NCERT Solutions for Class 9 Science Chapter 10 Work and Energy
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Page 1
NCERT
SOLUTIONS
CLASS - 9th
aglase .co
SOLUTIONS
CLASS - 9th
aglase .co
Page 2
Book : Science Ncert Solutions | Chapter-11 Science
Class : 9th
Subject : Science
Chapter : 11
Chapter Name : WORK AND ENERGY
Q1 A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the
force (Fig. 11.3). Let us take it that the force acts on the object through the displacement.
What is the work done in this case?
Answer. When a force F acts on an object to displace it through a distance S in its
direction,
then the work done W on the body by the force is given by:
Work done = Force × Displacement
W=F×S
F=7N
S=8m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J
Page : 148 , Block Name : Questions
Q1 When do we say that work is done?
Answer. Work is done whenever the given conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body caused by the applied force along the direction of
the applied force.
Page : 149 , Block Name : Questions
Q2 Write an expression for the work done when a force is acting on an object in the
Class : 9th
Subject : Science
Chapter : 11
Chapter Name : WORK AND ENERGY
Q1 A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the
force (Fig. 11.3). Let us take it that the force acts on the object through the displacement.
What is the work done in this case?
Answer. When a force F acts on an object to displace it through a distance S in its
direction,
then the work done W on the body by the force is given by:
Work done = Force × Displacement
W=F×S
F=7N
S=8m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J
Page : 148 , Block Name : Questions
Q1 When do we say that work is done?
Answer. Work is done whenever the given conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body caused by the applied force along the direction of
the applied force.
Page : 149 , Block Name : Questions
Q2 Write an expression for the work done when a force is acting on an object in the
Page 3
Book : Science Ncert Solutions | Chapter-11 Science
direction of its displacement.
Answer. When a force F displaces a body through a distance S in the direction of the
applied force, then the work done W on the body is given by the expression:
Work done = Force × Displacement
W=F×s
Page : 149 , Block Name : Questions
Q3 De�ne 1 J of work.
Answer. 1 J is the amount of work done by a force of 1 N on an object that displaces it
through a distance of 1 m in the direction of the applied force.
Page : 149 , Block Name : Questions
Q4 A pair of bullocks exerts a force of 140 N on a plough. The �eld being ploughed is 15
m long. How much work is done in ploughing the length of the �eld?
Answer. Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W=F×d
Applied force, F = 140 N
Displacement, d = 15 m
W = 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the �eld.
Page : 149 , Block Name : Questions
Q1 What is the kinetic energy of an object?
Answer. Kinetic energy is the energy possessed by a body by the virtue of its motion.
Every moving object possesses kinetic energy. A body uses kinetic energy to do work.
Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of
air is used to run wind mills, etc.
Page : 152 , Block Name : Questions
Q2 Write an expression for the kinetic energy of an object.
direction of its displacement.
Answer. When a force F displaces a body through a distance S in the direction of the
applied force, then the work done W on the body is given by the expression:
Work done = Force × Displacement
W=F×s
Page : 149 , Block Name : Questions
Q3 De�ne 1 J of work.
Answer. 1 J is the amount of work done by a force of 1 N on an object that displaces it
through a distance of 1 m in the direction of the applied force.
Page : 149 , Block Name : Questions
Q4 A pair of bullocks exerts a force of 140 N on a plough. The �eld being ploughed is 15
m long. How much work is done in ploughing the length of the �eld?
Answer. Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W=F×d
Applied force, F = 140 N
Displacement, d = 15 m
W = 140 × 15 = 2100 J
Hence, 2100 J of work is done in ploughing the length of the �eld.
Page : 149 , Block Name : Questions
Q1 What is the kinetic energy of an object?
Answer. Kinetic energy is the energy possessed by a body by the virtue of its motion.
Every moving object possesses kinetic energy. A body uses kinetic energy to do work.
Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of
air is used to run wind mills, etc.
Page : 152 , Block Name : Questions
Q2 Write an expression for the kinetic energy of an object.
Page 4
Book : Science Ncert Solutions | Chapter-11 Science
Answer. If a body of mass m is moving with a velocity v, then its kinetic energy
EK is given by the expression,
Ek = 12 mv2
Its SI unit is Joule (J).
Page : 152 , Block Name : Questions
Q3 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?
Answer. Expression for kinetic energy is Ek = 12 mv2
m = Mass of the object
v = Velocity of the object
Given that kinetic energy, Ek=25 J
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m s−1.
Therefore, its kinetic energy becomes 4 times its original value, because it is
proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its
original value, because it is proportional to the square of the velocity. Hence, kinetic
energy = 25 × 9 = 225 J.
Page : 152 , Block Name : Questions
Q1 What is power?
Answer. Power is the rate of doing work or the rate of transfer of energy. If W is the
amount of work done in time t, then power is given by the expression,
Energy
Power = Work
Time
= Time
P= W
T
It is expressed in watt (W).
Page : 156 , Block Name : Questions
Q2 De�ne 1 watt of power
Answer. A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s,
i.e.,
1W = 1J
1s
Answer. If a body of mass m is moving with a velocity v, then its kinetic energy
EK is given by the expression,
Ek = 12 mv2
Its SI unit is Joule (J).
Page : 152 , Block Name : Questions
Q3 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?
Answer. Expression for kinetic energy is Ek = 12 mv2
m = Mass of the object
v = Velocity of the object
Given that kinetic energy, Ek=25 J
(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m s−1.
Therefore, its kinetic energy becomes 4 times its original value, because it is
proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.
(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its
original value, because it is proportional to the square of the velocity. Hence, kinetic
energy = 25 × 9 = 225 J.
Page : 152 , Block Name : Questions
Q1 What is power?
Answer. Power is the rate of doing work or the rate of transfer of energy. If W is the
amount of work done in time t, then power is given by the expression,
Energy
Power = Work
Time
= Time
P= W
T
It is expressed in watt (W).
Page : 156 , Block Name : Questions
Q2 De�ne 1 watt of power
Answer. A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s,
i.e.,
1W = 1J
1s
Page 5
Book : Science Ncert Solutions | Chapter-11 Science
Page : 156 , Block Name : Questions
Q3 A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer. Power is given by the expression,
Work done
Power = Time
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
1000
Power = 10 = 100Js−1 = 100W
Page : 156 , Block Name : Questions
Q4 De�ne average power.
Answer. A body can do different amount of work in different time intervals. Hence, it is
better to de�ne average power. Average power is obtained by dividing the total amount of
work done in the total time taken to do this work.
Total work done
Average Power= Total time taken
Page : 156 , Block Name : Questions
Q1 Look at the activities listed below. Reason out whether or not work is done in the
light of your understanding of the term ‘work’.
Suma is swimming in a pond.
A donkey is carrying a load on its back.
A wind-mill is lifting water from a well.
A green plant is carrying out photosynthesis.
An engine is pulling a train.
Food grains are getting dried in the sun.
A sailboat is moving due to wind energy.
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma
swims in the forward direction caused by the forward reaction of water. Here, the force
causes a displacement. Hence, work is done by Seema while swimming.
Page : 156 , Block Name : Questions
Q3 A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer. Power is given by the expression,
Work done
Power = Time
Work done = Energy consumed by the lamp = 1000 J
Time = 10 s
1000
Power = 10 = 100Js−1 = 100W
Page : 156 , Block Name : Questions
Q4 De�ne average power.
Answer. A body can do different amount of work in different time intervals. Hence, it is
better to de�ne average power. Average power is obtained by dividing the total amount of
work done in the total time taken to do this work.
Total work done
Average Power= Total time taken
Page : 156 , Block Name : Questions
Q1 Look at the activities listed below. Reason out whether or not work is done in the
light of your understanding of the term ‘work’.
Suma is swimming in a pond.
A donkey is carrying a load on its back.
A wind-mill is lifting water from a well.
A green plant is carrying out photosynthesis.
An engine is pulling a train.
Food grains are getting dried in the sun.
A sailboat is moving due to wind energy.
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma
swims in the forward direction caused by the forward reaction of water. Here, the force
causes a displacement. Hence, work is done by Seema while swimming.
Page 6
Book : Science Ncert Solutions | Chapter-11 Science
(b) While carrying a load, the donkey has to apply a force in the upward direction.
But, displacement of the load is in the forward direction. Since, displacement is
perpendicular to force, the work done is zero.
(c) A wind mill works against the gravitational force to lift water. Hence, work is done by
the wind mill in lifting water from the well.
(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work
done is zero.
(e) An engine applies force to pull the train. This allows the train to move in the
direction of force. Therefore, there is a displacement in the train in the same direction.
Hence, work is done by the engine on the train.
(f) Food grains do not move in the presence of solar energy. Hence, the work done is zero
during the process of food grains getting dried in the Sun.
(g)Wind energy applies a force on the sailboat to push it in the forward direction.
Therefore, there is a displacement in the boat in the direction of force. Hence, work is
done by wind on the boat.
Page : 158 , Block Name : Exercise
Q2 An object thrown at a certain angle to the ground moves in a curved path and falls
back to the ground. The initial and the �nal points of the path of the object lie on the
same horizontal line. What is the work done by the force of gravity on the object?
Answer. Work done by the force of gravity on an object depends only on vertical
displacement. Vertical displacement is given by the difference in the initial and �nal
positions/heights of the object, which is zero.
Work done by gravity is given by the expression,
W = mgh
h = Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.
Page : 158 , Block Name : Exercise
Q3 A battery lights a bulb. Describe the energy changes involved in the process.
Answer. When a bulb is connected to a battery, then the chemical energy of the battery is
transferred into electrical energy. When the bulb receives this electrical energy, then it
converts it into light and heat energy. Hence, the transformation of energy in the given
situation can be shown as:
Chemical Energy → Electrical Energy → LightEnergy + Heatenergy
(b) While carrying a load, the donkey has to apply a force in the upward direction.
But, displacement of the load is in the forward direction. Since, displacement is
perpendicular to force, the work done is zero.
(c) A wind mill works against the gravitational force to lift water. Hence, work is done by
the wind mill in lifting water from the well.
(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work
done is zero.
(e) An engine applies force to pull the train. This allows the train to move in the
direction of force. Therefore, there is a displacement in the train in the same direction.
Hence, work is done by the engine on the train.
(f) Food grains do not move in the presence of solar energy. Hence, the work done is zero
during the process of food grains getting dried in the Sun.
(g)Wind energy applies a force on the sailboat to push it in the forward direction.
Therefore, there is a displacement in the boat in the direction of force. Hence, work is
done by wind on the boat.
Page : 158 , Block Name : Exercise
Q2 An object thrown at a certain angle to the ground moves in a curved path and falls
back to the ground. The initial and the �nal points of the path of the object lie on the
same horizontal line. What is the work done by the force of gravity on the object?
Answer. Work done by the force of gravity on an object depends only on vertical
displacement. Vertical displacement is given by the difference in the initial and �nal
positions/heights of the object, which is zero.
Work done by gravity is given by the expression,
W = mgh
h = Vertical displacement = 0
W = mg × 0 = 0 J
Therefore, the work done by gravity on the given object is zero joule.
Page : 158 , Block Name : Exercise
Q3 A battery lights a bulb. Describe the energy changes involved in the process.
Answer. When a bulb is connected to a battery, then the chemical energy of the battery is
transferred into electrical energy. When the bulb receives this electrical energy, then it
converts it into light and heat energy. Hence, the transformation of energy in the given
situation can be shown as:
Chemical Energy → Electrical Energy → LightEnergy + Heatenergy
Page 7
Book : Science Ncert Solutions | Chapter-11 Science
Page : 158 , Block Name : Exercise
Q4 Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1.
Calculate the work done by the force.
Answer. Kinetic energy is given by the expression,
(Ek )v = 12 mv2
Ek= Kinetic energy of the object moving with a velocity, v
m = Mass of the object
(i) Kinetic energy when the object was moving with a velocity 5ms−1
(Ek )5 = 12 × 20 × (5)2 = 250J
(ii) Kinetic energy when the object was moving with a velocity 2ms−1
(Ek )2 = 12 × 20 × (5)2 = 40J
Work done by force is equal to the change in kinetic energy.
Therefore, work done by force = (Ek )2 − (Ek )s
= 40 - 250 = -210 J
The negative sign indicates that the force is acting in the direction opposite to the
motion of the object.
Page : 158 , Block Name : Exercise
Q5 A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A
and B is horizontal, what is the work done on the object by the gravitational force?
Explain your answer.
Answer. Work done by gravity depends only on the vertical displacement of the body. It
does not depend upon the path of the body. Therefore, work done by gravity is given by
the expression,
W = mgh
Vertical displacement, h = 0
∴W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.
Page : 158 , Block Name : Exercise
Q6 The potential energy of a freely falling object decreases progressively. Does this
violate the law of conservation of energy? Why?
Page : 158 , Block Name : Exercise
Q4 Certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1.
Calculate the work done by the force.
Answer. Kinetic energy is given by the expression,
(Ek )v = 12 mv2
Ek= Kinetic energy of the object moving with a velocity, v
m = Mass of the object
(i) Kinetic energy when the object was moving with a velocity 5ms−1
(Ek )5 = 12 × 20 × (5)2 = 250J
(ii) Kinetic energy when the object was moving with a velocity 2ms−1
(Ek )2 = 12 × 20 × (5)2 = 40J
Work done by force is equal to the change in kinetic energy.
Therefore, work done by force = (Ek )2 − (Ek )s
= 40 - 250 = -210 J
The negative sign indicates that the force is acting in the direction opposite to the
motion of the object.
Page : 158 , Block Name : Exercise
Q5 A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A
and B is horizontal, what is the work done on the object by the gravitational force?
Explain your answer.
Answer. Work done by gravity depends only on the vertical displacement of the body. It
does not depend upon the path of the body. Therefore, work done by gravity is given by
the expression,
W = mgh
Vertical displacement, h = 0
∴W = mg × 0 = 0
Hence, the work done by gravity on the body is zero.
Page : 158 , Block Name : Exercise
Q6 The potential energy of a freely falling object decreases progressively. Does this
violate the law of conservation of energy? Why?
Page 8
Book : Science Ncert Solutions | Chapter-11 Science
Answer. No. The process does not violate the law of conservation of energy. This is
because when the body falls from a height, then its potential energy changes into kinetic
energy progressively. A decrease in the potential energy is equal to an increase in the
kinetic energy of the body. During the process, total mechanical energy of the body
remains conserved. Therefore, the law of conservation of energy is not violated.
Page : 158 , Block Name : Exercise
Q7 What are the various energy transformations that occur when you are riding a
bicycle?
Answer. While riding a bicycle, the muscular energy of the rider gets transferred into
heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic
energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular Energy = Kinetic Energy + Heat Energy
During the transformation, the total energy remains conserved.
Page : 158 , Block Name : Exercise
Q8 Does the transfer of energy take place when you push a huge rock with all your might
and fail to move it? Where is the energy you spend going?
Answer. When we push a huge rock, there is no transfer of muscular energy to the
stationary rock. Also, there is no loss of energy because muscular energy is transferred
into heat energy, which causes our body to become hot.
Page : 158 , Block Name : Exercise
Q9 A certain household has consumed 250 units of energy during a month. How much
energy is this in joules?
Answer.
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1kWh
1kWh = 3.6 × 106 J
Therefore, 250 units of energy = 250 × 3.6 × 106 = 9 × 108 J
Page : 158 , Block Name : Exercise
Q10 An object of mass 40 kg is raised to a height of 5 m above the ground. What is its
Answer. No. The process does not violate the law of conservation of energy. This is
because when the body falls from a height, then its potential energy changes into kinetic
energy progressively. A decrease in the potential energy is equal to an increase in the
kinetic energy of the body. During the process, total mechanical energy of the body
remains conserved. Therefore, the law of conservation of energy is not violated.
Page : 158 , Block Name : Exercise
Q7 What are the various energy transformations that occur when you are riding a
bicycle?
Answer. While riding a bicycle, the muscular energy of the rider gets transferred into
heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic
energy provides a velocity to the bicycle. The transformation can be shown as:
Muscular Energy = Kinetic Energy + Heat Energy
During the transformation, the total energy remains conserved.
Page : 158 , Block Name : Exercise
Q8 Does the transfer of energy take place when you push a huge rock with all your might
and fail to move it? Where is the energy you spend going?
Answer. When we push a huge rock, there is no transfer of muscular energy to the
stationary rock. Also, there is no loss of energy because muscular energy is transferred
into heat energy, which causes our body to become hot.
Page : 158 , Block Name : Exercise
Q9 A certain household has consumed 250 units of energy during a month. How much
energy is this in joules?
Answer.
1 unit of energy is equal to 1 kilowatt hour (kWh).
1 unit = 1kWh
1kWh = 3.6 × 106 J
Therefore, 250 units of energy = 250 × 3.6 × 106 = 9 × 108 J
Page : 158 , Block Name : Exercise
Q10 An object of mass 40 kg is raised to a height of 5 m above the ground. What is its
Page 9
Book : Science Ncert Solutions | Chapter-11 Science
potential energy? If the object is allowed to fall, �nd its kinetic energy when it is half-
way down.
Answer.
Gravitational potential energy is given by the expression,
W = mgh
h = Vertical displacement = 5m
m = Mass of the object = 40kg
g = Acceleration due to gravity = 9.8ms−2
∴ W = 40 × 5 × 9.8 = 1960J
Page : 158 , Block Name : Exercise
Q11 What is the work done by the force of gravity on a satellite moving round the earth?
Justify your answer.
Answer. At half-way down, the potential energy of the object will be
1960/2 = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due
to the law of conservation of energy. Hence, half-way down, the kinetic energy of the
object will be 980 J.
Page : 158 , Block Name : Exercise
Q12 Can there be displacement of an object in the absence of any force acting on it?
Think. Discuss this question with your friends and teacher.
Answer. Yes. For a uniformly moving object.Suppose an object is moving with constant
velocity. The net force acting on it is zero. But, there is a displacement along the motion
of the object. Hence, there can be a displacement without a force.
Page : 158 , Block Name : Exercise
Q13 A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he
done some work or not? Justify your answer
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
potential energy? If the object is allowed to fall, �nd its kinetic energy when it is half-
way down.
Answer.
Gravitational potential energy is given by the expression,
W = mgh
h = Vertical displacement = 5m
m = Mass of the object = 40kg
g = Acceleration due to gravity = 9.8ms−2
∴ W = 40 × 5 × 9.8 = 1960J
Page : 158 , Block Name : Exercise
Q11 What is the work done by the force of gravity on a satellite moving round the earth?
Justify your answer.
Answer. At half-way down, the potential energy of the object will be
1960/2 = 980 J.
At this point, the object has an equal amount of potential and kinetic energy. This is due
to the law of conservation of energy. Hence, half-way down, the kinetic energy of the
object will be 980 J.
Page : 158 , Block Name : Exercise
Q12 Can there be displacement of an object in the absence of any force acting on it?
Think. Discuss this question with your friends and teacher.
Answer. Yes. For a uniformly moving object.Suppose an object is moving with constant
velocity. The net force acting on it is zero. But, there is a displacement along the motion
of the object. Hence, there can be a displacement without a force.
Page : 158 , Block Name : Exercise
Q13 A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he
done some work or not? Justify your answer
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
Page 10
Book : Science Ncert Solutions | Chapter-11 Science
When a person holds a bundle of hay over his head, then there is no displacement in the
bundle of hay. Although, force of gravity is acting on the bundle, the person is not
applying any force on it. Hence, in the absence of force, work done by the person on the
bundle is zero.
Page : 159 , Block Name : Exercise
Q14 An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer. Energy consumed by an electric heater can be obtained with the help of the
expression,
P= W
T
Power rating of the heater, P = 1500 W = 1.5 kW
Time for which the heater has operated, T = 10 h
Work done = Energy consumed by the heater
Therefore, energy consumed = Power × Time
= 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.
Page : 159 , Block Name : Exercise
Q15 Illustrate the law of conservation of energy by discussing the energy changes which
occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the
bob eventually come to rest? What happens to its energy eventually? Is it a violation of
the law of conservation of energy?
Answer. The law of conservation of energy states that energy can be neither created nor
destroyed. It can only be converted from one form to another. Consider the case of an
oscillating pendulum.
When a pendulum moves from its mean position P to either of its extreme positions A or
B, it rises through a height h above the mean level P. At this point, the kinetic energy of
When a person holds a bundle of hay over his head, then there is no displacement in the
bundle of hay. Although, force of gravity is acting on the bundle, the person is not
applying any force on it. Hence, in the absence of force, work done by the person on the
bundle is zero.
Page : 159 , Block Name : Exercise
Q14 An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer. Energy consumed by an electric heater can be obtained with the help of the
expression,
P= W
T
Power rating of the heater, P = 1500 W = 1.5 kW
Time for which the heater has operated, T = 10 h
Work done = Energy consumed by the heater
Therefore, energy consumed = Power × Time
= 1.5 × 10 = 15 kWh
Hence, the energy consumed by the heater in 10 h is 15 kWh.
Page : 159 , Block Name : Exercise
Q15 Illustrate the law of conservation of energy by discussing the energy changes which
occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the
bob eventually come to rest? What happens to its energy eventually? Is it a violation of
the law of conservation of energy?
Answer. The law of conservation of energy states that energy can be neither created nor
destroyed. It can only be converted from one form to another. Consider the case of an
oscillating pendulum.
When a pendulum moves from its mean position P to either of its extreme positions A or
B, it rises through a height h above the mean level P. At this point, the kinetic energy of
Page 11
Book : Science Ncert Solutions | Chapter-11 Science
the bob changes completely into potential energy. The kinetic energy becomes zero, and
the bob possesses only potential energy. As it moves towards point P, its potential energy
decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches
point P, its potential energy becomes zero and the bob possesses only kinetic energy.
This process is repeated as long as the pendulum oscillates.
The bob does not oscillate forever. It comes to rest because air resistance resists its
motion. The pendulum loses its kinetic energy to overcome this friction and stops after
some time.
The law of conservation of energy is not violated because the energy lost by the
pendulum to overcome friction is gained by its surroundings. Hence, the total energy of
the pendulum and the surrounding system remain conserved.
Page : 159 , Block Name : Exercise
Q16 An object of mass, m is moving with a constant velocity, v. How much work should
be done on the object in order to bring the object to rest?
Answer. Kinetic energy of an object of mass, m moving with a velocity, v is given by the
expression,
Ek = 12 mv2
To bring the object to rest, 12 mv2 amount of work is required to be done on the object.
Page : 159 , Block Name : Exercise
Q17 Calculate the work required to be done to stop a car of 1500 kg moving at a velocity
of 60 km/h?
Answer. Kinetic energy, Ek = 12 mv2
Mass of car, m =1500Kg
5
Velocity of car. V = 60 km/h = 60 × 18 ms−1
2
∴ Ek = 12 × 1500 × (60 × 18
5
) = 20.8 × 104 J
Hence, 20.8 × 104 J of work is required to stop the car.
Page : 159 , Block Name : Exercise
Q18 In each of the following a force, F is acting on an object of mass, m. The direction of
displacement is from west to east shown by the longer arrow. Observe the diagrams
carefully and state whether the work done by the force is negative, positive or zero.
the bob changes completely into potential energy. The kinetic energy becomes zero, and
the bob possesses only potential energy. As it moves towards point P, its potential energy
decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches
point P, its potential energy becomes zero and the bob possesses only kinetic energy.
This process is repeated as long as the pendulum oscillates.
The bob does not oscillate forever. It comes to rest because air resistance resists its
motion. The pendulum loses its kinetic energy to overcome this friction and stops after
some time.
The law of conservation of energy is not violated because the energy lost by the
pendulum to overcome friction is gained by its surroundings. Hence, the total energy of
the pendulum and the surrounding system remain conserved.
Page : 159 , Block Name : Exercise
Q16 An object of mass, m is moving with a constant velocity, v. How much work should
be done on the object in order to bring the object to rest?
Answer. Kinetic energy of an object of mass, m moving with a velocity, v is given by the
expression,
Ek = 12 mv2
To bring the object to rest, 12 mv2 amount of work is required to be done on the object.
Page : 159 , Block Name : Exercise
Q17 Calculate the work required to be done to stop a car of 1500 kg moving at a velocity
of 60 km/h?
Answer. Kinetic energy, Ek = 12 mv2
Mass of car, m =1500Kg
5
Velocity of car. V = 60 km/h = 60 × 18 ms−1
2
∴ Ek = 12 × 1500 × (60 × 18
5
) = 20.8 × 104 J
Hence, 20.8 × 104 J of work is required to stop the car.
Page : 159 , Block Name : Exercise
Q18 In each of the following a force, F is acting on an object of mass, m. The direction of
displacement is from west to east shown by the longer arrow. Observe the diagrams
carefully and state whether the work done by the force is negative, positive or zero.
Page 12
Book : Science Ncert Solutions | Chapter-11 Science
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
In this case, the direction of force acting on the block is perpendicular to the
displacement. Therefore, work done by force on the block will be zero.
In this case, the direction of force acting on the block is in the direction of displacement.
Therefore, work done by force on the block will be positive.
In this case, the direction of force acting on the block is opposite to the direction of
displacement. Therefore, work done by force on the block will be negative.
Page : 159 , Block Name : Exercise
Q19 Soni says that the acceleration in an object could be zero even when several forces
are acting on it. Do you agree with her? Why?
Answer. Acceleration in an object could be zero even when several forces are acting on it.
This happens when all the forces cancel out each other i.e., the net force acting on the
object is zero. For a uniformly moving object, the net force acting on the object is zero.
Answer. Work is done whenever the given two conditions are satis�ed:
(i) A force acts on the body.
(ii) There is a displacement of the body by the application of force in or opposite to the
direction of force.
In this case, the direction of force acting on the block is perpendicular to the
displacement. Therefore, work done by force on the block will be zero.
In this case, the direction of force acting on the block is in the direction of displacement.
Therefore, work done by force on the block will be positive.
In this case, the direction of force acting on the block is opposite to the direction of
displacement. Therefore, work done by force on the block will be negative.
Page : 159 , Block Name : Exercise
Q19 Soni says that the acceleration in an object could be zero even when several forces
are acting on it. Do you agree with her? Why?
Answer. Acceleration in an object could be zero even when several forces are acting on it.
This happens when all the forces cancel out each other i.e., the net force acting on the
object is zero. For a uniformly moving object, the net force acting on the object is zero.
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Book : Science Ncert Solutions | Chapter-11 Science
Hence, the acceleration of the object is zero. Hence, Soni is right.
Page : 159 , Block Name : Exercise
Q20 Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer. Energy consumed by an electric device can be obtained with the help of the
expression for power,
P= W
T
Power rating of the device, P = 500 W = 0.50 kW
Time for which the device runs, T = 10 h
Work done = Energy consumed by the device
Therefore, energy consumed = Power × Time
= 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20
kWh = 20 Units.
Page : 159 , Block Name : Exercise
Q21 A freely falling object eventually stops on reaching the ground. What happenes to
its kinetic energy?
Answer. When an object falls freely towards the ground, its potential energy decreases
and kinetic energy increases. As the object touches the ground, all its potential energy
gets converted into kinetic energy. As the object hits the hard ground, all its kinetic
energy gets converted into heat energy and sound energy. It can also deform the ground
depending upon the nature of the ground and the amount of kinetic energy possessed by
the object.
Page : 159 , Block Name : Exercise
Hence, the acceleration of the object is zero. Hence, Soni is right.
Page : 159 , Block Name : Exercise
Q20 Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer. Energy consumed by an electric device can be obtained with the help of the
expression for power,
P= W
T
Power rating of the device, P = 500 W = 0.50 kW
Time for which the device runs, T = 10 h
Work done = Energy consumed by the device
Therefore, energy consumed = Power × Time
= 0.50 × 10 = 5 kWh
Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20
kWh = 20 Units.
Page : 159 , Block Name : Exercise
Q21 A freely falling object eventually stops on reaching the ground. What happenes to
its kinetic energy?
Answer. When an object falls freely towards the ground, its potential energy decreases
and kinetic energy increases. As the object touches the ground, all its potential energy
gets converted into kinetic energy. As the object hits the hard ground, all its kinetic
energy gets converted into heat energy and sound energy. It can also deform the ground
depending upon the nature of the ground and the amount of kinetic energy possessed by
the object.
Page : 159 , Block Name : Exercise
Document Details
| Board / Org | NCERT |
| Exam | Class 9 |
| Type | Solution |
| Pages | 13 |
| Updated | 22 Jul 2026 |