Mastering Physics Solutions Chapter 6 Applications Of Newton’s Laws
Chapter 6 Applications Of Newton’s Laws Q.1CQ
A clothesline always sags a little, even if nothing hangs from it. Explain.
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Chapter 6 Applications Of Newton’s Laws Q.1P
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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?
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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.
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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.
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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?
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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?
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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
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Chapter 6 Applications Of Newton’s Laws Q.9CQ
Water sprays off a rapidly turning bicycle wheel. Why?
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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?
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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
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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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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?
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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).
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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?
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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.)
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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.)
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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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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?
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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?
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Chapter 6 Applications Of Newton’s Laws Q.20CQ
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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?
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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.
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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?
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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.
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Chapter 6 Applications Of Newton’s Laws Q.24P
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Chapter 6 Applications Of Newton’s Laws Q.25P
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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.
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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).
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Chapter 6 Applications Of Newton’s Laws Q.28P
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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:
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?
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Chapter 6 Applications Of Newton’s Laws Q.31P
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Chapter 6 Applications Of Newton’s Laws Q.32P
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Chapter 6 Applications Of Newton’s Laws Q.33P
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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?
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Chapter 6 Applications Of Newton’s Laws Q.35P
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Chapter 6 Applications Of Newton’s Laws Q.36P
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Chapter 6 Applications Of Newton’s Laws Q.37P
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Chapter 6 Applications Of Newton’s Laws Q.38P
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Chapter 6 Applications Of Newton’s Laws Q.39P
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Chapter 6 Applications Of Newton’s Laws Q.40P
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Chapter 6 Applications Of Newton’s Laws Q.41P
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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?
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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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Chapter 6 Applications Of Newton’s Laws Q.45P
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Chapter 6 Applications Of Newton’s Laws Q.46P
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Chapter 6 Applications Of Newton’s Laws Q.47P
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Chapter 6 Applications Of Newton’s Laws Q.48P
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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?
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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?
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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?
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Chapter 6 Applications Of Newton’s Laws Q.52P
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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
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?
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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.
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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?
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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?
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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?
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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?
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Chapter 6 Applications Of Newton’s Laws Q.61P
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Chapter 6 Applications Of Newton’s Laws Q.62P
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Chapter 6 Applications Of Newton’s Laws Q.63P
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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?
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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.
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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
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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:
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:
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:
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:
Chapter 6 Applications Of Newton’s Laws Q.74GP
Solution:
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:
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:
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:
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:
Chapter 6 Applications Of Newton’s Laws Q.79GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.80GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.81GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.82GP
Solution:
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:
Chapter 6 Applications Of Newton’s Laws Q.84GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.85GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.86GP
Solution:
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:
Chapter 6 Applications Of Newton’s Laws Q.88GP
Solution:
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:
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:
Chapter 6 Applications Of Newton’s Laws Q.91GP
Solution:
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:
Chapter 6 Applications Of Newton’s Laws Q.93GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.94GP
Solution:
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:
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:
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:
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:
Chapter 6 Applications Of Newton’s Laws Q.99GP
Solution:
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:
Chapter 6 Applications Of Newton’s Laws Q.101GP
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:
Chapter 6 Applications Of Newton’s Laws Q.103GP
Solution:
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:
Chapter 6 Applications Of Newton’s Laws Q.105GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.106GP
Solution:
Chapter 6 Applications Of Newton’s Laws Q.107GP
Solution:
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:
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:
Chapter 6 Applications Of Newton’s Laws Q.110GP
Solution:
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?
Solution:
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?
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
Solution:
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.
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:
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:
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:
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:
