WORK POWER AND ENERGY
WORK :
1.
Work is said to be done on a body or system whenever a force acting on the body or system
causes some displacement.
Work is measured as the product of force and displacement.
2.
If the force and the displacement are in the same
direction then work W = Fs where F is the force and s is
the displacement caused by the force.
Force
displacement
If the direction of action of the force makes an angle  with direction of displacement then the
Force
work done is given by
W = F coss.
Cos has different values for different values of . The
following information will be helpful in working out
problems.

displacement
cos 0° = 1
3
= 0.866
2
4
cos 37° =  0.8
5
cos 30° =
1
cos 45° =
cos 60° =
2
 0.707
1
 0.5
2
cos 90° = 0
3.
When the force (or a component of the force) is in the direction of the displacement the work is
said to be positive. Whenever a force does positive work on the body the velocity of the body
increases. In this case it is said that “work is done on the body”. Example : You push some
object. As the object moves the direction of motion and the direction of application of the force
are the same. The velocity of the objective increases.
When the force of a component of the force is in the direction opposite to that of the
displacement, the work is said to be negative. We say the body does work against the body.
Whenever the work is negative the velocity of the body decrease. Example: You are moving on
a bicycle. When you apply breaks the retarding force is applied opposite to the direction of its
motion. It is said that the body does “work against the retarding force.” The velocity
decreases.
4.
The ability of a body to do work is called energy. Mechanical energy is of two types. They are :
i) kinetic energy and (ii) potential energy.
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5.
Work, Power and Energy
Kinetic energy : Kinetic energy is the energy possessed by a body by virtue of its motion. It is
1
denoted by E. E  mv 2
2
2
1
p2
 m   mv 
It can also be written as E  mv 2   

2
2m
2m
m
Its SI unit is joule (J). 1 J = (1 N) (1m) = 1 kg m 2 s–2.
2
–2
The cgs unit of work is 1 irg. 1 erg = 1 g cm s
6.
The energy possessed by a body by virtue of its position state or condition is called the potential
energy. In this chapter we deal only with the gravitational potential energy.
A body gains gravitational potential energy by going to some height above the surface of earth.
If a body of mass m is at a height h above the ground it has potential energy mgh.
This SI unit of potential energy is also J.
Its cgs unit is erg.
7.
Work done in taking a body to a height h abvoe the surface of earth is mgh. This does not
depend on the path taken by body, as long as the path has no friction.
Height h
Part -1
Part -2
Part -3
Part -4
ground
The diagram contains four paths. Path – 1 is a straight vertical path. Path –2 is along an inclined
plane. Path – 3 is along the arc of a circle. Path – 4 is an uneven path. But all the paths start at
the ground level and reach the height h.
The work done is reaching a height h is the same along all the paths if there is no friction
along the paths. That work is mgh.
We can say that whenever a body is taken along a frictionless path to a height h the work done
is mgh. All of this work becomes potential energy at that height.
Power is the rate of doing work.
work energy
It is calculated as : power, P =

.
time
time
It means that if a work W is done in time t, the power P =
W
t
If an energy E is gained or lost in a time t then the power P =
The unit of power is watt (W). 1W 
E
t
1J
1s
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8.
Work, Power and Energy
Law of conservation of energy :
This law states that in an isolated system the total content of energy always remains the same.
Energy can change from one form to the other but the total content cannot change.
Here isolated body or system means that body or that system which can neither give energy to
out side nor can take energy from outside.
In this class and in this chapter we deal only with systems possessing kinetic energy and
potential energy. These two energies are together called the mechanical energies.
The sum of the KE and PE is called the total mechanical energy (TME).
For example a body of mass m is dropped freely from a height h. As it falls its height decreases.
Potential energy decreases. But its velocity increases. Its kinetic energy increases.
At any point the potential energy + kinetic energy = total mechanical energy which is constant.
 TME = KE + PE constant.
Similarly a body is projected from ground with a velocity. It has only kinetic energy at the ground
level. As it climbs up it looses kinetic energy but gains potential energy. At any point the sum
of its potential energy and the kinetic energy shall be equal to total mechanical energy.
 TME = KE + PE = constant. 
9.
work – energy theorem :
This theorem says that whenever a uniform force acts on a body the change in its kinetic energy
is equal to the work done.
In symbols work = Energy = KEfinal – KEinitial
For example a force F acts on a body of mass moving with a velocity u. By the time the body
moved a distance s its velocity became v.
1
1
By work-energy theorem, FS  mv 2  mu2
2
2
10.
A machine gun fires n bullets, each of mass m, per second with velocity v each.
1
The gun imparts a kinetic energy mv 2 to each bullet.
2
The total energy imparted per second is = number of bullets fired per second  KE imparted to
1
1

each bullet =  mv 2  n  mnv 2 . The total energy imparted by a system in one second is the
2
2

power of the system.
The power of the machine gun is
11.
1
mv 2 
2
In the above discussion the gun imparts a momentum mv to each bullet. In one second it fires n
bullets. Each of the bullets has the same momentum mv imparted to it.
The total momentum imparted = (mv) (n) = mnv.
But the total momentum imparted in one second is called the rate of change of momentum and
this is the force exerted by the gun.
The force exerted by the machine on the gunner’s hand = mnv.
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Work, Power and Energy
WORKED EXAMPLES 
1.
A force of 20 N acts on a body and makes it move through 10 m, in the direction of the
force. Find the work done. Is the work positive or negative ?
Sol: Work = force  distance
 W = 20 N  10 m = 200 J
The work is positive because the displacement is in the direction of the force.
2.
A body of mass 5 kg is at rest. It is acted upon by a force of 25 N. What is the work done
in 4 s?
Sol: W = Fs
We are given F = 25 N
If we can find the displacement s, from the rest of the data, we can calculate W.
Body started from test .  u = 0
Body of mass 5 kg is acted upon by a force of 25 N. the acceleration a =
If it moves for time t the displacement undergone is given by s  ut 
 s0
F 25N

 5ms2
m 5kg
1 2
at
2
1
(5)(4)2  40m
2
W = Fs = 25 N  40 m = 1000 J
3.
A body of mass m is lying at rest on a smooth ground. A force F is applied on it at an
angle 30° from the horizontal. What is he work done in t s?
Sol: W = (F cos) (s)
We are given F and . If we find the value of s in terms of the rest of the data then we can
calculate W.
The horizontal displacement is caused by the component of the force Fcos.
F cos 
Acceleration of the body is a =
m
The body is initially at rest. u = 0
S  ut 
1 2
1  F cos   2
at  0  
t
2
2  m 
1  Fcos   2 F2 cos2  2
 W   Fcos   
t 
t
2  m 
2m
4.
A man of weight 80 kg carries a box of weight 20 kg on his head and climbs up 6 steps
each of height 30 cm. Find the work done by him
Sol: W = Fs.
H = 6h
Here the man is climbing some height.
Work done = (mass moved) (g) (H)
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Where H is the total height climber. H = 6h where h is the height of each step.
We are given that m has two parts. One part is the weight of the man which he has to carry.
The other part is the weight on his head.  The total weight moved = 80 kg + 20 kg = 100 kg
H = 6h = 6  30 cm = 180 cm = 1.8 m
–2
W = 100 kg  9.8 ms  1.8 m = 1764 J
5.
A bucket weights 1 kg. It can hold water of weight 9 kg. What is the work done in drawing
it from a well of depth 5m. (Neglect the mass of the rope).
Sol: A mass is moved vertically.  Work done W = mgh.
M = mass of the bucket + mass of water in it = 1 + 9 = 10 kg
h=5m
g = 9.8 ms–2
–2
 W = (10 kg) (9.8 ms ) (5m) = 490 J
6.
A man applies a force of 100 N at an angle 30° from the horizontal to push a roller of
mass 100 kg horizontally. What is the work done by him in moving the roller through 10
m horizontally.
Sol: Here the displacement is horizontal. The force makes an angle 30° with horizontal. The angle
between the force and the displacement is 30°.
Hence, the work done, W = F coss.
 W = (100 N) (cos 30°) (10 m)
 W = 1000  0.866 J = 866 J
7.
A motor can lift 9000- kg of iron ore from a mine of depth 120 m in one hour. If 25% of the
energy is wasted, find the power of the motor.
Sol: We are asked to find the power of the motor.
We are further given that 25% of the power supplied is wasted. This means that if the motor
supplied 100 units of power 25 units are wasted and 75 units alone becomes useful power.
If a mass m is lifted through a height h in time t then the work done is mgh and the power
mgh
developed is
.
t
We can build a logic as follows:
To get 75 units of useful work the motor shall have 100 units of power.
mgh
To get
useful power what shall be the power of the motor ?
t
mgh 100 4 mgh

t 75
3 t
It shall be
Given m = 9000 kg
–2
g = 9.8 ms
h = 120 m
9000  9.8  120
Power of the motor =
 2940 W
60  60
t = 1 hour = 60  60s = 3600 s
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8.
Work, Power and Energy
A machine gun fires 360 bullets per minute at a speed of 360 kmph. If each bullet weights
10 g find the power of the gun. Find he force on the gunner’s hand.
Sol: We know that when the machine gun fires n bullets per second and each of the bullets has
1
mass m speed v, the power of the machine gun is mnv 2 and the force on the gunner’s hand
2
is mnv.
1
Power : P  mnv 2
2
We are given m = 10 g = 10 10–3 kg = 10–2 kg
We are given that the gun fires 360 bullet per minute that is in 60 s.  The number of bullets
360
fires per second
 6 per second.
60
v = 360 kmph 360 
P 
5
–1
ms1 = 100 ms
18
1
2
102 (6) 100   300W
2


Force :
F = mnv = (10–2 kg) (6 s–1) (100 ms–1)
=6N
A body of mass 2kg has a velocity 9 ms–1. a) What is its kinetic energy ? B) What force is
needed to bring it to hat in 9 m?
1
1
Sol: Kinetic energy = mv 2  (2)(9)2  81J
2
2
9.
To find the force needed to stop it in 9 m.
Use the work energy theorem. The initial kinetic energy of the body is 81 J. Finally the body
comes to rest. The final kinetic energy = 0. Change in kinetic energy = 81 J.
By the work energy theorem the change in kinetic energy is equal to the work done.
We know that W = Fs.
 Fs = 81 J
81J
F
 9N
9m
10.
If the kinetic energy of a body of mass 0.5 kg is 100 J find its linear momentum.
Sol: We saw that kinetic energy can be expressed in terms of linear momentum as E 
 p2  2Em  p  2Em
 p  2 100  0.5   100  10kgms 1
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11.
When the velocity of a body is doubled what happens to its kinetic energy?
1
Sol: E  mv 2
2
The new velocity is double the previous one. v = 2v. Mass does not change.
The new kinetic energy. E 
1
1

m(2v 2 )  4  mv 2   4E
2
2

The kinetic energy becomes 4 times.
12.
If the momentum of a body is doubled, what happens to its kinetic energy ?
Sol: If p is the linear momentum, m the mass and E the kinetic energy of a body then E 
p2
.
2m
Now linear momentum is doubled and made 2p. Mass does not change. Let E be the new
kinetic energy.
 E 
 2p 
2
2m
4
p2
 4E
2m
Kinetic energy becomes 4 times.
13.
Two bodies of masses m1 and m2 have the same linear momentum. a) what is the ratio of
their kinetic energies? b) What is the ratio of their velocities?
p2
. When p is the same
2m
1
for any number of bodies the KE of each body depends on its mass as E  .
m
E1 m2


E2 m1
Sol: The relation between the linear momentum and kinetic energy is E 
The relation between the linear momentum and velocity is L = mv.
Since both bodies have the same linear momentum we have m1v1 = m2v2.
v
m
 1  2 
v 2 m1
14.
–1
A body of mass 2kg is thrown up with a velocity 8 ms . It can raise to a maximum height
h
h. At a height , what is its potential energy ?
2
Sol: Let us analyze the law of conservation of energy in this case.
Let the body be projected from A. Let B be a point in its path at a height x above the ground. Let
C be the highest point reached by it. Let us see the total mechanical energy (TME) at A, B and
C.
At A:
At the point of projection A, the body has kinetic energy EA 
1
mv 2 .
2
It is on ground. Its potential energy is zero. UA= 0
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Total mechanical energy (TMEA) = EA + UA =
Work, Power and Energy
1
1
mv 2  0  mv 2
2
2
At B:
Let the body reach a point B in its path at a height x
from the point of projection. Let its velocity at B be v b.
1
Its kinetic energy at B is mv 2b .
2
But
vb2
2
 v  2( g)x 
vb2
2
 v  2gx
Highest point height = h
C
h Point at height x
B
A
x
Point of projection
Its kinetic energy at B is :
1
EB  m v 2  2gx  EA  mgx
2


Its KE decreased by mgx.
Its potential energy = UB = mgx. It is increased from zero to mgx.
Its potential energy increased by mgx.
Decrease in kinetic energy is equal to the increase in potential energy.
But TMEB = EB + UB = EA – mgx + mgx = EA = TMEA
At C:
C is the highest point in its path. At C the velocity of the body is zero .
 Kinetic energy at B, EB= zero.
Potential energy at B = UB= mgh.
The body is projected with velocity v upward. The maximum height reached by it is given by :
v2
h
2g
 v2  1
 PE  UB   mg    mv 2  TME A 
 2g  2
 

We use this information to find the KE and the KE at the point at a height h/2 from the ground.
1
Its total mechanical energy = mv 2 at A and is equal to mgh at C.
2
At a height
15.
h
h 1 1
 1 1

its potential energy is mg   mv 2     2  82   32J
2
2 22
 22

A wooden block of mass m has length L, breadth B and height H. Find its potential
energy when it stands on the length breadth face. Find its potential energy when it
stands on the breadth –height face.
Sol: We take the entire mass, m, of the block to be concentrated at its
geometrical center. This point is called the centre of mass or the
centre of gravity.
H
L
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When the block is on its Length and breadth sides as shown the block is a point mass m at a height
H/2 from the ground.
mgH
Its potential energy is
2
When the block stands of the breadth – height face it is as shown :
It is now a point mass m at a height L/2 from the ground. Its potential
mgL
energy is
2
B
H
LEVEL – I
1.
Find the work done when a boy of 40 kg climbs 50 steps each of height 30 cm. (earlier EAMET)
2.
An electric motor exerts a force of 40 N on a cable and pulls it a distance of 30 m in one minute.
Find the power supplied by the motor. (earlier EAMCET)
3.
An elevator is designed to lift a load of 1000 kg through 6 floors of a building averaging 3.5 m
per floor, in 6 seconds. Find the power of the elevator motor. (earlier EAMCET)
4.
A body of mass 2 kg has kinetic energy 64 J. What is its linear momentum ? (earlier EAMCET)
5.
If the linear momentum of a body becomes 1.5 times, how many times will its kinetic energy
become?
6.
What is the power used by a motor in lifting a mass of 250 kg through 10 m in 49 s?
7.
An engine can lift 90 kg through 30 m in 60 s. If 40% of the power is wasted what is the power
of the engine?
8.
A bullet of mass 20 g strikes a block of wood with a velocity 500 ms–1 and passes through the
block. If it comes out of the block with a velocity 100 ms–1 find the work done by the bullet in
passing through the block.
9.
A porter stands for 20s holding a suit case of weight 40 kg on his head. What is the work done ?
10.
A body is projected vertically upward. During the ascent.
a) which energy decreases ?
b) which energy increases ?
c) which energy does not change ?
11.
A body is at rest and still has mechanical energy. What energy is it?
12.
Which energy is associated with a flying bird?
13.
Two bodies of different masses have the same momentum. Which has greater velocity?
14.
Two bodies of different masses have the same linear momentum. Which has greater energy?
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15.
A bullet is fired from a riffle. The bullet and the gun move. Which has greater kinetic energy?
16.
The mass of a body is halved and its velocity is doubled. What happens to its kinetic energy?
17.
A body of mass 1 kg at rest is subjected to a force of 5 N. In the first 4s:
i) Find the work done.
ii) Find the power.
18.
A bucket of water of weight 20 kg lifted through a height of 20 m in 4s. Find :
a) the work done and
b) the power.
19.
Two bodies of different masses have the same linear momentum. The same retarding force is
applied on both. Which stops after traveling greater distance?
20.
Two bodies of different masses have the same kinetic energy. When the same retarding force is
applied on both which stops after traveling greater distance?
LEVEL – II
1.
A body of mass 2 kg is thrown up vertically with a kinetic energy of 490 J. If the acceleration due
to gravity is 9.8 ms–2 find the height at which the kinetic energy of the body is equal to its
potential energy.
2.
When a body moves through air, air resists the motion of the body. The body has to spend
some of its kinetic energy to overcome this resistance. Such energy is lost. The law of
conservation of energy can be applied only to the remaining energy.
A 50 kg sphere is projected vertically upward with a speed of 200 ms–1. It rises to height of
1500 m. How much energy is used up in overcoming the air resistance.
3.
A girl is swinging in a swing. Her height, measured from the ground, is 4 m at the highest point
and 2 m at the lowest point. What is her maximum speed? (take g = 10 ms–2)
4.
A uniform rectangular marble slab of mass 20 kg has length 3.4 m and breadth 2 m. It is lying
flat on the ground. Find the work done in making it stand on the length side and also find the
work done in making it stand on the breadth side.
5.
A machine gun fires 240 bullets per minute. The mass of each bullet is 10g. If the power of the
gun is 7.2 kW find the velocity of each bullet.
6.
Two masses 5kg and 3 kg are separated by 16 m. Each is subjected to the a force of 8 N, at the
same instant , directed towards the other. Find :
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a) the time that elapses from the instant of applying the force to the instant they meet.
b) the distance covered by each mass before they meet.
7.
Water is pumped vertically upwards through a height of 50 m in 10 minutes to fill a tank of
capacity 600 litres. Find the power of the pump.
8.
–1
A bullet of mass 20 gram moving with a velocity of 200 ms
can penetrate into a fixed uniform
block of wood through a distance of 20 cm. Find the resistance offered by the block to the bullet.
9.
A bomb of mass 12 kg at rest explodes into two pieces of masses 4 kg and 8 kg. If the velocity
of the 8 kg mss is 6 ms–1 find the kinetic energy of the second piece.
10.
A force is applied on a mass of 1.5 kg in the vertically upward direction. The mass, starting from
rest, moved with acceleration of 1.2 ms–2.
a) Find the work done in 10s.
b) Find the power developed in this time.
11.
A body of 2kg falls from rest. What will be its kinetic energy during the fall at the end of 2 secs.
(g = 10 m/s2)
KEY TO LEVEL – I
1) 5880 J
2) 20 W
3) 34.3 kW
4) 16 kg ms–1
6) 500 W
7) 735 W
8) 2400 J
9) zero
11) PE
12) KE and PE
13) lighter body
14) lighter body
16) doubled
17) 200 J, 50 W
18) 3920 J, 980 W
5) 2.25
10) i) KE, ii) PE, iii) TME
15) bullet
19) lighter body
20) both cover the same distance
KEY TO LEVEL – II
1) 12.5 m
6) i)
2) 265  103 J
7.5 s, ii) 6m , iii) 10 m
10) 990 J, 99W
3) 2 10 ms–1
4) 196 J, 333.2 J
5) 600 ms–1
7) 490 W
8) 2000 N
9) 288 J
11) 400 J

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