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CHAPTER 4: WORK, ENERGY AND POWER
4.1 WORK
Work is defined as the product of magnitude of the displacement with the component of the force parallel
to the displacement.


Unit S.I for work – Nm @ Joule (J)
Formula work:
Where :
W
=
F x s
W = work done (J)
F
= Force (N)
S
= distance moved in the direction of the
force (s)
F
s
Example 1
A student lift a 25 kg bag onto his shoulders through a vertical distance of 3 m. what is the work done by
the student?
Solution:
F
=
=
( 25 ) ( 9.81 )
245.25 N
W
=
=
( 245.25 ) ( 3 )
735.75 Nm @ Joule
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CHAPTER 4: WORK, ENERGY AND POWER
When F is not parallel to s, then we must take the component of F which is parallel to the s.

F

F cos 
s
Formula:
Where :
W
=
F cos x s
W
F
S

= work done (J)
= Force (N)
= distance moved in the direction of the force (s)
= angle
Example 2:
A gardener pushes the lawn mower with a constant force of 90N at an angle of 40 o to the horizontal, how
much work does he do in pushing it a horizontal distance of 7.5m?
Solution:
W = F cos  x s
= 90 cos40° x 7.5
= 517 J
Work done raising an object against the force of gravity
F=w
≈
m
h
m
w = mg
The force required to lift an object of mass, m, near the surface of the earth is mg, that is, the weight of
the object, directed upwards. When it is raised to a height, h, the object is displaced by a distance h in the
direction of the force.
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CHAPTER 4: WORK, ENERGY AND POWER
Example 3:
A student lift a 25 kg bag onto his shoulders through a vertical distance of 3 m. what is the work done by
the student?
Solution:
F
=
=
( 25 ) ( 9.81 )
245.25 N
W
=
=
( 245.25 ) ( 3 )
735.75 Nm @ Joule
Exercise 1
How much work is done when a force of 5 kN is used to move a box through 600mm in the direction of
the force.
Exercise 2:
A box of 20 kg mass is dragged across a smooth floor using a rope which is makes an angle of 30° to the
horizontal. How much work is done if the box is being dragged for 12 m.
Exercise 3:
Find the work done in raising 100 kg of a box through a vertical distance of 3m.
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CHAPTER 4: WORK, ENERGY AND POWER
Exercise 4:
How much work done by a force of 40 N acting 37° above the horizontal in pulling a block 8 m along a
horizontal surface?
Exercise 5:
In raising 200 kg box 10 000 J of work is performed. How high is it raised ?
Exercise 6:
A force did 1500 Joules of work on an object in 7.5 m. How big was the force?
Exercise 7:
A man with a mass of 55 kg ascends a flight of 10 steps. Each step has a height of 0.25 m. Calculate the
work that has been done by the man.
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CHAPTER 4: WORK, ENERGY AND POWER
4.2 ENERGY
Energy is defined as the capacity to do work.
It is a scalar quantity and has the units of joules, (J).
Type of energy
Generally energy can be divided into two groups:
Mechanical Energy:
Energy due to position in a field force or energy due to movement
Non-mechanical Energy:
Energy that does not fall into the above category
Energy
Mechanical
Kinetic
Non-mechanical
Potential
Light
Linear
Gravitational
Rotational
Elastic
Electric
Magnetic
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Sound
Heat
CHAPTER 4: WORK, ENERGY AND POWER
4.2.1 KINETIK ENERGY
Definition: kinetic energy is the energy of motion
The kinetic energy of a body may be defined as the amount of work it can do before being brought to rest.
Formula:
Ek =
Where:
1
m v2
2
Ek = Kinetic energy (J)
m = mass (kg)
v = velocity (m/s2)
Example 4:
A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a constant breaking force
over a distance of 75m.
Find:
i.
ii.
iii.
The cars initial kinetic energy
The final kinetic energy
The breaking force
Solution:
i.
Initial kinetic energy = ½ mv2
= 500 x 302
= 450000J
ii.
Final Kinetik Energy = ½ mv2
= 500 x 102
= 500000J
iii.
Change in kinetic energy = 400 kJ
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Example 5:
The kinetic energy of a boat is calculated at 52,000 J. If the boat has a mass of 39,000 kg, with what
velocity is it moving?
Solution:
Ek= ½ mv²
v² = 2Ek
m
v² = 2(52000)
39000
v = 1.63 m/s
Example 6:
A rocket of mass 1.5x104 kg accelerates at 220 m/s2 for 29s from an initial speed of 5200 m/s.
i.
How fast will be rocket be travelling after the 29 s?
ii.
How much Kinetic Energy has the rocket gained?
Solution:
i.
v = u + at
= 5200 + (220 x 29)
= 5200 + 6380
= 11580m/s
ii.
Initial Kinetic Energy
Ek= ½ mv²
=0.5 x (1.5x104) x (5200)2
=2.028 x 1011 J
Final Kinetic Energy
Ek= ½ mv²
=0.5 x (1.5x104) x (11580)2
=1.006 x 1012J
so Kinetic Energy gained = final E - initial E
= 1.006 x 1012J - 2.028 x 1011J
= 8.032 x 1011J
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CHAPTER 4: WORK, ENERGY AND POWER
Exercise 8:
A pistol fires a 3g bullet with a velocity of 400 m/s. How much kinetic energy is given to the bullet.
Exercise 9:
The velocity of a 3 kg trolley is increased from 10 m/s to 16 m/s over a distance of 10 m. Find the
charge in the kinetic energy of the trolley.
Exercise 10:
The kinetic energy of a car is 20 kJ. If the car has a mass of 1,500 kg, with what velocity is it moving?
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CHAPTER 4: WORK, ENERGY AND POWER
4.2.2 POTENTIAL ENERGY
Definition: Potential energy, or stored energy, is the ability of a system to do work due to its position or
internal structure.
A body is at rest on the earth's surface. It is then raised a vertical distance h above the surface. The work
required to do this is the force required times the distance h.
Since the force required is it's weight, and weight, w = mg, then the work required is mgh.
The body now possesses this amount of energy - stored as potential energy - it has the capacity to do
this amount of work, and would do so if allowed to fall to earth.
Formula:
Ep
= mg h
Where :
Ep = Potential Energy (J)
m
= mass (kg)
g
= gravitational acceleration (9.81 m/s2)
h
= height (m)
.
Example 7:
What is the potential energy of a 10kg mass located 100m above the surface of the earth.
Solution:
m
Potential energy Ep = mgh
= 10 x 9.81 x 100
h
= 9810 J
= 9.81 kJ
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Example 8:
A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf. If the box gains
145J of Potential Energy (Ep), how high is the shelf?
Solution:
Potential energy Ep = mgh
h = Ep
mg
= 145
(5.8 x 9.81)
= 2.55 m
Example 9:
A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the
man?
Solution:
Potential energy Ep = mgh
m = Ep
gh
= 2268
( 9.81x3.6 )
= 64.22 kg
Exercise 11:
A 4 kg rock is sitting on a 3 m ledge. How much potential energy does it have?
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CHAPTER 4: WORK, ENERGY AND POWER
Exercise 12:
A 30 kg child climbs 15 meters up a tree. When he stops to have a look around, what is the
child’s potential energy?
Exercise 13:
A 2 kg box rests on a table top 0.6 m above the floor. What is the potential energy of the box with respect
to the floor? What is its potential energy with respect to a counter top 1.1 m above the floor?
4.2.4 CONSERVATION OF ENERGY
The principle of conservation of energy state that the total energy of a system remains constant. Energy
cannot be created or destroyed but may be converted from one form to another.
Without considering the air resistance, for the free fall object, at the top, from where the object is being
dropped, it has only potential energy and no kinetic energy as velocity is zero. When object is dropped it
starts falling down under effect of gravity and as it falls down, its potential energy is converted into kinetic
energy. As a result the kinetic energy of the object increases and potential energy decreases. At any
point the decrease in potential energy is equal to the increase in kinetic energy. The moment when object
hits the ground, the entire potential energy is converted into kinetic energy and this kinetic energy is equal
to the potential energy of the object at the top.
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CHAPTER 4: WORK, ENERGY AND POWER
Example 10:
A 0.5 kg ball is dropped from height of 10 m from floor level, calculate the velocity of the ball when it
reach the floor.
Solution:
Initial Potential energy, Ep = mgh1
E p = (0.5 x 9.81 x 10)
E p = 49.05 J
When reach the floor, Kinetic energy is equal to the initial potential energy
Ek = Ep = 49.05
Ek= ½ mv² = 49.05
v² = 49.05 x 2
0.5
V = 14 m/s
Example 11:
A boy drops a 1.2 kg of vase from height of 6.4 m.
(a) What is the kinetic energy of the vase when it is at a height of 4.2 m?
(b) With what speed will the vase hit the ground? (Neglect air resistance)
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Solution:
Given: m = 1.2 kg
ho = 6.4 m
h = 4.2 m
(vo = 0)
First, find the total mechanical energy since this quantity is conserved while it is falling. Initially, with vo =
0, the vase’s total mechanical energy is all potential. Taking the ground as the zero reference point.
At h = 6.4 meter
i.
Ep = mgh1
E p = (1.2 x 9.81 x 6.4)
E p = 75.34 J
At h = 4.2 meter
Ep2 = mgh2
Ep2 = (1.2 x 9.81 x 4.2)
Ep2 = 49.44 J
The kinetic energy is the change in potential energy, so
Ek = Ep1 - Ep2
= 75.34 – 49.44
= 25.9 J
ii.
Just before the vase strikes the ground (h = 0, EK = 0), the total mechanical energy is all kinetic.
E p  EK  12 mv2
Thus,
v
2Ek
m
v
275.34 J 
 125.57m / s
1.2kg
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CHAPTER 4: WORK, ENERGY AND POWER
Example 12:
A 2.0-kilogram ball rolling along a flat surface starts up a hill. If the ball reaches a height of 0.6 meters,
what was its initial speed?
Solution:
Ep = mgh1
E p = (2 x 9.81 x 0.6)
E p = 11.77 J
When the ball reaches 0.6 m of height, the initial kinetic energy is equal to the potential energy
Ek = Ep = 11.77 J
Ek= ½ mv² = 11.77
v² = 11.77 x 2
2
V = 3.43 m/s
Exercise 14:
A 600g ball is freely fall from a height of 8 meter, calculate the velocity of the ball when it reach the floor.
Exercise 15:
A ball of 2 kg of mass is dropped from a height of 5 m.
i.
What is the kinetic energy of the vase when it is at a height of 3 m?
ii.
With what speed will the vase hit the ground? (Neglect air resistance)
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CHAPTER 4: WORK, ENERGY AND POWER
Exercise 16:
The floor of the basement of a house is 3.0 m below ground level, and the floor of the attic is 4.5 m above
ground level.
i.
What are the respective potential energies of 1.5-kg objects in the basement and attic, relative
to ground level.
ii.
What is the change in potential energy if the object in the attic is brought to the basement?
4.3 POWER
Power is defined as the rate of work done.
P = W
t
Where :
P = Power (Watt)
W = work done ( J)
t = time taken (s)
Units of power are watts (W).


1 watt = 1 joule per second.
Also horsepower (hp). (1hp = 745.6 W.)
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CHAPTER 4: WORK, ENERGY AND POWER
4.3.1 Power and velocity
Power can also be expressed in terms of force and velocity.
Where :
P = Power (Watt)
F = Force (N)
V = velocity (m/s2)
P = Fv
Example 13:
It takes 20 seconds to push the box in example 4.4 across the distance. What is the power?
Solution:
Work done = 3000 J
Power = 3000 (J)
20
= 150 W
Example 14:
An electric locomotive is pulling a train at a constant speed of 30 m/s. The train has a rolling resistance of
100 kN. Calculate the power of the electric locomotive.
Solution:
Power = force x speed = 100 000 x 30
= 3 000 000 W
= 3 MW
Example 15:
A constant force of 2kN pulls a crate along a level floor a distance of 10 m in 50s.
What is the power used?
Solution:
Work done = force x distance
= 2000 x 10
= 20000 J
Power = work done
time taken
= 20000
50
= 400 W
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CHAPTER 4: WORK, ENERGY AND POWER
Example 16:
A hoist operated by an electric motor has a mass of 500 kg. It raises a load of 300 kg vertically at a
steady speed of 0.2 m/s. Frictional resistance can be taken to be constant at 1200 N. What is the power
required?
Solution:
Total mass m = 800 kg
Weight , w
= 800 x 9.81
= 7849 N
Total force = 7848 + 1200
= 9048 N
Power
= force F x speed v
= 9048 x 0.2
=1810 W
Exercise 17:
A woman runs a kilometer using a force of 210N and a power output of 500W. How long in minutes does
it take this woman to complete 1 kilometer?
Exercise 18:
How much power is used if a force of 35N is used to push a box a distance of 10 meters in 5 seconds?
Exercise 19:
What is the power of a kitchen blender if it can perform 3,750 joules of work in 15 seconds?
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CHAPTER 4: WORK, ENERGY AND POWER
4.3.2 EFFICIENCY
Definition – Efficiency is defined as the ratio of the useful output power or work to the total power or work
put into the machine.
This ratio is normally expressed as a percentage.
Where:
=

Pout x 100%
Pin
= efficiency (%)
Pout = power output (W)
Pin = power input (W)
Example 17:
Calculate the power output of a pump if its efficiency is 65 % and its the power input is 900 kW.
Solution:
 = Pout x 100%
Pin
Pout = Pin 
100%
Pout = 9000 x 65%
100%
= 5850 KW
Example 18:
If a machine has a 100W power input and 50W power output, calculate the efficiency
Solution:

Pout
 100% 
Pin

50W
 100% 
100W
  0.5100%
  50%
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CHAPTER 4: WORK, ENERGY AND POWER
Example 19:
An electric motor with a 2 hp output drives a machine with an efficiency of 75%. What is the energy output
of the machine per second? (1hp = 745.6 Watt)
Solution:
Pin = 2 hp = 1491.2 W
 = Pout x 100%
Pin
Pout = 75% x 1491.2
100%
= 1118.4 W
Exercise 20:
Calculate the power output of a pump if its efficiency is 65 % and its the power input is 900 kW.
Exercise 21:
A pulley system operates with 80% efficiency. If the work input is 500 joules, how much useful works
produced?
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CHAPTER 4: WORK, ENERGY AND POWER
Exercise 22:
The motor of an electric drill with efficiency of 90% has a power input of 3 hp. Calculate:
i.
Power output, if given 1 hp = 745.6 Watt.
ii.
Work is done by the drill in a time of 1 minute.
END OF CHAPTER (Questions)
Question 1
Give the definition, formula and SI unit for:
i.
Work
ii.
Energy
iii.
Power
iv.
Kinetic Energy
v.
Potential Energy
Question 2
A 70 kg athlete can run 100m within 10 seconds. Calculate his Kinetic Energy:
Question 3
A 200 grams bullet move through the air with a velocity of 150 m/s at 250 m height. Calculate the total
Potential energy and Kinetic Energy of the bullet.
Calculate in Joule the work done in each of the following cases:
i.
A mass of 4 kg is lifted to a height of 20 m.
ii.
A trolley is pulled with a force of 30 N for a distance of 9m
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CHAPTER 4: WORK, ENERGY AND POWER
Question 4
A man pulls a 135 kg trolley with a force of 675 N for 8 m at an angle 35° from horizontal. Calculate
the work done of the trolley (Neglect friction).
Question 5
A 10000J of work is raised up a box vertically. If the mass of the box is 20kg, calculate the height that
can be reached by the box.
Question 6
A car is moving with a velocity of 20 m/s. If the mass of the car is 3000kg, calculate the kinetic energy
possessed by the car. (Given g = 9.81 m/s2)
Question 7
A student with a weight of 650N is going up a staircase of 60 m height within 1.5 minutes. How much
power is needed by the student to go up the stairs?
Question 8
Calculate the work done by 5 kW powered motor for 7 minutes.
Question 9
A pump can lift 150 kg water of height in 40 seconds from it source. Calculate the work done and the
power of the pump.
Question 10
A ball of mass 10 kg falls from a height of 2.5m. Calculate the potential energy and kinetic energy
possessed by the ball.
Question 11
Calculate the height of box of mass of 200kg can be lifted if the work done is 10000 J.
Question 12
Calculate the work done by the force; refer to the diagram as shown as figure Q13 below:
F = 30 kN
20°
S = 50 m
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CHAPTER 4: WORK, ENERGY AND POWER
Question 13
A steel ball of mass 2 kg is released from a height of 8 m from the ground. On hitting the ground, the
ball rebounds to a height of 3.2 m as shown in Figure Q14.
If air resistance can be neglected and the acceleration due to gravity g = 9.81 m/s 2, find
i.
The kinetic energy of the ball as it reaches the ground,
ii.
The velocity of the ball on reaching the ground,
iii.
The kinetic energy of the ball as it leaves the ground on rebound.
8.0 m
3.2 m
Figure Q14
Question 14
The motor of an electric drill with efficiency of 80% has a power input of 1.5 hp. Calculate:
iii.
Power output, if given 1 hp = 746 Watt.
iv.
Work is done by the drill in a time of 30 sec.
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