SECTION 5-4 NEWTON’S THIRD LAW OF MOTION
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion13ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion13ps1

Chapter 5 Newton’s Laws Of Motion Q.14CQ
Is it possible for an object at rest to have only a single force acting on it? If your answer is yes, provide an example. If your answer is no, explain why not.
Solution:
No.
If only a single force acts on the object, it will not stay at rest but will accelerate in the direction of the force.

Chapter 5 Newton’s Laws Of Motion Q.14P
CE Predict/Explain A small car collides with a large truck, (a) Is the magnitude of the force experienced by the car greater than, less than, or equal to the magnitude of the force experienced by the truck? (b) Choose the best explanation from among the following:
I. Action-reaction forces always have equal magnitude.
II. The truck has more mass, and hence the force exerted on it is greater.
III. The massive truck exerts a greater force on the lightweight car.
Solution:
(a) The magnitude of the force experienced by the car is equal to magnitude of the force experienced by the truck.
(b) This is because action reaction forces are always equal in magnitude according to ’s third law. So, option I is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.15CQ
Is it possible for an object to be in motion and yet have zero net force acting on it? Explain.
Solution:
Yes.
When the object is moving with a constant velocity, its acceleration is zero. Thus, the force acting on the object is also zero.

Chapter 5 Newton’s Laws Of Motion Q.15P
CE Predict/Explain A small car collides with a large truck, (a) Is the acceleration experienced by the car greater than, less than, or equal to the acceleration experienced by the truck? (b) Choose the best explanation from among the following:
I. The truck exerts a larger force on the car, giving it the greater acceleration.
II. Both vehicles experience the same magnitude of force, therefore the lightweight car experiences the greater acceleration.
III. The greater force exerted on the truck gives it the greater acceleration.
Solution:
(a) The acceleration experienced by the car is greater than the acceleration experienced by the truck.
(b) This is because of both cars experience same magnitude of force, therefore the lightweight car experiences greater acceleration. So, option II is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.16CQ
A bird cage, with a parrot inside, hangs from a scale. The parrot decides to hop to a higher perch. What can you say about the reading on the scale (a) when the parrot jumps, (b) when the parrot is in the air, and (c) when the parrot lands on the second perch? Assume that the scale responds rapidly so that it gives an accurate reading at all times.
Solution:
(a) The scale goes down at the moment the parrot jumps.
This happens because the parrot pushes down on its perch in order to jump, and as a result the scale reads a larger value.
(b) When the bird is in the air, the scale just reads the weight of the cage.
(c) When the parrot lands on a higher perch, the scale goes down and then reads a higher value again. This happens because the perch exerts an upward force on the bird in order to bring the bird to rest.

Chapter 5 Newton’s Laws Of Motion Q.16P
You hold a brick at rest in your hand. (a) How many forces act on the brick? (b) Identify these forces, (c) Are these forces equal in magnitude and opposite in direction? (d) Are these forces an action-reaction pair? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion16ps
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Chapter 5 Newton’s Laws Of Motion Q.17CQ
Suppose you jump from the cliffs of Acapulco and perform a perfect swan dive. As you fall, you exert an upward force on the Earth equal in magnitude to the downward force the Earth exerts on you. Why, then, does it seem that you are the one doing all the accelerating? Since the forces are the same, why aren’t the accelerations?
Solution:
The acceleration of a body is inversely proportional to its mass. Earth has a huge mass compared to a man, and so has negligible acceleration towards the man. Thus, the forces are the same but not the accelerations.

Chapter 5 Newton’s Laws Of Motion Q.17P
Referring to Problem 16, you are now accelerating the brick upward, (a) How many forces act on the brick in this case? (b) Identify these forces, (c) Are these forces equal in magnitude and opposite in direction? (d) Are these forces an action-reaction pair? Explain.
Solution:
(A) There are two forces.
(B) Gravitational force and the upward force applied by your hand.
(C) No, these forces are not equal and opposite since the brick is moving up.
(D) No.

Chapter 5 Newton’s Laws Of Motion Q.18CQ
A friend tells you that since his car is at rest, there are no forces acting on it. How would you reply?
Solution:
The car is at rest, and the net force acting on the car is zero.
However, it is wrong to say that no force is acting on the car. When the car is at rest, gravitational force acts on the car, which is balanced by the normal force reaction of the ground upon the car.

Chapter 5 Newton’s Laws Of Motion Q.18P
On vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85 m/s2, (a) What is the net force exerted on the trailer? (b) What force does the traiter exert on the car? (c) What is the net force acting on the car?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.19CQ
Since all objects are “weightless” in orbit, how is it possible for an orbiting astronaut to tell if one object has more mass than another object? Explain.
Solution:
If an astronaut pushes the object, the acceleration can indicate the object’s mass. The heavier mass will have lower acceleration compared to the lighter mass.

Chapter 5 Newton’s Laws Of Motion Q.19P
IP A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push, (a) Is the force experienced by the child more than, less than, or the same as the force experienced by the parent? (b) Is the acceleration of the child more than, less than, or the same as the acceleration of the parent? Explain, (c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the magnitude of the parent’s acceleration?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion19ps
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Chapter 5 Newton’s Laws Of Motion Q.20CQ
To clean a rug, you can hang it from a clothesline and beat it with a tennis racket. Use Newton’s laws to explain why beating the rug should have a cleansing effect.
Solution:
As you hit the rug with the tennis racket you cause it to accelerate rapidly.
The dust from the rug, if it is not attached too firmly, will be left behind as the rug accelerates away from it.

Chapter 5 Newton’s Laws Of Motion Q.20P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion20p
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion20ps
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Chapter 5 Newton’s Laws Of Motion Q.21CQ
If you step off a high board and drop to the water below, you plunge into the water without injury. On the other hand, if you were to drop the same distance onto solid ground, you might break a leg. Use Newton’s laws to explain the difference.
Solution:
When we jump on the ground, we plunge a very small distance compared to jumping into water. The smaller the distance, the larger the acceleration, and thus the greater the force.

Chapter 5 Newton’s Laws Of Motion Q.21P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion21p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.22CQ
A moving object is acted on by a net force. Give an example of a situation in which the object moves (a) in the same direction as the net force, (b) at right angles to the net force, or (c) in the opposite direction of the net force.
Solution:
(a) When a cart is pushed forward and set into motion, the direction of the force acting on it and the direction of its motion are equal.
(b) If a ball or any object is thrown up, then, at the top of its flight, its motion is in the horizontal direction. Additionally, the force of gravity acting on the ball (or any object) is in the vertical direction. Here, the direction of force and the direction of motion are perpendicular to each other. In this example, air resistance is neglected.
(c) While you are riding a bicycle, the frictional force acts in the direction opposite to the direction of motion of the bicycle.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion22cqs
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Chapter 5 Newton’s Laws Of Motion Q.22P
IP Two boxes sit side-by-side on a smooth horizontal surface. The lighter box has a mass of 5.2 kg; the heavier box has a mass of 7.4 kg. (a) Find the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (b) If the 5.0-N force is applied to the heavy box instead, is the contact force between the boxes the same as, greater than, or less than the contact force in part (a)? Explain, (c) Verify your answer to part (b) by calculating the contact force in this case.
Solution:

Chapter 5 Newton’s Laws Of Motion Q.23CQ
Is it possible for an object to be moving in one direction while the net force acting on it is in another direction? If your answer is yes, provide an example. If your answer is no, explain why not.
Solution:
Yes, during the journey of a ball thrown upward, the direction of force is always downward. Thus, the force is in an opposite direction until the ball reaches its maximum height, and then it comes back down.

Chapter 5 Newton’s Laws Of Motion Q.23P
CE A skateboarder on a ramp is accelerated by a nonzero net force. For each of the following statements, state whether it is always true, never true, or sometimes true, (a) The skateboarder is moving in the direction of the net force, (b) The ac-celeration of the skateboarder is at right angles to the net force. (c) The acceleration of the skateboarder is in the same direction as the net force. (d) The skateboarder is instantaneously at rest.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion23ps
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Chapter 5 Newton’s Laws Of Motion Q.24CQ
Since a bucket of water is “weightless” in space, would it hurt to kick the bucket? Explain.
Solution:
Answer: Yes
Explanation:
It is given that the bucket of water is weightless in space, but every object has a certain mass, and also has inertia. As inertia is the tendency of a body to resist any change in its state of motion or rest, so when you kick the bucket, it resists a change in its state of motion by exerting an equal and opposite force, which gives the hurt to your foot.

Chapter 5 Newton’s Laws Of Motion Q.24P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion24p
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion24ps

Chapter 5 Newton’s Laws Of Motion Q.25CQ
In the movie The Rocketeer, a teenager discovers a jet-powered backpack in an old barn. The backpack allows him to fly at incredible speeds. In one scene, however, he uses the backpack to rapidly accelerate an old pickup truck that is being chased by “bad guys.” He does this by bracing his arms against the cab of the pickup and firing the backpack, giving the truck the acceleration of a drag racer. Is the physics of this scene “Good,” “Bad,” or “Ugly?” Explain.
Solution:
This scene is an example of bad physics. It is true that the jet-powered backpack produces enough force to push the truck, but this force is imparted to the truck through the arms of the teenager. This is a very unrealistic situation, because force that is great enough to accelerate an old truck would likely crush the teenager’s arms.

Chapter 5 Newton’s Laws Of Motion Q.25P
A farm tractor tows a 3700-kg trailer up an 18° incline with a steady speed of 3.2 m/s. What force does the tractor exert on the trailer? (Ignore friction.)
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion25ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion25ps1

Chapter 5 Newton’s Laws Of Motion Q.26CQ
List three common objects that have a weight of approximately 1 N.
Solution:
Any object whose weight is approximately 100g has a weight equal to 1N
(i) An apple weighs approximately 1N
(ii) A chocolate bar
(iii) A pack of crayons

Chapter 5 Newton’s Laws Of Motion Q.26P
A surfer “hangs ten,” and accelerates down the sloping face of a wave. If the surfer’s acceleration is 3.25 m/s2 and friction can be ignored, what is the angle at which the face of the wave is inclined above the horizontal?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion26ps

Chapter 5 Newton’s Laws Of Motion Q.27P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion27p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.28P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion28p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.29P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion29p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.30P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion30p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.31P
IP Before practicing his routine on the rings, a 67-kg gymnast stands motionless, with one hand grasping each ring and his feet touching the ground. Both aims slope upward at an angle of 24° above the horizontal, (a) If the force exerted by the rings on each arm has a magnitude of 290 N, and is directed along the length of the arm, what is the magnitude of the force exerted by the floor on his feet? (b) If the angle his arms make with the horizontal is greater that 24°, and everything else remains the same, is the force exerted by the floor on his feet greater than, less than, or the same as the valne found in part (a)? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion31ps
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Chapter 5 Newton’s Laws Of Motion Q.32P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion32p
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion32ps
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Chapter 5 Newton’s Laws Of Motion Q.33P
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion33ps
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Chapter 5 Newton’s Laws Of Motion Q.34P
· A train is traveling up a 3.73° incline at a speed of 3.25 m/s when the last car breaks free and begins to coast without friction, (a) How long does it take for the last car to come to rest momentarily? (b) How far did the last car travel before mo-mentarily coming to rest?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.35P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion35p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.36P
You pull upward on a stuffed suitcase with a force of 105 N, and it accelerates upward at 0.705 m/s2. What are (a) the mass and (b) the weight of the suitcase?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion36ps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion36ps1

Chapter 5 Newton’s Laws Of Motion Q.37P
BIO Brain Growth A newborn baby’s brain grows rapidly. In fact, it has been found to increase in mass by about 1.6 mg per minute, (a) How much does the brain’s weight increase in one day? (b) How long does it take for the brain’s weight to increase by 0.15 N?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion37ps
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Chapter 5 Newton’s Laws Of Motion Q.38P
Suppose a rocket launches with an acceleration of 30.5 m/s2. What is the apparent weight of an 92-kg astronaut aboard this rocket?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion38ps

Chapter 5 Newton’s Laws Of Motion Q.39P
At the bow of a ship on a stormy sea, a crewman conducts an experiment by standing on a bathroom scale. In calm waters, the scale reads 182 lb. During the storm, the crewman finds a maximum reading of 225 lb and a minimum reading of 138 lb. Find (a) the maximum upward acceleration and (b) the maximum downward acceleration experienced by the crewman.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.40P
IP As part of a physics experiment, you stand on a bathroom scale in an elevator. Though your normal weight is 610 N, the scale at the moment reads 730 N. (a) Is the acceleration of the elevator upward, downward, or zero? Explain. (b) Calculate the magnitude of the elevator’s acceleration. (c) What, if anything, can you say about the velocity of the elevator? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion40ps
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Chapter 5 Newton’s Laws Of Motion Q.41P
When you weigh yourself on good old terra firma (solid ground), your weight is 142 lb. In an elevator your apparent weight is 121 lb. What are the direction and magnitude of the elevator’s acceleration?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.42P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion42p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.43PWhen you lift a bowling ball with a force of 82 N, the ball accelerates upward with an acceleration a. If you lift with a force of 92 N, the ball’s acceleration is 2a. Find (a) the weight of the bowling ball, and (b) the acceleration a.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion43ps

Chapter 5 Newton’s Laws Of Motion Q.44P
A 23-kg suitcase is being pulled with constant speed by a handle that is at an angle of 25° above the horizontal. If the normal force exerted on the suitcase is 180 N, what is the force F applied to the handle?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion44ps

Chapter 5 Newton’s Laws Of Motion Q.45P
(a) Draw a free-body diagram for the skier in Problem 32. (b) Determine the normal force acting on the skier.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion45ps

Chapter 5 Newton’s Laws Of Motion Q.46P
· A 9.3-kg child sits in a 3.7-kg high chair. (a) Draw a free-body diagram for the child, and find the normal force exerted by the chair on the child. (b) Draw a free-body diagram for the chair, and find the normal force exerted by the floor on the chair.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.47P
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Solution:
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Chapter 5 Newton’s Laws Of Motion Q.48P
A 5.0-kg bag of potatoes sits on the bottom of a stationary shopping cart. (a) Sketch a free-body diagram for the bag of potatoes. (b) Now suppose the cart moves with a constant velocity. How does this affect your free-body diagram? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion48ps

Chapter 5 Newton’s Laws Of Motion Q.49P
IP (a) Find the normal force exerted on a 2.9-kg book resting on a surface inclined at 36° above the horizontal. (b) If the angle of the incline is reduced, do you expect the normal force to increase, decrease, or stay the same? Explain.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.50P
IP A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 35° with the surface of the lawn. (a) If a 219-N force is applied along the handle of the 19-kg mower, what is the normal force exerted by the lawn on the mower? (b) If the angle between the surface of the lawn and the handle of the mower is increased, does the normal force exerted by the lawn increase, decrease, or stay the same? Explain.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.51P
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion51p
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.52GP
·CE Predict/Explain Riding in an elevator moving upward with constant speed, you begin a game of darts. (a) Do you have to aim your darts higher than, lower than, or the same as when you play darts on solid ground? (b) Choose the best explanation from among the following:
I. The elevator rises during the time it takes for the dart to travel to the darfboard.
II. The elevator moves with constant velocity. Therefore, Newton’s laws apply within the elevator in the same way as on the ground.
III. You have to aim lower to compensate for the upward speed of the elevator.
Solution:
(a) You have to aim your darts same as when you play darts on solid ground.
(b) This is because the elevator is moving with constant velocity. Therefore ’s laws apply within the elevator in the same way as on earth.
Therefore option II is best explanation.

Chapter 5 Newton’s Laws Of Motion Q.53GP
CE Predict/Explain Riding in an elevator moving with a constant upward acceleration, you begin a game of darts. (a) Do you have to aim your darts higher than, lower than, or the same as when you play darts on solid ground? (b) Choose the best explanation from among the following:
I. The elevator accelerates upward, giving its passengers a greater “effective” acceleration of gravity.
II. You have to aim lower to compensate for the upward acceleration of the elevator.
III. Since the elevator moves with a constant acceleration, Newton’s laws apply within the elevator the same as on the ground.
Solution:
In the game of darts a player have to aim and throw the dart on the board to hit the bulls- eye.
(a)
When elevator is going upward with constant acceleration the board hanged on wall of the elevator will also move upward with same acceleration. When a person will throw a dart it will immediately start accelerating downward on the other hand board will keep accelerating upward. Therefore dart will not hit the aimed position. It will hit lower than the aimed position.
Hence in the accelerating elevator a person has to aim his darts higher than the normal.
(b)
When the elevator is accelerating upward then due to force exerted by the floor of the elevator effective acceleration of the passenger increases. Therefore first explanation is right.
As discussed in previous section passenger has aim higher to compensate for the upward acceleration. Therefor second explanation is wrong.
Newton’s law can be applied directly only for inertial reference frame (stationary reference point or moving with constant reference point). Since elevator is moving with constant acceleration therefore newton’s law applied within the elevator is not same as on the ground. Therefore third explanation is wrong
Hence option I is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.54GP
CE Give the direction of the net force acting on each of the following objects. If the net force is zero, state “zero.” (a) A car accelerating northward from a stoplight. (b) A car traveling southward and slowing down. (c) A car traveling westward with constant speed. (d) A skydiver parachuting downward with constant speed. (e) A baseball during its flight from pitcher to catcher (ignoring air resistance).
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion54gps

Chapter 5 Newton’s Laws Of Motion Q.55GP
·CE Predict/Explain You jump out of an airplane and open your parachute after an extended period of free fall. (a) To decelerate your fall, must the force exerted on you by the parachute be greater than, less than, or equal to your weight? (b) Choose the best explanation from among the following:
I. Parachutes can only exert forces that are less than the weight of the skydiver.
II. The parachute exerts a force exactly equal to the skydiver’s weight.
III. To decelerate after free fall, the net force acting on a skydiver must be upward.
Solution:
(a) To decelerate your fall the force exerted by the parachute must be greater than your weight.
(b) To decelerate after free fall, the net force acting on a sky diver must be upward. This can be achieved only when the force exerted by the parachute is greater than your weight
So option III is the best explanation.

Chapter 5 Newton’s Laws Of Motion Q.56GP
In a tennis serve, a 0.070-kg ball can be accelerated from rest to 36 m/s over a distance of 0.75 m. Find the magnitude of the average force exerted hy the racket on the ball during the serve.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion56gp

Chapter 5 Newton’s Laws Of Motion Q.57GP
A 51.5-kg swimmer with an initial speed of 1.25 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.20 m, what was the force exerted on her by the water?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.58GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion58gp
Solution:
The three identical pucks are acted upon by a force of 3N.
F = ma
Since the mass and the force of the 3 pucks are the same, then the acceleration produced by this force must also be the same. The acceleration produced by a force does not depend on the speed of the object, so the acceleration of A, B, and C are all equal.

Chapter 5 Newton’s Laws Of Motion Q.59GP
IP The VASIMR Rocket NASA plans to use a new type of rocket, a Variable Specific Impulse Magnetoplasma Rocket (VASIMR), on future missions. A VASIMR can produce 1200 N of thrust (force) when in operation. If a VASIMR has a mass of 2.2 × 105 kg, (a) what acceleration will it experience? Assume that the only force acting on the rocket is its own thrust, and that the mass of the rocket is constant. (b) Over what distance must the rocket accelerate from rest to achieve a speed of 9500 m/s? (c) When the rocket has covered one-quarter the acceleration distance found in part (b), is its average speed 1/2, 1/3, or 1/4 its average speed during the final three-quarters of the acceleration distance? Explain.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.60GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion60gp
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.61GP
At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.62GP
· IP BIO The Force of Running Biomechanical research has shown that when a 67-kg person is running, the force exerted oh each foot as it strikes the ground can be as great as 2300 N. (a) What is the ratio of the force exerted on the foot by the ground to the person’s body weight? (b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person’s weight, what are the magnitude and direction of the person’s acceleration? (c) If the acceleration found in part (b) acts for 10.0 ms, what is the resulting change in the vertical component of the person’s velocity?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.63GP
IP BIO Grasshopper Liftoff To become airborne, a 2.0-g grasshopper requires a takeoff speed of 2.7 m/s. It acquires this speed by extending its hind legs through a distance of 3.7 cm. (a) What is the average acceleration of the grasshopper during takeoff? (b) Find the magnitude of the average net force exerted
on the grasshopper by its hind legs during takeoff. (c) If the mass of the grasshopper increases, does the takeoff acceleration increase, decrease, or stay the same? (d) If the mass of the grasshopper increases, does the required takeoff force increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion63gps
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Chapter 5 Newton’s Laws Of Motion Q.64GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion64gp
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion64gps

Chapter 5 Newton’s Laws Of Motion Q.65GP
IP An archer shoots a 0.024-kg arrow at a target with a speed of 54 m/s. When it hits the target, it penetrates to a depth of 0.083 m. (a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion65gps
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Chapter 5 Newton’s Laws Of Motion Q.66GP
An apple of mass m = 0.13 kg falls out of a tree from a height h = 3.2 m. (a) What is the magnitude of the force of gravity, mg, acting on the apple? (b) What is the apple’s speed, v, just before it lands? (c) Show that the force of gravity times the height, mgh, is equal to . (We shall investigate the significance of this result in Chapter 8.) Be sure to show that the dimensions are in agreement as well as the nu-merical values.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion66gps
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Chapter 5 Newton’s Laws Of Motion Q.67GP
An apple of mass m = 0.22 kg falls from a tree and hits the ground with a speed of v = 14 m/s. (a) What is the magnitude of the force of gravity, mg, acting on the apple? (b) What is the time, t, required for the apple to reach the ground? (c) Show that the force of gravity times the time, mgt, is equal to mv. (We shall investigate the significance of this result in Chapter 9.) Be sure to show that the dimensions are in agreement as well as the numerical values.
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion67gps

Chapter 5 Newton’s Laws Of Motion Q.68GP
BIO The Fall of T. rex Paleontologists estimate that if a Tyrnnnosaunis rex were to trip and fall, it would have experienced a force of approximately 260,000 N acting on its torso when it hit the ground. Assuming the torso has a mass of 3800 kg, (a) find the magnitude of the torso’s upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g.) (b) Assuming the torso is in free fall for a distance of 1.46 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion68gps
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Chapter 5 Newton’s Laws Of Motion Q.69GP
Deep Space I The NASA spacecraft Deep Space I was shut down on December 18, 2001, following a three-year journey to the asteroid Braille and the comet Borrelly. This spacecraft used a solar-powered ion engine to produce 0.064 ounces of thrust (force) by stripping electrons from neon atoms and accelerating the resulting ions to 70,000 mi/h. The thrust was only as much as the weight of a couple sheets of paper, but the engine operated continuously for 16,000 hours. As a result, the speed of the spacecraft increased by 7900 mi/h. What was the mass of Deep Space I? (Assume that the mass of the neon gas is negligible.)
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.70GP
Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 51.5 N is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of 1.25 m/s2?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion70gps
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Chapter 5 Newton’s Laws Of Motion Q.71GP
IP While waiting at the airport for your flight to leave, you observe some of the jets as they take off. With your watch you find that it takes about 35 seconds for a plane to go from rest to takeoff speed. In addition, you estimate that the distance re-quired is about 1.5 km. (a) If the mass of a jet is 1.70 × 105 kg, what force is needed for takeoff? (b) Describe the strategy you used to solve part (a).
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.72GP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion72gp
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.73GP
Your groceries are in a bag with paper handles. The handles will tear off if a force greater than 51.5 N is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of 1.25 m/s2?
Solution:
The Newton’s second law of motion defined the acceleration of an object produced by a net force is directly proportional to the magnitude of net force and inversely proportional to the mass of object.
The force acting on mass m moves with acceleration a is given as follows:
F=ma
Here, m is the mass and a is the acceleration.
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion73gps

Chapter 5 Newton’s Laws Of Motion Q.74GP
IP Responding to an alarm, a 102-kg fireman slides down a pole to the ground floor, 3.3 m below. The fireman starts at rest and lands with a speed of 4.2 m/s. (a) Find the average force exerted on the fireman by the pole. (b) If the landing speed is half that in part (a), is the average force exerted on the fireman by the pole doubled? Explain. (c) Find the average force exerted on the fireman by the pole when the landing speed is 2.1 m/s.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.75GP
For a birthday gift, you and some friends take a hot-air balloon ride. One friend is late, so the balloon floats a couple of feet off the ground as you wait. Before this person arrives, the combined weight of the basket and people is 1220 kg, and the balloon is neutrally buoyant. When the late arrival climbs up into the basket, the balloon begins to accelerate downward at 0.56 m/s2. What was the mass of the last person to climb aboard?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.76GP
D:\APLUS images\Mastering Physics Solutions Chapter 5 Newton’s Laws Of Motion75gps1.png
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion76gps

Chapter 5 Newton’s Laws Of Motion Q.77GP
When two people push in the same direction on an object of mass m they cause an acceleration of magnitude a1. When the same people push in opposite directions, the acceleration of the object has a magnitude a2. Determine the magnitude of the force exerted by each of the two people in terms of m, a1 and a2.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.78GP
· · · An air-track cart of mass m1 = 0.14 kg is moving with a speed v0 = 1.3 m/s to the right when it collides with a cart of mass m2 = 0.25 kg that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is F = 1.5 N during the collision. (a) What is the acceleration of cart 1? Give direction and magnitude. (b) What is the acceleration of cart 2? Give direction and magnitude. (c) How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.) (d) What is the final speed of the carts, vf? (e) Show that m1v0 is equal to (m1 + m2)vf. (We shall investigate the significance of this result in Chapter 9.)
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.79PP
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion79gp
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion79gps

Chapter 5 Newton’s Laws Of Motion Q.80PP
A driver who does not wear a seatbelt continues to move forward with a speed of 18.0 m/s (due to inertia) until something solid like the steering wheel is encountered. The driver now comes to rest in a much shorter distance—perhaps only a few centimeters. Find the magnitude of the net force acting on a 65.0-kg driver who is decelerated from 18.0 m/s to rest in 5.00 cm.
A. 3240 N
B. 1.17 × 104 N
C. 2.11 × 105 N
D. 4.21 × 105 N
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion80pps

Chapter 5 Newton’s Laws Of Motion Q.81PP
Suppose the initial speed of the driver is doubled to 36.0 m/s. If the driver still has a mass of 65.0 kg, and comes to rest in 1.00 m, what is the magnitude of the force exerted on the driver during this collision?
A. 648 N
B. 1170 N
C. 2.11 × 104 N
D. 4.21 × 104 N
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion81pps

Chapter 5 Newton’s Laws Of Motion Q.82PP
If both the speed and stopping distance of a driver are doubled, by what factor does the force exerted on the driver change?
A. 0.5
B. 1
C. 2
D. 4
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion82pps
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion82pps1

Chapter 5 Newton’s Laws Of Motion Q.83IP
IP Referring to Example 5-4 Suppose that we would like the contact force between the boxes to have a magnitude of 5.00 N, and that the only thing in the system we are allowed to change is the mass of box 2—the mass of box 1 is 10.0 kg and the applied force is 20.0 N. (a) Should the mass of box 2 be increased or decreased? Explain. (b) Find the mass of box 2 that results in a contact force of magnitude 5.00 N. (c) What is the acceleration of the boxes in this case?
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.84IP
Referring to Example 5-4 Suppose the force of 20.0 N pushes on two boxes of unknown mass. We know, however, that the acceleration of the boxes is 1.20 m/s2 and the contact force has a magnitude of 4.45 N. Find the mass of (a) box 1 and (b) box 2.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.85IP
IP Referring to Figure 5-9 Suppose the magnitude of is increased from 41 N to 55 N, and that everything else in the system remains the same. (a) Do you expect the direction of the satellite’s acceleration to be greater than, less than, or equal to 32°? Explain. Find (b) the direction and (c) the magnitude of the satellite’s acceleration in this case.
Solution:
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Chapter 5 Newton’s Laws Of Motion Q.86IP
IP Referring to Figure 5-9 Suppose we would like the acceleration of the satellite to be at an angle of 25°, and that the only quantity we can change in the system is the magnitude of . (a) Should the magnitude of be increased or decreased? Explain. (b) What is the magnitude of the satellite’s acceleration in this case?
Solution:
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion86ips
Mastering Physics Solutions Chapter 5 Newton's Laws Of Motion86ips1
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Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics

Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics

Mastering Physics Solutions

Chapter 4 Two-Dimensional Kinematics Q.1CQ
What is the acceleration of a projectile when it reaches its highest point? What is its acceleration just before and just after reaching this point?
Solution:
Projectile motion, ignoring air resistance, always acts downward. Thus, during the entire motion projectile, acceleration remains constant.

Chapter 4 Two-Dimensional Kinematics Q.1P
CE Predict/Explain As you walk briskly down the street, you toss a small ball into the air. (a) If you want the ball to land in your hand when it comes back down, should you toss the ball straight upward, in a forward direction, or in a backward direction, relative to your body?
(b) Choose the best explanation from among the following:
I. If the ball is thrown straight up you will leave it behind.
II. You have to throw the ball in the direction you are walking.
III. The ball moves in the forward direction with your walking speed at all times.
Solution:
(a) If you want the ball to land in your hand when it comes back down, you should toss the ball straight upward.
(b) When a person tosses a ball upward while walking, the horizontal component of velocity of the ball and the person will be the same. So the ball moves in the forward direction with your walking speed at all times when you toss the ball straight upwards.
Therefore option III is correct.

Chapter 4 Two-Dimensional Kinematics Q.2CQ
A projectile is launched with an initial speed of v0 at an angle θ above the horizontal. It lands at the same level from which it was launched. What was its average velocity between launch and landing? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics2cqs

Chapter 4 Two-Dimensional Kinematics Q.2P
A sailboat runs before the wind with a constant speed of 4.2 m/s in a direction 32° north of west. How far (a) west and (b) north has the sailboat traveled in 25 min?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics2ps
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Chapter 4 Two-Dimensional Kinematics Q.3CQ
A projectile is launched from level ground. When it Sands, its direction of motion has rotated clockwise through 60°. What was the launch angle? Explain.
Solution:
The projectile was launched at an angle of 30°, so its direction of motion has rotated through 60º.

Chapter 4 Two-Dimensional Kinematics Q.3P
As you walk to class with a constant speed of 1.75 m/s, you are moving in a direction that is 18.0° north of east. How much time does it take to change your displacement by (a) 20.0 m east or (b) 30.0 m north?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics3ps

Chapter 4 Two-Dimensional Kinematics Q.4CQ
In a game of baseball, a player hits a high fly ball to the outfield. (a) Is there a point during the flight of the ball where its velocity is parallel to its acceleration? (b) Is there a point where the ball’s velocity is perpendicular to its acceleration? Explain in each case.
Solution:
(a) No, the velocity has two components. The vertical component is tangential to the flight, and the horizontal component is always to the right. Velocity is not parallel to the acceleration in this case.
(b) The ball’s velocity is perpendicular to its acceleration when it is at maximum height.
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics4cqs

Chapter 4 Two-Dimensional Kinematics Q.4P
Starting from rest, a car accelerates at 2.0 m/s2 up a hill that is inclined 5.5° above the horizontal, How far (a) horizontally and (b) vertically has the car traveled in 12 s?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics4ps

Chapter 4 Two-Dimensional Kinematics Q.5CQ
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics5cq
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics5cqs
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics5ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics5ps1

Chapter 4 Two-Dimensional Kinematics Q.5P
IP A particle passes through the origin with avelocity of (6.2m/s)ŷ. If the particle’s acceleration is (-4.4m/s2) , (a) what are its x and y positions after 5.0 s? (b) What are vx and uy at this time? (c) Docs the speed of this particle increase with time, decrease with time, or increase and then decrease? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics5ps

Chapter 4 Two-Dimensional Kinematics Q.6CQ
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics6pcq
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics6pcqs
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics6cqs
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Chapter 4 Two-Dimensional Kinematics Q.6P
An electron in a cathode-ray tube is traveling horizontally at 2.10 × 109 cm/s when deflection plates give it an upward acceleration of 5.30 × 1017 cm/s2. (a) How long does it take for the electron to cover a horizontal distance of 6.20 cm? (b) What is its vertical displacement during this time?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics6ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics6ps1

Chapter 4 Two-Dimensional Kinematics Q.7CQ
Do projectiles for which air resistance is nonnegligible, such as a bullet fired from a rifle, have maximum range when the launch angle is greater than, less than, or equal to 45°? Explain.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.7P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics7p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.8CQ
Two projectiles are launched from the same point at the same angle above the horizontal. Projectile 1 reaches a maximum height twice that of projectile 2. What is the ra tio of the initial speed of projectile 1 to the initial speed of projectile 2? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics8cqs

Chapter 4 Two-Dimensional Kinematics Q.8P
CE Predict/Explain Two divers run horizontally off the edge of a low cliff. Diver 2 runs with twice the speed of diver 1. (a) When the divers hit the water, is the horizontal distance covered by diver 2 twice as much, four times as much, or equal to the horizontal distance covered by diver 1? (b) Choose the best explanation from among the following:
I. The drop time is the same for both divers.
II. Drop distance depends on t2.
III. All divers in free fall cover the same distance.
Solution:
(a) The distance horizontal covered by the diver 2 is twice the horizontal distance covered by the diver 1.
(b) As the vertical distance traveled by the two divers is same, also both the divers has zero initial vertical component of velocity, therefore the time of travel for both the divers is same. Therefore diver 2 travels more time than the diver 1.
So option I is correct.

Chapter 4 Two-Dimensional Kinematics Q.9CQ
A child rides on a pony walking with constant velocity. The boy leans over to one side and a scoop of ice cream falls from his ice cream cone. Describe the path of the scoop of ice cream as seen by (a) the child and (b) his parents standing on the ground nearby.
Solution:
(A) The child sees the scoop falling straight downward.
(B) However, the father sees from the ground, that the trajectory of path of scoop is parabolic.

Chapter 4 Two-Dimensional Kinematics Q.9P
CE Predict/Explain Two youngsters dive off an overhang into a lake. Diver 1 drops straight down, and diver 2 runs off the cliff with an initial horizontal speed v0.(a) is the splashdown speed of diver 2 greater than, less than, or equal to the splashdown speed of diver 1? (b) Choose the best explanation from among the following:
I. Both divers are in free fall, and hence they will have the same splashdown speed.
II. The divers have the same vertical speed at splashdown, but diver 2 has the greater horizontal speed.
III. The diver who drops straight down gains more speed than the one who moves horizontally.
Solution:
(a) The splashdown speed of diver 2 is greater than the splash down speed of diver 1.
(b) The divers have same vertical speed at splashdown, but diver 2 has the greater horizontal speed. Therefore the diver 2 has the greater speed than the diver 1.
Therefore option II is the best.

Chapter 4 Two-Dimensional Kinematics Q.10CQ
Drivingdown the highway, you find yourself behind a heavily loaded tomato truck. You follow close behind the truck, keeping the same speed. Suddenly a tomato falls from the back of the truck. Will the tomato hit your car or land on the road, assuming you continue moving with the same speed and direction? Explain.
Solution:
When a tomato suddenly falls from the truck, it will land on the road because the horizontal speed is the same during the entire duration of the fall.

Chapter 4 Two-Dimensional Kinematics Q.10P
An archer shoots an arrow horizontally at a target 15 m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower. What was the initial speed of the arrow?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics10ps
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Chapter 4 Two-Dimensional Kinematics Q.11CQ
A projectile is launched from the origin of a coordinate system where the positive x axis points horizontally to the right and the positive y axis points vertically upward. What was the projectile’s launch angle with respect to the x axis if, at its highest point, its direction of motion has rotated (a) clockwise through 50° or (b) counterclockwise through 30°? Explain.
Solution:
(A) At the highest point, only the horizontal velocity exists. So it is launched from 50º to the positive x-axis.
(B) It is launched from 30º below the negative x-axis.

Chapter 4 Two-Dimensional Kinematics Q.11P
Victoria Falls The great, gray-green, greasy Zambezi River flows over Victoria Falls in south central Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.60 m/s just before going over the falls, what is the speed of the water when it hitsthe bottom? Assume the water is in free fall as it drops.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics11ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics11ps1

Chapter 4 Two-Dimensional Kinematics Q.12P
A diver runs horizontally off the end of a diving board wi th an initial speed of 1.85 m/s. if the diving board is 3.00 m above the water, what is the diver’s speed just before she enters the water?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics12ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics12ps1

Chapter 4 Two-Dimensional Kinematics Q.13P
An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.95 m/s. The rock falls through a vertical distance of 1.40 m and lands a horizontal distance of 8.75 m from the astronaut. What is the acceleration of gravity on Zircon?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics13ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics13ps1

Chapter 4 Two-Dimensional Kinematics Q.14P
IP Pitcher’s Mounds Pitcher’s mounds are raised to compensate for the vertical drop of the ball as it travels a horizontal distance of 18 hi to the catcher, (a) If a pitch is thrown horizontally with an initial speed of 32 m/s, how far does it drop by the time it reaches the catcher? (b) If the speed of the pitch is increased, does the drop distance increase, decrease, or stay the same? Explain, (c) If this baseball game were to be played on the Moon, would the drop distance increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics14ps

Chapter 4 Two-Dimensional Kinematics Q.15P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics15p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.16P
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Solution:
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(C) The speed of the clam increases horizontally. However, the vertical velocity won’t change since only g and the time are counted in order to determine y.

Chapter 4 Two-Dimensional Kinematics Q.17P
Amountain climber jumpsa 2.8-m-wide crevasse by leaping horizontally with a speed of 7.8 m/s. (a) If the climber’s direction of motion on landing is -45°, what is the height difference between the two sides of the crevasse? (b) Where does the climber land?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.18P
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.19P
IP A white-crowned sparrow flying horizontally with a speed of 1.80 m/s folds its wings and begins to drop in free fall, (a) How far does the sparrow fall after traveling a horizontal distance of 0.500 m? (b) If the sparrow’s initial speed is increased, does the distance of fall increase, decrease, or stay the same?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics19p
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics19ps

Chapter 4 Two-Dimensional Kinematics Q.20P
If, in the previous problem, a jack-o-lantern is given an initial horizontal speed of 3.3 m/s, what are the direction and magnitude of its velocity (a) 0.75 s later, and (b) just before it lands?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics20ps
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Chapter 4 Two-Dimensional Kinematics Q.21P
Fairgoers ride a Ferris wheel with a radius of 5.00 m (Figure 4-16). The wheel completes one revolution every 32.0 s. (a) What is the average speed of a rider on this Ferris wheel? (b) If a rider accidentally drops a stuffed animal at the top of the wheel, where does it land relative to the base of the ride? (Note: The bottom of the wheel is 1.75 m above the ground.)
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics21ps

Chapter 4 Two-Dimensional Kinematics Q.22P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics22p
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics22ps

Chapter 4 Two-Dimensional Kinematics Q.23P
Baseball and the Washington Monument On August 25, 1894, Chicago catcher William Schriver caught a baseball thrown from the top of the Washington Monument (555 ft, 898 steps), (a) If the ball was thrown horizontally with a speed of 5.00 m/s, where did it land? (b) What were the ball’s speed and direction of motion when caught?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.24P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics24p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.25P
IP A ball rolls off a table and falls 0.75 m to the floor, landing with a speed of 4.0 m/s. (a) What is the acceleration of the ball just before it strikes the ground? (b) What was the initial speed of the ball? (c) What initial speed must the ball have if it is to land with a speed of 5.0 m/s?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics25ps

Chapter 4 Two-Dimensional Kinematics Q.26P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics26p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.27P
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.28P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics28p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.29P
A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 18.0 m/s at an angle of 37.5° above the horizontal, (a) What is the horizontal component of the ball’s velocity just before it is caught? (b) How long is the ball in the air?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics29ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics29ps1

Chapter 4 Two-Dimensional Kinematics Q.30P
Referring to the previous problem, what are the y component of the ball’s velocity and its direction of motion just before it is caught?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics30ps

Chapter 4 Two-Dimensional Kinematics Q.31P
A cork shoots out of a champagne bottle at an angle of 35.0° above the horizontal. If the cork travels a horizontal distance of 1.30 m in 1.25s, what was its initial speed?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics31ps

Chapter 4 Two-Dimensional Kinematics Q.32P
D:\APLUS images\Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics31ps.png
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics32ps

Chapter 4 Two-Dimensional Kinematics Q.33P
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.3 m/s at an angle of 15° below the horizontal. Tt is released 0.80 m above the floor. What horizontal distance does the ball cover before bouncing?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.34P
Repeat the previous problem for a bounce pass in which the ball is thrown 15° above the horizontal.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics34ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics34ps1

Chapter 4 Two-Dimensional Kinematics Q.35P
IP Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m above the ground. Snowball A is thrown straight down-
ward; snowball B is thrown in a direction 25° above the horizontal, (a) Is the landing speed of snowball A greater than, less than, or the same as the landing speed of snowball B? Explain. (b) Verify your answer to part (a) by calculating the landing speed of both snowballs.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics35ps
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Chapter 4 Two-Dimensional Kinematics Q.36P
In the previous problem, find the direction of motion of the two snowballs just before they land.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics36ps
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics36ps1

Chapter 4 Two-Dimensional Kinematics Q.37P
A golfer gives a ball a maximum initial speed of 34.4 m/s. (a) What is the longest possible hole-irt-one for this golfer? Neglect any distance the ball might roll on the green and assume that the tee and the green are at the same level, (b) What is the minimum speed of the ball during this hole-in-one shot?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics37ps
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Chapter 4 Two-Dimensional Kinematics Q.38P
What is the highest tree the ball in the previous problem could clear on its way to the longest possible holc-in-one?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics38ps

Chapter 4 Two-Dimensional Kinematics Q.39P
The “hang time” of a punt is measured to be 4.50 s. If the ball was kicked at an angle of 63.0° above the horizontal and was caught at the same level from which it was kicked, what was its initial speed?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics39ps

Chapter 4 Two-Dimensional Kinematics Q.40P
In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 18 m/s at an angle of 32° above the horizontal, (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics40ps
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Chapter 4 Two-Dimensional Kinematics Q.41P
IP Tn the previous problem, (a) what are the magnitude and direction of the ball’s velocity when it stilkes the wall? (b) Has the ball reached the highest point of its trajectory at this time? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics41ps
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Chapter 4 Two-Dimensional Kinematics Q.42P
A passenger on the Ferris wheel described in Problem 21 drops his keys when he is on the way up and at the 10 o’clock position. Where do the keys land relative to thebase of the ride?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.43P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics43p
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics43ps

Chapter 4 Two-Dimensional Kinematics Q.44P
A certain projectile is launched with an initial speed v0. At its highest point its speed is v0/4- What was the launch angle?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics44p

Chapter 4 Two-Dimensional Kinematics Q.45P
Punkin Chunkin In Sussex County, Delaware, a post-Halloween tradition is “Punkin Chunkin,” in which contestants build cannons, catapults, trebuchets, and other devices to launch pumpkins and compete for the greatest distance. Though hard to believe, pumpkins have been projected a distance of 4086 feet in this contest. What is the mnirmum initial speed needed for such a shot?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics45ps

Chapter 4 Two-Dimensional Kinematics Q.46P
A dolphin jumps with an initial velocity of 12.0 m/s at an angle of 40.0° above the horizontal. The dolphin passes through the center of a hoop before returning to the water. If the dolphin is moving horizontally when it goes through the hoop, how high above the water is the center of the hoop?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics46ps

Chapter 4 Two-Dimensional Kinematics Q.47P
A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1 m/s, and it travels a distance of 4.6 m. What were (a) the initial direction of the ball and (b) its time of flight?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics47ps

Chapter 4 Two-Dimensional Kinematics Q.48P
A golf ball is struck with a five iron on level ground. It lands 92.2 m away 4.30 s later. What were (a) the direction and (b) the magnitude of the initial velocity?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.49P
A Record Toss Babe Didrikson holds the world record for the longest baseball throw (296 ft) by a woman. For the following questions, assume that the ball was thrown at an angle of 45.0° above the horizontal, that it traveled a horizontal distance of 296 ft, and that it was caught at the same level from which it was thrown, (a) What was the ball’s initial speed? (b) How long was the ball in the air?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics49ps

Chapter 4 Two-Dimensional Kinematics Q.50P
In the photograph to the left on page 87, suppose the cart that launches the ball is 11 cm high. Estimate (a) the launch speed of the ball and (b) the time interval between successive stroboscopic exposures.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics50ps

Chapter 4 Two-Dimensional Kinematics Q.51P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics51p
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.52P
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics52p
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics52ps

Chapter 4 Two-Dimensional Kinematics Q.53P
IP A soccer ball is kicked with an initial speed of 10.2 m/s in a direction 25.0° above the horizontal. Find the magnitude and direction of its velocity (a) 0.250 s and (b) 0.500 s after being kicked, (c) is the ball at its greatest height before or after 0.500 s? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics53ps
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Chapter 4 Two-Dimensional Kinematics Q.54P
A second soccer ball is kicked with the same initial speed as in Problem 53. After 0.750 s it is at its highest point. What was its initial direction of motion?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.55P
IP A golfer tees off on level ground, giving the ball an initial speed of 46.5 m/s and an initial direction of 37.5° above the horizontal, (a) How far from the golfer does the ball land? (b) The next golfer in the group hits a ball with the same initial speed but at an angle above the horizontal that is greater than 45.0°. Tf the second ball travels the same horizontal distance as the first ball, what was its initial direction of motion? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics55ps
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Chapter 4 Two-Dimensional Kinematics Q.56P
IP One of the most popular events at Highland games is the hay toss, where competitors use a pitchfork to throw a bale of hay over a raised bar. Suppose the initial velocity of a bale of hay is . (a) After what minimum time is its speed equal to 5,00 m/s? (b) How long after the hay is tossed is it moving in a direction that is 45.0° below the horizontal? (c) If the bale of hay is tossed with the same initial speed, only this time straight upward, will its time in the air increase, decrease, or stay the same? Explain.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics56ps
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Chapter 4 Two-Dimensional Kinematics Q.57GP
CE Child 1 throws a snowball horizontally from the top of a roof; child 2 throws a snowball straight down. Once in flight, is the acceleration of snowball 2 greater than, less than, or equal to the acceleration of snowball 1?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics57gps

Chapter 4 Two-Dimensional Kinematics Q.58GP
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics58gp
Solution:
The penguin lands at an elevation that is above the water then the speed of the penguin gets decreased before it lands on the ice.
For a projectile initial and final point will be on the same level therefore velocity remains the same where as in the case of penguins jump from water to ice levels are different the speed before it land on ice becomes less.

Chapter 4 Two-Dimensional Kinematics Q.59GP
CE Predict/Explain A person flips a coin into the air and it lands on the ground a few feet away, (a) If the person were to perform an identical coin flip on an elevator rising with constant speed, would the coin’s time of flight be greater than, less than, or equal to its time of flight when the person was at rest? (b) Choose the best explanation from among the following:
I. The floor of the elevator is moving upward, and hence it catches up with the coin in mid flight.
II. The coin has the same upward speed as the elevator when it is tossed, and the elevator’s speed doesn’t change during the coin’s flight.
III. The coin starts off with a greater upward speed because of the elevator, and hence it reaches a greater height.
Solution:
(a) The time of flight of the coin in the lift moving up with constant speed is same as the time of flight on ground.
(b) This is because of the coin has the same upward speed as the elevator when it is tossed, and the elevator’s speed doesn’t change during the coin’s flight.
So option II is the best explanation.

Chapter 4 Two-Dimensional Kinematics Q.60GP
CEPredict/Explain Suppose the elevator in the previous problem is rising with a constant upward acceleration, rather than constant velocity, (a) In this case, would the coin’s time of flight be greater than, less than, or equal to its time of flight when the person was at rest? (b) Choose the best explanation from among the following:
I. The coin has the same acceleration once it is tossed, whether the elevator accelerates or not.
II. The elevator’s upward speed increases during the coin’s flight, and hence it catches up with the coin at a greater height than before.
III. The coin’s downward acceleration is less than before because the elevator’s upward acceleration partially cancels it.
Solution:
(a) The time of flight of the coin in the lift moving up with constant acceleration is less than the time of flight on ground.
(b) In this case the coin will have greater acceleration relative to the floor of the elevator. Or the elevator’s upward speed increases during the coin’s flight, and hence it catches up with the coin at greater height than before. So, the option II is the best explanation.

Chapter 4 Two-Dimensional Kinematics Q.61GP
A train moving with constant velocity travels 1.70 m north in 12 s and an undetermined, distance to the west. The speed of the train is 32 m/s. (a) Find the direction of the train’s motion relative to north, (b) How far west has the train traveled in this time?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics61gps

Chapter 4 Two-Dimensional Kinematics Q.62GP
Referring to Example 4-2, find (a) the x component and (b) the y component of the hummingbird’s velocity at the time t = 0.72 s. (c) What is the bird’s direction of travel at this time, relative to the positive x axis?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics62gps

Chapter 4 Two-Dimensional Kinematics Q.63GP
A racket ball is struck in such a way that it leaves the racket with a speed of 4.87 m/s in the horizontal direction. When the ball hits the court, it is a horizontal distance of 1.95 m from the racket. Find the height of the racket ball when it left the racket.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.64GP
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics64gp
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.65GP
Repeat the previous problem, this time assuming that the balloon is descending with a speed of 2.00 m/s.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics65gps
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Chapter 4 Two-Dimensional Kinematics Q.66GP
IP A soccer ball is kicked from the ground with an initial speed of 14.0 m/s. After 0.275 s its speed is 12.9 m/s. (a) Give a strategy that will allow you to calculate the ball’s initial direction of motion, (b) Use your strategy to find the initial direction.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.67GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.68GP
When the dried-up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.62 m/s at an angle of 60.5° above the horizontal. If the seed pod is 0.455 m above the ground, (a) bow long does it take for the seed to land? (b) What horizontal distance does it cover during its flight?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.69GP
Referring to Problem 68, a second seed shoots out from the pod with the same speed but with a direction of motion 30.0° below the horizontal, (a) How long does it take for the second seed to land? (b) What horizontal distance does it cover during its flight?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.70GP
A shot-putter throws the shot with an initial speed of 12.2 m/s from a height of 5.15 ft above the ground. What is the range of the shot if the launch angle is (a) 20.0°, (b) 30.0°, or (c) 40.0°?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.71GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.72GP
Aball thrown straight upward returns to its original level in 2.75 s. A second ball is thrown at an angle of 40.0° above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.75 s?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.73GP
IP A cannon is placed at the bottom of a cliff 61.5 m high. If the cannon is fired straight upward, the cannonball just reaches the top of the cliff, (a) What is the initial speed of the cannonball? (b) Suppose a second cannon is placed at the top of the cliff. This cannon is fired horizontally, giving its cannonbails the same initial speed found in part(a). Show that the range of this cannon is the same as the maximum range of the cannon at the base of the cliff. (Assume the ground at the base of the cliff is level, though the result is valid even if the ground is not level.)
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.74GP
IP A cannon is placed at the bottom of a cliff 61.5 m high. If the cannon is fired straight upward, the cannonball just reaches the top of the cliff, (a) What is the initial speed of the cannonball? (b) Suppose a second cannon is placed at the top of the cliff. This cannon is fired horizontally, giving its cannonbails the same initial speed found in part(a). Show that the range of this cannon is the same as the maximum range of the cannon at the base of the cliff. (Assume the ground at the base of the cliff is level, though the result is valid even if the ground is not level.)Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics74gps

Chapter 4 Two-Dimensional Kinematics Q.75GP
Sliot Put Record The men’s world record for the shot put, 23.12 in, was set by Randy Barnes of the United States on May 20,1990. If the shot was launched from 6.00 ft above the ground at an initial angle of 42.0°, what was its initial speed?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.76GP
Referring to Conceptual Checkpoint 4-3, suppose the two snowballs are thrown from an elevation of 15 m with an initial speed of 12 m/s. What is the speed of each ball when it is 5.0 m above the ground?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.77GP
IP A hockey puck just clears the 2.00-m-high boards on its way out of the rink. The base of the boards is 20.2 m from the pqint where the puck is launched, (a) Given the launch angle of the puck, θ, outline a strategy that you can use to find its initial speed, u0 (b) Use your strategy to find u0 for 0 = 15.0°.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.78GP
Referring to Active Example 4-2, suppose the ball is punted from an initial height of 0.750 m. What is the initial speed of the ball in this case?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics78gps

Chapter 4 Two-Dimensional Kinematics Q.79GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.80GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.81GP
As discussed in Example 4-7, the archcrfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish’s mouth. Suppose the archerfish squirts water with a speed of 2.15 m/s at an angle of 52.0° above the horizontal, and aims for a beetle on a leaf 3.00 cm above the water’s surface, (a) At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time? (b) How much time will the beetle have to react?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.82GP
(a) What is the greatest horizontal distance from which the archerfish can hit the beetle, assuming the same squirt speed and direction as in Problem 81? (b) How much time docs the beetle have to react in this case?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.83GP
Find the launch angle for which the range and maximum height of a projectile are the same.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics83gps

Chapter 4 Two-Dimensional Kinematics Q.84GP
A mountain climber jumps a crevasse of width W by leaping horizontally with speed u0.(a) if the height difference between the two sides of the crevasse is h, what is the minimum value of u0 for the climber to land safely on the other side? (b) In this case, what is the cumber’s direction of motion on landing?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.85GP
Prove that the landing speed of a projectile is independent of launch angle for a given height of launch.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.86GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.87GP
Landing on a Different Level A projectile fired from y = 0 with initial speed u0 e:\04-02-2016\chapter 4\1403\9781111788452\exercisesand initial angle θ lands on a different level, y = h. Show that the time of flight of the projectile is where T0 is the time of flight for h = 0 and H is the maximum height of the projectile.
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.88GP
A mountain climber jumps a crevasse by leaping horizontally with speed u0. If the climber’s direction of motion on landing is θ below the horizontal, what is the height difference h between the two sides of the crevasse?
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.89GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.90GP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.91PP
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Solution:
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Chapter 4 Two-Dimensional Kinematics Q.92PP
How much time elapses between the first and second bounces?
A. 1.38 s
B. 2.58 s
C. 5.15 s
D. 5.33 s
Solution:
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Chapter 4 Two-Dimensional Kinematics Q.93PP
How far does a rover travel in the horizontal direction between its first and second bounces?
A. 13.2 m
B. 49.4 m
C. 51.1m
D. 98.7 m
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics93pps

Chapter 4 Two-Dimensional Kinematics Q.94PP
What is the average velocity of a rover between its first and second bounces?
A. 0
B. 2.57 m/s in the x direction
C. 9.92 m/s at 75.0° above the x axis
D. 9.58 m/s in the y direction
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics94pps

Chapter 4 Two-Dimensional Kinematics Q.95IP
Referring to Example 4-5 (a) At what launch angle greater than 54.0° does the golf ball just barely miss the top of the tree in front of the green? Assume the ball has an initial speed of 13.5 m/s, and thatthe tree is 3.00 m high and is a horizontal distance of 14.0 m from the launch point, (b) Where does the ball land in the case described in part (a)? (c) At what launch angle less than 54.0° does the golf ball just barely miss the top of the tree in front of the green? (d) Where does the ball land in the case described in part (c)?
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics95ips
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Chapter 4 Two-Dimensional Kinematics Q.96IP
Referring to Example 4-5 Suppose that the golf ball is launched with a speed of 15.0 m/s at an angle of 57.5° above the horizontal, and that it lands on a green 3.50 m above the level where it was struck, (a) What horizontal distance does the ball cover during its flight? (b) What increase in initial speed would be needed to increase the horizontal distance in part (a) by 7.50 m? Assume everything else remains the same.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics96ips
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Chapter 4 Two-Dimensional Kinematics Q.97IP
Referring to Example 4-6 Suppose the ball is dropped at the horizontal distance of 5.50 m, but from a new height of 5.00 m. The dolphin jumps with the same speed of 12.0 m/s. (a) What launch angle must the dolphin have if it is to catch the ball? (b) At what height does the dolphin catch the ball in this case? (c) What is the minimum initial speed the dolphin must have to catch the ball before it hits the water?
Solution:
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Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics97ips1
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics97ips2

Chapter 4 Two-Dimensional Kinematics Q.98IP
IP Referring to Example 4-6 Suppose we change the dolphin’s launch angle to 45.0°, but everything else remains the same. Thus, the horizontal distance to the ball is 5.50 m, the drop height is 4.10 m, and the dolphin’s launch speed is 12.0 m/s. (a) What is the vertical distance between the dolphin and the ball when the dolphin reaches the horizontal position of the ball? We refer to this as the “miss distance.” (b) If the dolphin’s launch speed is reduced, will the miss distance increase, decrease, or stay the same? (c) Find the miss distance for a launch speed of 10.0 m/s.
Solution:
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics98ips
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics98ips1
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics98ips2
Mastering Physics Solutions Chapter 4 Two-Dimensional Kinematics98ips4