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Form 4 Physics – Chapter 2 – Lesson 10
Objective:
1. Student will be able to define work W  as the product of an applied
force F  and displacement s  of an object in the direction of the applied
force, i.e. W  Fs .
2. Student will be able to state when work is done, energy is transferred
from one object to another.
3. Student will be able to define kinetic energy and state that E K 
1 2
mv .
2
4. Student will be able to define gravitational potential energy and state
that EP  mgh .
5. Student will be able to state the principle of conservation of energy.
6. Student will be able to define power P and state that P 
W
.
t
7. Student will be able to explain what efficiency of a device is.
8. Student will be able to solve problems involving work, energy, power
and efficiency.
________________________________________________________________________
2.10 Work, Energy, Power and Efficiency
2.10.1 Work
1. Work is done when a force causes an object to move in the direction of the force.
2. The work done, W is defined as the product of the force, F and the distance, s in
the direction of the force.
W = F  s
Where, W = Work done
F = Force
s = Displacement in the direction of the force
3. The SI unit for work is the joule J  . 1 J is the work done when a force of 1 N
moves an object through a distance of 1 m in the direction of the force.
4. Work is a scalar quantity.
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Form 4 Physics – Chapter 2 – Lesson 10
5. Work done can be viewed in three situations.
(a)
The displacement, s of the object is in the direction of the force, F.
Example
A constant force of 5 N pulls an object a distance of 2 m.
Work done, W = F  s = 5  2 = 10 J
(b)
The displacement, s of the object is not in the direction of the force, F.
The object does not move in the direction of F. The horizontal component
of the force, F cos  moves the object on the surface of the floor.
Work done, W = Force  Distance in the direction of the force
= F cos   s
 W  Fs cos 
Example
A constant force of 5 N at an angle of 30 pulls an object a distance of
2 m.
F = 5 N , s = 2 m ,   30
Work done, W = Fs cos
= 5  2  cos 30
= 8.7 J
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Form 4 Physics – Chapter 2 – Lesson 10
(c)
The displacement, s of the object is perpendicular to the direction of the
force, F.
Example
Work done, W = Force  Distance in the direction of the force
= F  0 (The object does not move in the direction of F)
 W 0
Example 2.10.1:
Figure shows a boy pushing his bicycle with a force of 25 N through a distance of
3 m.
Calculate the work done by the boy.
Solution:
Applied force, F = 25 N
Displacement in the direction of the force, s = 3 m
Work done, W = Fs = 25  3 = 75 J
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Form 4 Physics – Chapter 2 – Lesson 10
Example 2.10.2:
A fishmonger pulling a crate of fish along the floor with a force of 40 N through a
distance of 6 m.
What is the work done in pulling the crate?
Solurion:
Horizontal component of force = 40 cos 50 = 25.71 N
Displacement = 6 m
Work done, W = 6  25.71 = 154.26 J
Example 2.10.3:
A student walks a distance of 5 m holding a book that has a weight of 10 N. How
much work has the student done on the book?
Solution:
Using the formula in 5 (b):
W = Fs cos 
F = 10 N , s = 5 m ,   90
 W  Fs cos   10  5  0  0
This means that the student has not done any work on the book.
The student exerts an upwards force of 10 N while he is holding the book. When
the student walks forward a distance of 5 m, the book is not displaces upwards or
downwards – that is, the displacement in the direction of the force is zero.
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Form 4 Physics – Chapter 2 – Lesson 10
2.10.2 Area Under a Force – Distance Graph
1. Assume that a constant force, F is acting on an object and causes it to move a
distance of s metres.
Work done, W = F s
2. The force – distance graph in figure below represents the above situation.
Area under a force – distance graph = Work done
3. It can thus be concluded:
Area under a force – distance graph = Work done
4. The above relationship is also true for a non-uniform force.
Example 2.10.4:
A force value is increased from 0 to 30 N is applied to a spring which is compressed
by 12 cm as shown in figure below.
Calculate the work done.
Solution:
Work done, W = Area under the F – s graph
=
1
 30  0.12
2
= 1.8 J
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Form 4 Physics – Chapter 2 – Lesson 10
2.10.3 Work done and Gravitational Potential Energy
1. Gravitational potential energy is the energy of an object due to its higher position
in the gravitational field.
2. Consider an object of mass, m, being raised through a vertical distance, h.
3. The weight of the object = mg.
4. The force, F, required to raise the object steadily must have a magnitude to the
weight, mg.
5. Work done on the object, W = Fs = mgh.
6. The work done is transferred to the object as gravitational potential energy.
7. Therefore, gravitational potential energy is given by EP  mgh .
Example 2.10.5:
In a diving competition, a boy of a mass 40 kg stands on a 3 m high springboard.
What is the gravitational potential energy of the boy?
Solution:
Mass, m = 40 kg
Height, h = 3 m
Gravitational potential energy, E P  mgh  40  10  3  1200 J
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Form 4 Physics – Chapter 2 – Lesson 10
2.10.4 Work Done and The Change in Kinetic Energy
1. Kinetic energy is the energy of an object due to its motion.
2. Consider a force, F, acting on an object at rest of mass, m. The object move with
an acceleration, a, over a displacement, s.
Work done on the object, W=Fs.
Using the equation of uniformly accelerated motion, v 2  u 2  2as .
Since, u  0 , v 2  2as .
Hence, s 
v2
2a
Substituting F  ma and s 
v2
into W=Fs,
2a
 v2  1
We obtain, W  ma   mv 2
 2a  2
The work done is the energy transferred to the object as kinetic energy.
Therefore, kinetic energy of the object: E K 
1 2
mv .
2
Example 2.10.6:
A 0.6 kg trolley across the floor at a velocity of 0.5 m s 1 . What is the kinetic
energy of the trolley?
Solution:
Kinetic energy, E K 
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1 2 1
2
mv   0.6  0.5  0.075 J
2
2
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Form 4 Physics – Chapter 2 – Lesson 10
2.10.5 Power
1. When a force is applied on an object and there is motion in the direction of the
force, work is said to be done by the force.
2. The object can be moved by the force from one place to another place over a short
period of time or a longer period of time.
3. Consider a child being lifted to the same height by the parents. The mother takes a
longer time to lift the girl. Since the girl was lifted to the same height by both
parents, the father and mother did the same amount of work. The father who did
the work in a shorter period of time is said to have generated a higher power when
lifted the girl.
4. Power is defined as the rate at which work is done, or the amount of work done
per second.
Power 
Work done
Time taken
P
W
t
5. The unit of power is watt (W). A power of 1W is generated when 1 J of work is
done in 1 s.
6. The table below shows typical values of power in some common situations.
Situations
Power
Person walking up stairs
180 W
Athlete running
360 W
Power output of a washing machine
240 W
Engine of a moving car
50 kW
Jet engine
80 MW
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Form 4 Physics – Chapter 2 – Lesson 10
Example 2.10.7:
In the snatch event of a weightlifting competition, a weightlifter lifts 140 kg from
the floor to a height of 1.2 m above the floor in one complete movement in a time
of 0.8 s. What is the power generated by the weightlifter during this time?
Solution:
Force used to lift the weights, F  ma  140 10  1400N
Displacement, s  1.2m
Time taken, t  0.8s
Work done, W  1400 1.2  1680J
Power generated, P 
W 1680 J

 2100W
t
0.8s
Example 2.10.8:
A crane with output power of 12 kW is used to lift a steel bar of mass 400 kg to a
height of 20 m. What is the time taken by the crane to do this work?
Solution:
Force to lift the steel bar, F  400 10  4000N
Work done, W  4000  20  80000J
Power output, P 
12000W 
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W
t
80000 J
t
t
80000 J
 6.67 s
12000W
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Form 4 Physics – Chapter 2 – Lesson 10
2.10.6 Efficiency
1. Figure shows vehicles used in various models of transportation. The engine of a
vehicle transforms the chemical energy in the petrol to kinetic energy of the
vehicle as it moves.
2. The engine is unable to change all the chemical energy in the petrol to become the
kinetic energy of the vehicle. Other forms of energy such as thermal energy and
sound energy are also obtained from the operation of the engine.
3.
Figure shows the transformation of energy as a car moves.
4. The kinetic energy of the car is the useful energy output of the car engine. The
thermal and sound energy are the unwanted energy.
5. Figure above shows a diagrammatic representation of the energy transformation.
6. The efficiency of a device is defined as the percentage of the energy input that is
transformed into useful energy.
Efficiency 
Useful energy transferred
 100%
total energy sup plied
7. The efficiency of a device can also be calculated in terms of useful power output
and power input.
Efficiency 
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Useful power output
 100%
Power input
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Form 4 Physics – Chapter 2 – Lesson 10
8. Table below shows typical values of efficiency in some common situation.
Situation
Efficiency
Petrol engine
25%
Diesel engine
35%
Electric motor
75%
Light bulb
20%
Example 2.10.9:
A petrol engine has a work output of 96 kJ per minute. What is the power output if
the engine efficiency is 20%?
Solution:
Power output 
Efficiency 
20% 
9600 J
 1600W
60s
Useful power output
 100%
Power input
1600W
 100%
Power input
Power input 
160000
 8000W
20
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