Work Done

How to Calculate the Energy Used

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What is Energy

Definition: Energy is the ability to do work. The amount of energy possessed by a body is equal to the amount of work it can do when its energy is released. Thus, energy is defined as the capacity of doing work. Energy is a scalar quantity and it exists in various forms.

Units of energy: The units of energy are the same as that of work. In SI system, the unit of energy is joule (J). In CGS system, the unit of energy is erg.

  • 1 Joule = 107 ergs
  • Other units of energy in common use are watt-hour and kilowatt hour.
  • 1 watt-hour = 1 watt × 1 hour
    = 1 watt × 60 × 60 sec
    = 3600 J
  • 1 kilowatt-hour (kWh) = 3.6 × 106 Joule
  • Heat energy is usually measured in calorie or kilocalorie such that
    1 calorie = 4.18 J
  • A very small unit of energy is electron volt(eV).
    1 eV = 1.6 × 10-19 J

The energy possessed by a body due to its state of rest or state of motion is called mechanical energy.
Mechanical energy is of two types
(A) Kinetic Energy            (B) Potential Energy.

Principle of Conservation of Energy

  1. Principle of conservation of energy states that energy cannot be created or destroyed but can be changed from one form to another.
  2. The total energy of the universe is constant. The total energy of an isolated system is constant.
  3. Figure shows the transformation of energy from one form to another.
    Principle of Conservation of EnergyPrinciple of Conservation of Energy 1

Principle of Conservation of Energy Experiment

Aim: To study the principle of conservation of energy.
Materials: Ticker tape, a polystyrene sheet, string, a 300 g slotted mass, a pulley, cellophane tape
Apparatus: Ticker timer, a.c. power supply, two retort stands with clamps, trolley, electronic balance, plane
Method:

  1. A friction-compensated plane is arranged as shown in Figure.
  2. The mass of the trolley, m1 is measured with an electronic balance.
  3. A slotted mass of mass, m2 = 300 g is tied to one end of a non-elastic string.
  4. The other end of the string is tied to one end of the trolley.
  5. The string is placed over the pulley and held at 0.5 m above the polystyrene sheet.
  6. The ticker timer is switched on and the slotted mass is released so that it falls downward, pulling the trolley down the runway.
  7. The ticker timer is analysed to determine the final velocity of the trolley.
    Principle of Conservation of Energy 2

Results:
Principle of Conservation of Energy 3
Discussion:

  1. The plane is friction-compensated to minimise energy loss due to friction.
  2. When the slotted mass drops, it loses gravitational potential energy. The trolley and slotted mass gain kinetic energy.
  3. Ideally the gravitational potential energy loss, Ep equals to the kinetic energy gain, Ek. This is in accordance to the principle of conservation of energy.
  4. However, the experimental results shows that Ek is slightly less than Ep. This is because of unavoidable energy loss due to friction of the trolley as well as air friction.

 

How do you find Power in Physics

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What is Power

PowerFigure shows two electric motors, A and B respectively. Each motor lifts an identical load from the floor.
Motor A can lift the load more quickly than motor B. Hence, motor A can do the same amount of work in a shorter time.
Motor A is said to be more powerful than motor B.
Definition: Power is defined as the rate of doing work.
\( \text{Power}=\frac{\text{Work}\,\text{done}}{\text{Time}\,\text{taken}} \)
\( \text{P}=\frac{\text{W}}{\text{t}} \)
In other words, power is the work done per unit time, power is a scalar quantity.
Since W = F.S therefore
\( \text{P}=\frac{\text{W}}{\text{t}}=\frac{\text{FS}}{\text{t}}=F\times V=\text{force}\times \text{velocity} \)
Unit of power: The S.I. unit of power is watt and it is the rate of doing work at 1 joule per second.
\(1\text{ watt}=\frac{\text{1}\,\text{joule}}{\text{1}\,\text{seconds}}\)
1 kilowatt = 1 kW = 1000 W
1 Horse power = 1 H.P. = 746 W

Activity 1

Aim: To measure the power generated by a student running up the stairs.
Apparatus: Stopwatch, metre rule, weighing machine
Method:
Power 1

  1. The mass, m of a student is measured.
  2. The student is asked to run up the stairs from one floor to the next floor of a building in the school.
  3. The vertical height, h gained by the student in running up to the next floor is measured by multiplying the number of steps by the height of each step.
  4. The time taken, t for the run is recorded by another student using a stopwatch.

Analysis of Data:
Mass of student = m kg
Vertical height = h m
Time taken = t s
Work done = Gain in gravitational potential energy = mgh, where g = 9.8 m s-2
Power generated, P = W/t watt
Discussion:

  1. For a more accurate result, more than one person should time the run and the average time should be calculated.
  2. This activity assumes that all the work done goes to the increase in gravitational potential energy only. However, in reality, some work is also used in overcoming frictional forces.

What is the Efficiency?

  1. Useful energy is the energy that can be used to do a certain work.
  2. Wasted energy is the energy that is lost to the surrounding and cannot be used to do useful work.
  3. Figure shows an efficient device. Most of the input energy is converted into useful energy. Very little of the input energy is wasted.
    Efficiency 1
  4. Figure shows an inefficient device. Most of the input energy is wasted. Only a small portion of the input energy is converted into useful energy.
    Efficiency 2
  5. The efficiency of a device is defined as the percentage of the input energy that is transformed into useful energy.
    Efficiency 5
  6. The efficiency of a device can also be calculated in terms of power.
    Efficiency 6
  7. In the engine of a vehicle as shown in Figure, the energy needed is obtained by burning fuels like petrol and diesel. The energy obtained is converted into kinetic energy of the car. However, a large portion of the energy is wasted.
    Efficiency 7
  8. There are two types of engines that are commonly used in vehicles; the petrol engine and the diesel engine. The efficiency of a diesel engine is higher than that of a petrol engine. However, diesel engines are heavier and more costly to construct. Diesel engines are generally noisier than petrol engines. Diesel engines are preferred by heavy vehicles like lorries, buses, tractors and locomotives.
  9. Electric vehicles are more efficient because less energy is wasted as heat. However, electric vehicles are not common because of the high cost of technology required to build them.

Power and Efficiency Example Problems with Solutions

Example 1. A machine raises a load of 750 N through a height of 15 m in 5s. Calculate:
(i) the work done by the machine.
(ii) the power at which the machine works.
Solution:     (i) Work done is given by W = F.s
Here      F = 750 N; s = 15 m
∴ W = 750 × 15 = 11250 J
= 11.250 kJ
(ii) Now, power of the machine is given by
\( \text{P}=\frac{\text{W}}{\text{t}} \)
Here,    W = 11250 J; t = 5 s
∴ Power P = \(\frac { 11250J }{ 5s }\) = 2250 W = 2.250 kW

Example 2. A weight lifter lifted a load of 100 kg to a height of 3 m in 10 s. Calculate the following:
(i) amount of work done
(ii) power developed by him
Solution:     (i) Work done is given by
W = F . s
Here, F = mg = 100 × 10 = 1000 N
W = 1000 N × 3 m = 3000 joule
(ii) Now, P = \(\frac { W }{ t }\) , where W = 3000 J and t = 10 s
∴ P = \(\frac { 3000J }{ 10s }\) = 300 W

Example 3. A water pump raises 60 liters of water through a height of 20 m in 5 s. Calculate the power of the pump. (Given: g = 10 m/s2, density of water = 1000 kg/m3)
Solution:     Work done, W = F.s             …(1)
Here, F = mg             …(2)
But, Mass = volume × density
Volume = 60 liters = 60 × 10-3 m3
Density = 1000 kg/m3
∴ Mass, m = (60 × 10-3 m3) × (1000 kg/m3) = 60 kg
∴ Equation (2) becomes
F = 60 kg × 10 m/s2 = 600 N
Now, W = F . s = 600 N × 20 m = 12000 J
\( \text{P}=\frac{\text{W}}{\text{t}} \)
= \(\frac { 12000J }{ 5s }\) = 2400 W

Example 4. A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10 s. Calculate the power used by her.
Solution:     Given that m = 5 kg; h = 10 m; t = 10 s
g = 10 m/s2
\( \text{P}=\frac{\text{W}}{\text{t}} \)
\(=\frac{\text{mgh}}{\text{t}}=\frac{\text{5 }\!\!\times\!\!\text{ 10 }\!\!\times\!\!\text{ 10}}{\text{10}}=\text{50W}\)

Example 5. A 40 kg boy climbs a vertical height of 3.5 m in 4 s.
Power 2
(a) Calculate the gain in gravitational potential energy of the boy.
(b) Determine the power generated by the boy. [Take g=9.8m s-2]
Solution:
Power 3

Example 6. A crane lifts a heavy bucket to a height of 2.5 m from the ground in 3.5 s.
Efficiency 3
(a) Calculate the power generated by the crane in lifting the bucket if the mass of the bucket is 840 kg.
(b) Explain why the power generated by the crane is actually higher than the value calculated in (a), [g = 9.8 m s-2]
Solution:
Efficiency 4
(b) This is because besides lifting the bucket, work is also done to overcome frictional forces between the cable and the pulley and other parts of the crane.

Example 7. An electric motor has an input power of 120 W. It lifts a 20 kg load to a vertical height of 1.5 m in 5 s. What is the efficiency of the electric motor? [g = 9.8 m s-2]

Efficiency 8

How do you find work in physics?

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What is the formula for work?

Definition: In our daily life “work” implies an activity resulting in muscular or mental exertion. However, in physics the term ‘work’ is used in a specific sense involves the displacement of a particle or body under the action of a force. “work is said to be done when the point of application of a force moves.
Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force.
Work = Force × Distance moved in the direction of force.

The work done by a force on a body depends on two factors
(i) Magnitude of the force, and
(ii) Distance through which the body moves (in the direction of force)

Unit of Work
Work-and-Energy
When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule.
Work = Force × Displacement
1 joule = 1 N × 1 m
or 1 J = 1 Nm (In SI unit)

Work Done Analysis

Work done when force and displacement are along same line:
Work done by a force: Work is said to be done by a force if the direction of displacement is the same as the direction of the applied force.
Work done against the force: Work is said to be done against a force if the direction of the displacement is opposite to that of the force.
Work done against gravity: To lift an object, an applied force has to be equal and opposite to the force of gravity acting on the object. If ‘m’ is the mass of the object and ‘h’ is the height through which it is raised, then the upward force
(F) = force of gravity = mg
If ‘W’ stands for work done, then
W = F . h = mg . h
Thus W = mgh
Therefore we can say that, “The amount of work done is equal to the product of weight of the body and the vertical distance through which the body is lifted.

Work done when force and displacement are inclined (Oblique case):
Consider a force ‘F’ acting at angle θ to the direction of displacement ‘s’ as shown in fig.
Work-done
Work done when force is perpendicular to Displacement
θ = 90º
W = F.S × cos 90º = F.S × 0 = 0
Thus no work is done when a force acts at right angle to the displacement.

 

Special Examples

  1. When a bob attached to a string is whirled along a circular horizontal path, the force acting on the bob acts towards the centre of the circle and is called as the centripetal force. Since the bob is always displaced perpendicular to this force, thus no work is done in this case.
  2. Earth revolves around the sun. A satellite moves around the earth. In all these cases, the direction of displacement is always perpendicular to the direction of force (centripetal force) and hence no work is done.
  3. A person walking on a road with a load on his head actually does no work because the weight of the load (force of gravity) acts vertically downwards, while the motion is horizontal that is perpendicular to the direction of force resulting in no work done. Here, one can ask that if no work is done, then why the person gets tired. It is because the person has to do work in moving his muscles or to work against friction and air resistance.

Work Done By A Force Example Problems With Solutions

Example 1:    How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?
Solution:    The work done is calculated by using the formula:
W = F × S
Here,      Force F = 10 N
And, Distance, S = 1 m
So, Work done, W = 10 × 1 J
= 10 J
Thus, the work done is 10 joules

Example 2:    Find the work done by a force of 10 N in moving an object through a distance of 2 m.
Solution:     Work done = Force × Distance moved
Here,         Force = 10 N
Distance moved = 2 m
Work done, W = 10 N × 2 m
= 20 Joule = 20 J

Example 3:    Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.
Solution:    Force, F = 120 N; Distance, s = 100 m
Using the formula, we have
W = Fs = 120 N × 100 m = 12,000 J

Example 4:    A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s2. Calculate the work done.
Sol. Given: mass, m = 5 kg
acceleration, a = 3 m/s2
Force acting on the body is given by
F = ma = 5 × 3 = 15 N
Now, work done is given by
W = Fs = 15 N × 4 m = 60 J

Example 5:    Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. (Take g = 9.8 ms-2).
Solution:    Force of gravity
mg = 200 × 9.8 = 1960.0 N
h = 5 m
Work done, W = mgh
or W = 1960 × 5 = 9800 J

Example 6:    A boy pulls a toy cart with a force of 100 N by a string which makes an angle of 60º with the horizontal so as to move the toy cart by a distance horizontally. Calculate the work done.
Solution:    Given F = 100 N, s = 3 m, θ= 60º.
Work done is given by
W = Fs cos θ= 100 × 2 × cos 60º
= 100 × 3 × 1/2 = 150 J (∵ cos 60º = 1/2 )

Example 7:    An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.
Solution:    Given W = 64,000 J; F = 8,000 N
Work done is given by W = Fs
or    64000 = 8000 × s
or    s = 8 m

Example 8:    A horse applying a force of 800 N in pulling a cart with a constant speed of 20 m/s. Calculate the power at which horse is working.
Solution:    Power, P is given by force × velocity, i.e.
P = F . v
Here F = 800 N; v = 20 m/s
∴ P = 800 × 20 = 16000 watt
= 16 kW

Example 9:    A boy keeps on his palm a mass of 0.5 kg. He lifts the palm vertically by a distance of 0.5 m. Calculate the amount of work done.
Use g = 9.8 m/s2.
Solution:    Work done, W = F . s
Here, force F of gravity applied to lift the mass, is given by
F = mg
= (0.5 kg) × (9.8 m/s2)
= 4.9 N
and s = 0.5 m
Therefore, W = (4.9) . (0.5m) = 2.45 J.

Example 10:    A truck of mass 2500 kg is stopped by a force of 1000 N. It stops at a distance of 320 m. What is the amount of work done ? Is the work done by the force or against the force?
Solution:    Here the force, F = 1000 N
Displacement, s = 320 m
∴ Work done, W = F . s
= (1000N) . (320 m)
= 320000 J
In this case, the force acts opposite to the direction of displacement. So the work is done against the force.

Example 11:    A car weighing 1200 kg and travelling at a speed of 20 m/s stops at a distance of 40 m retarding uniformly. Calculate the force exerted by the brakes. Also calculate the work done by the brakes.
Solution:    In order to calculate the force applied by the brakes, we first calculate the retardation.
Initial speed, u = 20 m/s; final speed,
v = 0, distance covered, s = 90 m
Using the equation, v2 = u2 + 2as, we get
02 = (20)2 + 2 × a × 40
or   80a = –400
or   a = –5 m/s2
Force exerted by the brakes is given by
F = ma
Here m = 1200 kg; a = – 5 m/s2
∴ F = 1200 × (–5) = – 6000 N
The negative sign shows that it is a retarding force.
Now, the work done by the brakes is given by
W = Fs
Here F = 6000 N; s = 40 m
∴ W = 6000 × 40 J = 240000 J
= 2.4 × 105 J
∴ Work done by the brakes = 2.4 × 105 J

Example 12. A worker uses a horizontal force of 400 N to push a rubbish cart.
work 1
Calculate the work done to push the cart a distance of 180 m on a horizontal surface.
Solution:
Since the direction of the movement of the cart is the same as the direction of the applied force, therefore W = F x s
= 400 x 180 = 72 000 J

Example 13. A boy uses a force of 120 N to pull a crate along a straight corridor. The applied force is at an angle of 30° with the horizontal floor.
work 2
Calculate the work done by him after pulling the crate a distance of 90 m.
Solution:
W = F cos θ x s
= (120 cos 30°) x 90 = 9353 J

Example 14. Jin uses a force of 16 N to mop a floor. The applied force makes an angle of 28° with the floor.
work 3
What is the work done by Jin when he pushes the mop a horizontal distance of 2.5 m?
Solution:
W = F cos θ x s
= (16 cos 28°) x 2.5 = 35.3 J