Chapter 10 Work and Energy class 9 pdf Download

WORK, ENERGY AND POWER
4.1 INTRODUCTION
In our daily life we talk about the terms work, energy, power etc.,
we need energy to do work. In normal life any physical or mental
activity is termed, as work done. We also say that the one who can
do more work has more energy. Thus the definition of energy is
dependent on work. Living beings have to perform several basic
activities to survive. The energy for these processes comes from
food.We needenergyforother activitieslikeplaying,singing,reading,
writing, thinking, jumping, cycling and running.Activities that are
strenuous require more energy.
Question based on basic knowledge required to understand this
chapter
1. A force of 20 N is acting on a body of mass 2 kg. What is its
acceleration?
(A) 2 m/s2
(B) 40 m/s2
(C) 22 m/s2
(D)10 m/s2
2. Velocity of body changes from 3 m/sto 7 m/s in 2 seconds. What is
the acceleration which acted on the body.
(A) 2 m/s2
(B) 4 m/s2
(C) 8 m/s2
(D)–2 m/s2
3. A body of mass 20 kg is moving with a velocity of 45 m/s. What is
its momentum?
(A) 90 kg m/s (B)
4
9
kg m/s
(C) 900 kg m/s (D) 100 kg m/s
4. A body of mass 15 kg is moving with a velocity of 36 km/h. It hits
another body of mass 3 kg at rest. The two bodies combine &
move with velocity V. What is the value of V in km/h.
(A) 24 (B) 30 (C) 60 (D)none
5. A body is dropped froma height of 80 m, it reaches to earths surface
in:
(A) 4 sec. (B) 2 sec. (C) 16 sec. (D)8 sec.
6. Rate of change of momentum is equal to
(A) force (B) impulse
(C) acceleration (D) velocity
4.1 Introduction
4.2 Work
4.3 Energy
4.4 Kinetic Energy
4.6 Work Energy Theorem
4.7 Potential Energy
4.8 Potential Energy of any
object at any height (h)
4.9 Power
“IIT-JEE Foundation”
*4.5 Relation between
Kinetic energy and
Momentum
*4.10The equivalence of
mass and energy
*4.11 Collision
VAVA CLASSES/PHY/9TH
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7. Rate of change of velocity is equal to
(A) distance (B) displacement
(C) force (D) acceleration
8. A body is moving in a circular path of radius 49 m. It completes 1 revolution in 14 seconds. The angular
velocityis:
(A)
22
49
rad/s (B)
3
7
rad/s (C) 49 rad/s (D) 35 rad/s
4.2 WORK
In our day-to-day life, the word work means any kind of mental and physical activity. For example , we
say that we are doing work while reading a book, walking a level road with a box on our head, pushing a
wall etc. But in scientific terms no work is done in these cases.
In physics work is done if a force applied on a body displaces the body in its own direction.
In other words the conditions, which must be satisfied for the work done, are:
(a) A force must act on the body
(b) The body must be displaced from one position to another.
“Work is said to be done by a force on a body if the force applied causes a displacement in the
body.”
Work done is directly proportional to the applied force (F).
W  F
Work is also directly proportional to the distance travelled by the body (in direction of applied force) or
displacement
W  S
4.2.1 Work done by constant force
W = F × S
Unit = N × m or joule. It is a scalar quantity.
S
F
1 J = 1 N × 1 m
Thus 1 joule of work is done when a force of 1 N displaces
a body by 1m.
So, we see that work done is equal to the product of the magnitude of the applied force and the distance
travelled by the body.
Illustration 1
A force of 10 N acts on an object. The displacement is 4 m in the direction of the force. What is the
work done?
Solution
Work done W = F × S
10 N × 4m = 40 N – m or 40 Joule
*4.2.2 Work done when force is acting at some angle
If F is the force acting on any body at any angle  with the horizontal. The horizontal component of the
force is F cos  and the component in vertical direction is F Sin , since the block moves in the horizontal
direction so only horizontal component of force contributes to the work done.

W = F cos  × S
Now let us consider the following cases.
Case–1: If 0   < 90º  cos  is positive
Work done is positive
S
F
Case–2: If  = 90º  cos 90º = 0
Work done is zero
Case–3: If 90° <   180°  cos  is negative
Work done is negative.
Illustration 2
A force of 40 N acts on a toy at an angle of 60°. The toy moves for a distance of 12 m calculate the
amount of work done.
Solution
Component of force on horizontal direction = 40 cos 60°
Displacement = 12 m
Work done = 40 cos 60° × 12
= 40 (1/2) × 12 = 240 J
Try yourself
1 A work of 720 J is done when a body is displaced by 36 cm. Calculate the force responsible for this.
2 A force F = 90 N is applied in horizontal direction. Calculate the work done by it is vertical direction.
3 A force 30 N is applied on a body at an angle of 45º with the horizontal. It does a work of 900 N & 450 N
is horizontal & vertical directions respectively. Calculate the distance travelled in each direction.
4.3 ENERGY
We often say that one man is stronger then the other or has more energy than the other. Thus we compare
the amount of work both can do. The person with more energy can do more work.
“The capacity to do work is known as energy”. Anything which can do work has energy. The energy
possessed by an object is measured in terms of its capacity of doing work. The unit of energy is, therefore,
the same as that of work, that is, joule (J). 1 J is the energy required to do 1 J of work. The amount of
energy possessed by a body is equal to the amount of work it can do. Energy is a scalar quantity.The
various types of energy are
(i) Mechanical Energy: The energy by the virtue of which a body can do some mechanical work
directly is known as mechanical energy. It is the sum of the kinetic and potential energy.
(ii) Heat or Thermal Energy: The energy possessed by a body due to temperature is known as heat or
thermal energy.
(iii) Chemical energy: The energy released in chemical reactions is known as chemical energy.
(iv) Sound energy: The energy of a vibrating body producing sound is known as sound energy.
(v) Electrical energy; The energy of moving electrons in a conductor connected with a battery is known
as electrical energy.
(vi) Solar energy: The energy raidated by the sun is known as solar energy.
(vii) Nuclear energy: The energy released when 2 nuclei of light elements combine with each other to
form a heavy nucleus or when heavy nucleus breaks into 2 light nuclei is known as nuclear energy.

4.4 KINETIC ENERGY (K.E.)
“The energy possessed by a body virtue of its motion is known Kinetic energy”. Flowing water
from a dam can run a turbine to generate electricity, the flowing wind can uproot big trees or can run the
blades of windmill and can be used for producing electricity or for doing some mechanical work. When a
moving object hits another stationary object, it is displaced from its position. It means, the moving object
has done work on the stationary block. So, the moving object has energy which is known as kinetic
energy.
Consider a body of mass m at rest. Let a force F is applied on the body so that the body attains a velocity
v after travelling a distance s.
Work done W = Fs .....(i)
Also the velocity of body changes from 0 to v, so the body is accelerated. Then by Newton’s II law
F = ma
Therefore
W = (ma)s .....(ii)
Now
v2
– u2
= 2as
as the body starts from rest,  u = 0
m
u = 0
(rest)
m
V
We get v2
– 0 = 2as
or s =
2
v
2a
.....(iii)
From the above three equations, we get
W =
2
v
ma ×
2a
=
2
1
mv
2
This work done is the kinetic energy of the body.
 Kinetic energy = ½ mv2
= ½ (mass of body) (velocity of body)2
Here note that
KE  v2
 higher the velocity of an object the greater is its KE.
also KE  m
 heigher the mass, greater is the KE
Illustration 3
A body can do 40 J work. What is its energy?
Solution
Energy is capacity to do work
 Energy = Work done
= 40 J
Illustration 4
A body of mass 10 kg is moving with a velocity of 3 m/s. What is its K.E.?
Solution
KE =
2
1
mv
2

=
1
2
× 10 × 3 × 3
= 5 × 3 × 3
= 45 Joule.
Illustration 5
The KE of a body is 1800 joule & its mass is 9 kg. What is its velocity?
Solution
KE =
2
1
mv
2
100 =
1
2
× 9 × v2
v2
=
1800 × 2
9
= 400
v = 400 = 20 m/s
Try yourself
4. A body whose velocity is 90 km/h has a KE of 25 J. Find its mass.
5. A body of weight 300 N has a velocity 10 m/s. Find its kinetic energy.
*4.5 RELATION BETWEEN K.E. & MOMENTUM
Momentum (p) = m × v
 v =
p
m
Now KE =
2
1
mv
2
=
1
m
2
×
2
p
m
 
 
 
p2
= 2(m)(KE)
p = 2 × mass × kinetic energy
Illustration 6
Kinetic energy of a body of mass 100 g is 80 J. Find its momentum & velocity.
Solution
p 2m K.E.
 
100
2 80 16
1000
    =4 kg m/s
mv = p
v = p/m
= 4/0.1 = 40 m/s

Try yourself
6. The momentum of a body of mass 2 kg is 100 kg m/s, find its KE.
4.6 WORK ENERGY THEOREM
Work energy theorem states that–
“Work done by a force on a body is equal to change in kinetic energy of the body.”
Consider a body of mass m moving with initial velocity ‘u’. Let a force F be applied so that velocity
changes to ‘v’ while travelling a distance s.
Work done W = F × s .....(i)
Since velocity of body changes while travelling.
 acceleration acts on it
Then by newton’s II law
F = ma .....(ii)
From (i) and (ii), we get
W = (ma)s .....(iii)
also v2
– u2
= 2as
s =
2 2
v –u
2a
.....(iv)
putting(iv) in (iii)
W = ma ×
2 2
v –u
2a
W =
2
1
mv
2
–
2
1
mu
2
W = Final, K.E. – Initial K.E. = change in K.E.
=  K.E.
Thus work done by a force is equal to change in KE of the body.
4.7 POTENTIAL ENERGY (U)
Anything which possesses energy has a capability to do work. When a hammer is lifted to a height & then
allowed to fall on a nail on a wooden surface we see that the nail goes inside the wood.
This shows that the hammer gained some energy when it was lifted.
This energy was required to drag the nail inside the wooden block.
“The energy possessed by a body by virtue of its position or shape or configuration is known as
potential energy.
4.7.1 Gravitation Potential Energy
The potential energy of a body by virtue of its height above ground level is called gravitational potential
energy.
4.7.2 Elastic Potential Energy
The potential energy of a body by virtue of its configuration is

known as elastic potential energy.
Eg: Energy attained by an arrow when it is stretched backwords.
4.8 POTENTIAL ENERGY OF ANY OBJECT AT ANY HEIGHT (h)
The work done in lifting an object of mass ‘m’ against force of gravity through a certain height (h) is equal
to the potential energy of the object at height h.
consider a body of a mass m at a height ‘h’.
Force on the body = weight
 F = mg
distance travelled s = h (height)
Now W = F × s = mg × h
ground
h
 Gravitational potential energy = mgh
Note that:
GPE  h (height of the object)
GPE  m (mass of the object)
4.8.1 Law of conservation of energy
“Energy can neither be created nor destroyed. It can only be changed from one form to another”.
It means energy in a system is always conserved.
Now we know that
Mechanical Energy = Kinetic Energy + Potential Energy
Now if no external force is acting, mechanical energy remains
conserved in a system.
i.e. M.E.i
= M.E.f
 K.E.i
+ P.E.i
= K.E.f
+ P.E.f
Simplest example is when a body at height ‘h’ is released.
Consider a body at height ‘h’ at rest shown in (figure-1).
ground
PE = mgh
KE = 0
h
(figure-1)
PE of body = mgh
KE = 0 (body is at rest)
Total energy = mgh + 0 = mgh ..(i)
Now, the body is released from rest
Let the object falls through a height x (< h)
as shown in (figure-2)
height above ground= h–x
 PE = mg (h – x)
Now velocity of body
v2
– u2
= 2gh
v2
– 0 = 2gx
v2
= 2gx

KE =
2
1
mv
2
=  
1
m 2gx
2
= mgx
PE = mg(h – x)
KE = ½ mV2
V
h
x
h–x
(figure-2)
Total mechanical energy = PE + KE
= mg(h – x) + mgx
= mgh .....(ii)
Now body is just above ground i.e. about to hit the ground as shown in (figure-3)
PE of the body.
= mg h
= mg (0)
h
(figure-3)
= 0
Also,
v2
– u2
= 2gh  v2
– 0 = 2gh
 v2
= 2gh
KE = ½ mv2
=  
1
m 2gh
2
= mgh
Total mechanical energy = PE + KE
= 0 + mgh
= mgh .....(iii)
Thus, from equation (i), (ii) and (iii), we see that the total mechanical energy of the body remains same at
all positions of its motion i.e. the energy is always conserved.
*4.8.1 Potential Energy stored in a spring
Consider a spring of stiffness k
Now, if the spring is stretched or compressed by any distance x.
The spring applies force in the direction opposite to the direction in which it is
displaced from its normal position.
Force F = – kx
Also the spring aquries elastic potential energy
k, x = 0
Normal
position
Which is U = ½kx2
where k = stiffness of spring
x
x
mean
position
Illustration 7
The velocity of a body of mass 200 g changes from 36 km/h to 90 km/h. What is the work done?
Solution
Work done = change in KE = KE
= K.E.f – K.E.i

=
2 2
f i
1 1
mv mv
2 2

Vi = 36 km/h
= 36 ×
5
18
= 10 m/s
Vf = 90 ×
5
18
= 25 m/s
KE =    
2 2
1
m 25 – 10
2
 
 
=
1
2
×
200
1000
 
625–100 =
525
10
= 52.5 J
Work done = 52.5 J
Illustration 8
An object of mass 5 kg is kept at a height of 100 m. What is its potential energy? If it is dropped,
what will be its kinetic energy just before hitting the ground?
Solution
At height 100 m
P.E. = mgh = 5 × 10 × 100  5000 J
Now by law of conservation of energy.
K.E.i + P.E.i = K.E.f + P.E.f
K.E.i = 0 as u = 0
& P.E.f = 0 as h = 0
0 + P.E.i = K.E.f + 0
 0 + 5000 = K.E.f + 0
K.E.f = 5000 J
Illustration 9
A spring of stiffness k = 20 is compressed to a distance of 30 cm. What is the potential energy
stored in it?
Solution
PE of spring =
2
1
kx
2
=  
1 30 30
20
2 100 100
  
  
  
= 10 ×
3
10
×
3
10
=
9
10
J
Try yourself
7. Work done by a body is 4000 J. Its weight is 200 N. If the body is initially at rest, find the final velocity.
8. Potential energy of a body of weight 550 N is 1100 J. Find its height. It is then released. Find its velocity
just before it hits the ground & time taken to come to ground.

4.9 POWER (P)
It is usually said that one man is more powerful than the other. It means that first one has more energy
than the other. It also means that he can do the same work in lesser amount of time than the other. Thus
power is described not only in terms of work done but also in terms of time taken to do it.
“Power is defined as the rate of doing work.”
Power (P) =
work done(W)
time taken(t)
=
energy supplied
time
also W = F × s & Velocity = displacement / time  V = s/t
 P =
F × s
t
 P = F × v
 Power = applied force × velocity attained.
Unit of power =
work
time
= J/s = Watt
1 Watt = 1 J/s
Thus, 1 Watt of power is delivered when 1J work is done in 1s.
The unit of power in british engineering system is horse power, hp.
1 hp = 746 W
4.9.1 Commercial unit of energy: Kilowatt hour (kwh)
The commercial unit of energy is kilowatt hour. It is the energy consumed by a machine in 1 hour.
1 kw H = 1 kw × 1 hr = 1000 Watt × 1 hour
= 1000 J/s × 60 s × 60 s
= 3600000 J = 3.6 × 106
J
Illustration 10
A motor can do 72000 J of work in 1 hour. Calculate the power delivered?
Solution
Power =
work done
time taken
time = 1 hr = 60 × 60 seconds
Power =
72000
3600
= 20 Watt
Illustration 11
A body can do 22380 J work in 1 minute. Calculate the power delivered in watt & horse power.
Solution
We know
Power (P) =
work (W)
time (t)

P =
22380 J
1 minute
=
22380 J
60 s
= 373 Watt
Now 746 W = 1 horse power
 373 W =
1
746
× 373 = 0.5 hp
Try yourself
9. Power delivered by a motor is 72 W. Calculate the work done in 1 hour and power in hp also.
*4.10 THE EQUIVALENCES OF MASS & ENERGY
Einstien proposed a theory which relates mass to energy. He proposed that in every physical & chemical
process, the mass of an isolated system is conserved. He showed that mass & energy are equivalent &
are related by
E = mc2
C = speed of light in vacuum
= 3 × 108
m/s
This explains the huge amount of energy released in nuclear processes fission and fusion.
Even 1 gm of matter produces huge amount of energy
E = mc2
(1 × 10–3
) (3 × 108
)2
 9 × 1016
× 10–3
= 9 × 1013
J
Illustration 12
Mass of an object changes from 200 gm to 198 gm in a process. What is the amount of energy
released.
Solution
E = mc2
Mass involved in energy = 200 gm – 198 gm = 2 gm
E =
2
1000
× (3 × 108
)2
= 1.8 × 1014
J
*4.11 COLLISION
In physics we study motion.At the same time, we try to discover physical quantities, which do not change
in a physical process. The laws of momentum & energy conservation are typical examples. In this section
we shall study & apply these laws to a commonly encountered phenomenon namely collisions. Several
games such as billiards, marbles or carrom involve collisions.
Consider two bodies of masses m1
& m2
moving with velocities V1
& V2
respectively.
m1
V1
m2
V2
(Before collision)
We assume that V1
> V2
so that collision may occur.
Now let their velocities change to V1
´
& V2
´
after collisions.
m1
V ´
1
m2
V2´
(After collision)

Now by law of conservation of momentum initial momentum = final momentum.
m1
V1
+ m2
V2
= m1
V1
´ + m2
V2
´ .....(i)
 m1
V1
– m1
V1
´ = m2
V2
´ – m1
V2
 m1
(V1
– V1
´) = m2
(V2
´
– V2
) .....(ii)
Now, If the collision is perfectly elastic
then K.E.i
= K.E.f

2 2 ´2 ´2
1 1 2 2 1 1 2 2
1 1 1 1
m V m V m V m V
2 2 2 2
  
=
2
2 ´
1 1 1 1
m V – m V =
2
´ 2
2 2 2 2
m V – m V
  
2
2 ´
1 1 1
m V –V =  
2
´ 2
2 2 2
m V –V .....(iii)
Dividing(iii) by(ii)
V1
+ V1
´ = V2
+ V2
´
V1
– V2
= V2
´ – V1
´ .....(iv)
Here we see that (V1
– V2
) is the velocity with which the two objects are approaching each other &
(V2
´–V1
´) is the velocity with which the 2 objects get separated from each other.
Therefore, from equation (iv),
velocity of approach before collision = velocity of separation after collision
4.11.1 Coefficient of restitution (e)
e =
velocity of separation
velocity of approach
Note that for perfectly elastic collision
e = 1
and for perfectly inelastic collisions
e = 0
for partially elastic collisions
0 < e < 1
Now multiply equation (iv) from m2
and subtract it from equation (i) we get
1 2 2
1 1 2
1 2 1 2
m –m 2m
V´= V + V
m +m m +m
   
   
   
.....(A)
Also multiply equation (iv) from m1
& add to equation (i)
 
1 2
1
2 1 2
1 2 1 2
m –m
2m
V ´= V – V
m +m m +m
 
 
 
 
   
.....(B)
Equation Aand B give the final velocities in terms of the initial velocities & the masses

Illustration 13
Two bodies of masses 3 kg & 6 kg are moving with velocities of 30 m/s and 20 m/s in same
direction. They collide in a perfectly elastic way. Find their final velocities.
Solution
Since the bodies collide perfectly elastic
 e = 1
 KE = 0
Now momentum conservation
m1V1 + m2V2 = ´ ´
1 1 2 2
m V + m V
(3)(30) + (6)(20) = ´ ´
1 2
3V + 6V
210 = ´ ´
1 2
3V + 6V
 70 = ´ ´
1 2
V + 2V .....(i)
Also by energy conservation
2 2
2 2 ´ ´
1 1 2 2 1 1 2 2
1 1 1 1
m V + m V m V + m V
2 2 2 2

         
2 2
2 2 ´ ´
1 2
1 1 1 1
3 30 + 6 20 3 V + 6 V
2 2 2 2

900 + 800 = ´2 ´2
1 2
V 2V

1700 = ´2 ´2
1 2
V 2V
 .....(ii)
from (i) & (ii)
V1´ =
150
9
m/s; V2´ =
80
3
m/s
*****

EXERCISE-I
1. A porter is standing holding a suitcase. Is he doing any work?
2. When an arrow is released from where does it receive the kinetic energy.
3. What is the potential energy of an object of mass m at a height h above the surface of the earth.
4. What is the work done by a body moving in a circular orbit in one round trip?
5. What should the angle between the force and the displacement for work to be (i) maximum, (ii) minimum?
6. Which type of energy is possessed in a system due to its configuration?
7. A hammer of mass 1 kg falls freely on a nail placed on a piece of wood. If it falls from a height of 1 m, how
much kinetic energy will it have just before hitting the nail? (g = 10 m/s2
)
8. When is the work done by a force positive and negative? Give one example of each?
9. A porter walks on a platform while carrying a box on his head. What is the work done by him against
gravity?
10. On what factors does the potential energy of a body depend?
11. The kinetic energy of a body moving with velocity of 5m/s is 125 J. What is the mass of the body?
12. What should be done to the velocity a body in order to make its kinetic energy half of its initial value?
13. Write units of (i) work, (ii) energy and (iii) power.
14. What is the elastic potential energy? Give one example.
15. What is power? What is the difference between work and power?
16. A body of mass 5 kg is allowed to fall from a height of 10 m. Calculate the change in its potential energy.
17. To what height should book of mass 0.5 kg be lifted so that its energy changes by 1J.
EXERCISE-II
1. Define work, Give the relation between force and work. Obtain the necessary formula for work done.
2. What is mechanical energy. Deduce the formula for kinetic energy.
3. Explain the term ‘potential energy.’Deduce the expression for gravitational potential energy of a body.
4. What is meant by a conservative force? Give one example of such a force. The bob of a simple
pendulum is displaced from the mean position and is then released. Show by drawing the necessary
diagram that the law of energy conservation is valid for the swinging pendulum.
5. Find the work done in lifting 100 kg of water to a height of 10 m. (g = 10 m/s2
).
6. The initial and final velocities of a body of mass 0.2 kg are 3 m/s and 7 m/sec.Assuming its motion to be
in a straight line calculate the work done.
7. A car is moving with a uniform velocity of 54 km/h. What is the kinetic energy of a body of mass 40 kg
situated in the car?
8. A man whose mass is 56 kg climbs up steps of a stair of total height 4.5 meter in 3.5 seconds. What is
the average power required by this man?
9. Calculate the power of an engine which is capable of raising an object of 200 kg to a height of 50 m in
100s.
10. Abody of mass 25 kg is dropped from a height of 5 m. Find its (a) kinetic energy and (b) velocity, just as
it reaches the ground. (g = 10 m/s2
)
11. A gardener pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10s. Calculate
the work done and the power used by him.

12. A person whose mass is 50 kg climbs up a hill of height 100 m. Calculate the work done by him against
the gravity.
13. An object of mass 10 kg is initially at rest. What is the work required to be done to develop a velocity of
20 m/s in it.
EXERCISE-III
SECTION-A
 Fill in the blanks
1. Work done by a force acting at an angle  is W F.S. ______________.
2. Work done on a body is change in its _______________ energy.
3. Energy stored in a stretched bow is _____________ potential energy.
4. Rate of doing work is known as _______________.
5. 1 horse power = ____________ watt.
6. For _____________ collisions the vlaue of e = 0
SECTION-B
 Multiple choice question with one correct answers
1. A body undergoes a displacement of 8 m when a force of 6N is applied. The direction of force makes an
angle of 60o
with displacement. The work done by the force on the body is-
(A) 24 J (B) 24 3 J (C) 48 J (D) 24 5 J
2. If the angle between force F and displacement s is 60o
, then the work done is-
(A) Fs (B) F/s (C)
2
s
F
(D) 2Fs
3. A stone of mass falls freely through a vertical distance d, its kinetic energy is-
(A) mgd (B) ½ ma2
(C) mg/d (D) zero
4. Two bodies with kinetic energies in the ratio 4 : 1 are moving with equal liner momentum. The ratio of their
masses is
(A) 1 : 2 (B) 1 : 1 (C) 4 : 1 (D) 1 : 4
5. A mass of 12 kg at rest expleds into two pieces of masses 4 kg and 8 kg which move in opposite directions.
If the velocity of 8 kg piece is 6 m/s, then the kinetic energy of the other piece is (in joules).
(A) 64 (B) 128 (C) 144 (D) 288
6. The Power of a pump which can pump 200 kg of water to a height of 200 m in 10 s is (g = 10 m/s2
)
(A) 40 KW (B) 80 KW (C) 4000 KW (D) 960 KW
7. A mass m falls freely rest. The linear momentum after it has fallen through a height h is (g = acceleration
due to gravity)
(A) mgh (B) m 2gh (C) m gh (D) zero
8. A body of mass m1
moving with a velocity 10 ms–1
collides with another body at rest of mass m2
. After
collision the Velocities of the two bodies are 2 ms–1
and 5 ms–1
respectively along the direction of motion of
m1
. The ration
2
1
m
m
is:

(A)
12
5
(B)
8
5
(C)
5
8
(D)
5
12
9. Two bodies, of equal masses along the same straight line with velocites + 3 m/s and – 5 m/s respectively,
collide elastically. Their velocities after the collision will be respectively:
(A) 0.3 m/s and – 0.5 m/s (B) – 0.4 m/s and 0.3
(C) – 5 m/s and + 3 m/s (D) –0.3 m/s and 0.5 m/s
10. A bomb of mass m = 1 kg thrown vertically upwards with a speed u = 100 m/sec. explodes into two parts
after t = 5 sec. A fargment of mass m1
= 400 gm moves downwards with a speed v1
= 25 m/sec., then
speed v2
and direction of another mass m2
will be:
(A) 40 m/sec. downwards (B) 40 m/sec. upwards
(C) 60 m/sec. upwards (D) 100 m/sec. upwards
11. A mass of 20 kg moving with a speed of 10 m/s collides with another stationary mass of 5 kg. As a result
of the collision, the two masses stick together. The kinetic energy of the composite mass will be:
(A) 600 joule (B) 800 joule (C) 1000 joule (D) 1200 joule
12. The work done against gravity in taking 10 kg. mass at 1 m height in 1 sec. will be:
(A) 49 J (B) 98 J (C) 196 J (D) None of these
13. A heavy body moving with a velocity 20 ms–1
and another small object at rest undergo an elastic collision.
The latter will move with a velocity of:
(A) 20 m/s (B) 40 m/s (C) 60 m/s (D) zero
14. A gun of weight of 10 kg fires a shot of 0.5 g with a velocity 230 m/s. Velocity of recoil gun is:
(A) 1.51 cm/sec (B) 1.15 cm/sec(C) 1.5 cmsec (D) 1.10 cm /sec
15. A bullet of mass m and moving with velocity v is fired into a block of mass M and stick to it. The final
velocitywould be:
(A)
M
mv
(B)
m
m
M 
v (C)
M
m
m

v (D)
m
M
M

v
16. The K.E. of a body of mass 2kg and momentum of 3Ns is:
(A) 1 J (B) 2 J (C) 3 J (D) 4 J
17. A collision is said to be perfectly inelastic when :
(A) coefficient of restitution = 0 (B) coefficient of restitution = 1
(C) coefficient of restitution =  (D) coefficient of restitution < 1
18. In an explosion a body breaks up into two pieces of unequal masses. In this:
(A) Both parts will have numerically equal momentum
(B) Lighter part will have more momentum
(C) Heavier part will have mover momentum (D) Both parts will have equal kinetic energy
SECTION-C
 Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.
(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true

1. Assertion: When a body is released from height ‘h’ its KE just before hitting the ground is ½mV2
where
V is the velocity on reaching the ground.
Reason: Energy can change its form one form to another.
2. Assertion: No work is done by a coolie carrying a heavy load on his head.
Reason: Work done is the product of force & displacent in direction of force.
3. Assertion: Unit of power is Joule.
Reason: Power is rate of doing work.
SECTION-D
 Match the following (one to one)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. Only One entries of column-I may have the matching with the some entries of column-
II and one entry of column-II Only one matching with entries of column-I
1. Column I Column II
(A) mgh (P) Energy stored in spring
(B)
1
2
mV2
(Q) Potential energy
(C)
1
2
kx2
(R) Einstien equation
(D) E = mc2
(S) kinetic energy
(A) work done (P) Momentum
(B) e (Q) coefficient of restitution
(C)
work done
time taken
(R) change in kinetic energy
(D) 2 × m × KE (S) Power
EXERCISE-IV
SECTION-A
 Multiple choice question with one correct answers
1. For a body failling freely under gravity, its kinetic energy –
(A) remains constant (B) goes on increasing (C) goes on decreasing (D) zero
2. A solid wooden block resting on a frictionless surface is hit by a bullet. The bullet gets embedded. During
this process.
(A) only kinetic energy is conserved
(B) only momentum is conserved
(C) both momentum and kinetic energy are conserved
(D) neither momentum nor kinetic energy is conserved

3. A stone is projected vertically up to reach maximum height ‘h’. The ratio of its kinetic energy to potential
energy, at a height
4th
5
will be:
(A) 5 : 4 (B) 4 : 5 (C) 1 : 4 (D) 4 : 1
4. A ball is projected vertically down with an initial velocity from a height of 20 m on to a horizontal floor.
During the impact it loses 50% of its energy and rebounds to the same height, the initial velocity of its
projection is
(A) 20 ms–1
(B) 15 ms–1
(C) 10 ms–1
(D) 5 ms–1
5. A ball falling from a height of 5 m rebounds to 1.8 m height. Then the ratio of velocities of the ball after and
before rebound is:
(A)
4
5
(B)
1
5
(C)
2
5
(D)
3
5
6. A ball released form certain height which loses 50% of its K.E. on striking the ground. It will attain a height
again:
(A) one fourth the initial (B) half the initial (C) three fourth initial (D) none
7. Kinetic energies of two bodies of 1 kg and 4 kg are same. The ratio of their momenta is:
(A) 1 : 16 (B) 1 : 2 (C) 2 : 1 (D) 4 : 1
8. A scooter (m = 40 kg) having velocity 4 m/s collides with another scooter (m = 60 kg) having velocity 2 m/
s. If the collisionis inelastic, then loss in kinetic energy is:
(A) 48 J (B) 110 J (C) 392 J (D) 440 J
9. If the momentum of a certain body is increased by 50% its kinetic energy will be increased by:
(A) 25% (B) 50% (C) 100% (D) 125%
10. A body of mass 5 kg has momentum of 10 kg m/s. When a force of 0.2 N is applied on it for 10 seconds,
what is the change in its kinetic energy?
(A) 1.1 J (B) 2.2 J (C) 3.3 J (D) 4.4 J
SECTION-B
 Multiple choice question with one or more than one correct answers
1. A body of mass 50 kg is from a height 150 m. After 5 seconds
(A) PE = 12500 J (B) KE = 62500 J
(C) height = 125 m (D) momentum = 2500 kg m/s
2. A bucket of mass 20 kg (with water) is pulled to a height of 10 m in 20 seconds. Then
(A) Work done = 200 J (B) Power = 100 watt
(C) Power = 4000 watt (D) Work done = 2000 J
3. A spring of stiffness h = 100 is compressed by 2 cm. Then
(A) spring force = –200N (B) spring force = –2N (C) energy = 0.02 J (D) energy = 200 J
4. Velocity of a body of weight 98N changes from 30 m/s to 60 m/s in 20 seconds by application of an external
force. (g = 9.8 m/s)
(A) force applied F = 25N (B) change in momentum = 300 kg m/s
(C) work done = 12500 J (D) power delivered = 625 watt

5. Two bodies of mass 2 kg & 4 kg are moving with initial velocities 15 m/s & 10 m/s respectively. Let V1
&
V2
be the final velocities after collision (perfectly in elastic). Then
(A) V1
= 11.67 m/s (B) V2
= 23.33 m/s (C) e = 0 (D) KE = 16.67 J
SECTION-C
 Comprehension
Passage-1
A body of mass 2 kg moving with velocity 10 m/s callides elastically with another body of mass 2000 kg at
rest.
1. Velocity of 2 kg mass after collision
(A) 0 m/s (B) 10 m/s (C) –10 m/s (D) 20 m/s
2. Velocity of 2000 kg mass after collision
(A) 10 m/s (B) 20 m/s (C) 0 m/s (D) infinite
3. Change in momentum of first body:
(A) 40 kg m/s (B) 20 kg m/s (C) 0 kg m/s (D) datainsufficient
Passage-2
A crane lifts a car of mass 100 kg to a height 50 m in 10 seconds.
4. Work done by crane
(A) 50 kJ (B) 50 J (C) 2 J (D) 5 kJ
5. Power of crane in HP is
(A) 6.7 HP (B) 0 HP (C) 67 HP (D) 76 HP
6. If power delivered is 6700 W. The time taken is:
(A) 10seconds (B) 15seconds (C) 7.462seconds (D) datainsufficient
SECTION-D
 Match the following (one to many)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. One or more than one entries of column-I may have the matching with the same entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
(A) Circular motion (P) W = F.S. cos 
(B) Coolie carrying a load on his head (Q) Work done = 0 J
(C) Force acting at an angle ‘’ (R)  = 90º
(D) Kinetic energy (S) Depends on horizontal displacement
EXERCISE-V
1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it
that the force acts on the object through the displacement. What is the work done in this case?
2. When do we say that work is done?
3. Write an expression for the work done when a force is acting on an object in the direction of its displace-
ment.

4. Define 1 J of work.
5. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much
work is done in ploughing the length of the field?
6. What is the kinetic energy of an object?
7. Write an expression for the kinetic energy of an object.
8. The kinetic energy of an object of mass, m moving with a velocity of 5 m s–1
is 25 J. What will be its kinetic
energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three
times?
9. What is power?
10. Define 1 watt of power.
11. A lamp consumes 1000 J of electrical energy in 10 S. What is its power?
12. Define average power.
EXERCISE-VI
1. A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force. If the
force acts on the object all through the displacement, then calculate work done.
2. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 above the ground. Calculate the
work done by him on the luggage.
3. An object of mass 15 kg is moving with a uniform velocity of 4 m s–1
. What is the kinetic energy possessed
by the object?
4. What is the work to be done to increase the velocity of a car from 30 km h–1
to 60 km h–1
if the mass of the
car is 1500 kg?
5. Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given,
g = 9.8 m s–2
.
6. An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480
J, find the height at which the object is with respect to the ground. Give, g = 10 m s–2
.
7. Two girls, each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and
the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by
each girl?
8. Aboy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power.
Take g = 10 m s–2
.
9. An electric bulb of 60 W is used for 6 h per day. Calculate the ‘units’of energy consumed in one day by the
bulb.

Answers
KNOWLEDGEBASEQUESTIONS
1. (D) 2. (A) 3. (C) 4. (B) 5. (A)
6. (A) 7. (D) 8. (A)
TRYYOURSELF
1. 2000 N 2. 0 N
3. 30 m in horizontal, 15 m in vertical 4. 80 g
5. 1500 J
6. 2500 J 7. 20 m/s
8. h = 2m, v 2 10
 m/s, t 2 / 5 s
 9. 259200 J, 0.0965 hp
EXERCISE-I
11. 10 kg 12.
V
V´
2

16. 5 × 102
J 17. 0.2 m
EXERCISE-II
5. 104
J 6. 4 J
7. 4500 J 8. 720 Watt
9. 104
Watt 10. 1250 J, 10 m/s
11. 500 J, 50 Watt 12. 6 × 105
J
13. 2000 J
EXERCISE-III
SECTION-A
1. cos  2. kinetic 3. elastic 4. power 5. 746 6. inelastic
SECTION-B
1. (A) 2. (C) 3. (A) 4. (D) 5. (D)
6. (A) 7. (B) 8. (B) 9. (A) 10. (D)
11. (B) 12. (B) 13. (B) 14. (B) 15. (C)
16. (A) 17. (A) 18. (A)
SECTION-C
1. (B) 2. (A) 3. (D)

SECTION-D
1. (A–Q), (B–S), (C–P), (D–R) 2. (A–R), (B–Q), (C–S), (D–P)
EXERCISE-IV
SECTION-A
1. (B) 2. (B) 3. (C) 4. (A) 5. (D)
6. (B) 7. (B) 8. (A) 9. (D) 10. (D)
SECTION-B
1. (A,B,D) 2. (B,D) 3. (B,C) 4. (B,C,D) 5. (A,C,D)
SECTION-C
1. (C) 2. (C) 3. (A) 4. (A) 5. (A)
6. (C)
SECTION-D
1. (A)-(P,Q,R), (B)-(P,Q,R), (C)-(P,S), (D)-(none)
EXERCISE-V
1. 56 NS(J) 5. 2100 J 8. 100 J, 225 J 11. 100 W
EXERCISE-VI
1. 10 J 2. 225 J 3. 120 J 4. 156375 J
5. 588 J 6. 4 m 7. A-160 W, B-64 W 8. 375 W
9. 0.36 Units
*****

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