NCERT Solutions for Class 9 Science Chapter 10 :- Work and Energy :- PDF
In this article, we will provide you with NCERTExerciseSolutions for Class 9 Science Physics Chapter 10 Work and Energy to help students in their CBSE exam preparation. This solution provides them with answers to the Exercise questions provided in the NCERT Class 9 textbooks.
| Subject | Science (Physics) |
|---|---|
| Class | 9 |
| Chapter No. | 10 |
| Chapter Name | Work and Energy |
| Type | Exercise Solution |
Class 9 Science Ch 9 :- Work and Energy Exercise Solution :- PDF
WORK AND ENERGY EXERCISE QUESTION ANSWER
Ques 1 :- Look at the
activities listed below. Reason out whether or not work is done in the light of
your understanding of the term ‘work’.
Ans :- Work is done whenever the given two conditions are satisfied :-
(i) A force acts on the body.
(ii) There is a displacement of the body by applying force in or opposite to the direction of force.
(a) Suma is swimming in a pond.
--- Yes
work is being done in this case but the work done is negative since the
force is being applied backwards but the displacement (motion of Suma) is in
the forward direction.
Work
done = Force X Displacement
W
= - (F x S) (Negative sign because Force and displacement is in opposite
direction)
(b) A donkey is carrying a
load on its back.
--- 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 is lifting
water from a well.
Yes
work is being done in this case because the force is being
applied upwards and the water is also being lifted upward.
(d) A green plant is carrying out photosynthesis.
--- No force is required when a
green plant is carrying out photosynthesis. The plant does not exert any force to move. Since there is no displacement or force. Hence no work is done.
(e) An engine is pulling a train.
-- When an engine is pulling a train, it is applying a force in the forward direction. So, it is moving in the forward direction. Since displacement and force are in the same direction. Hence work is done by the engine.
(f) Food grains are getting
dried in the sun.
--- There is no force involved in the process of drying food grains in the sun and the grains do not move. Since there is no force or displacement. Hence no work is done.
(g) A sailboat is moving due to wind energy.
--- When a sailboat is moving due
to wind energy it is applying force in the forward direction. So it is moving
in the forward direction. Since displacement and force are in the same
direction. Hence work is done.
Ques 2 :- 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 final 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?
Ans:- 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 final positions or
heights of the object which is zero.
Work
done against gravity = m x g x h
Where h = Vertical
displacement = 0
∴ W = m x
g × 0 =
0 Joule
Therefore the work done by gravity on the given object is zero joule.
The
mass and acceleration due to gravity remains constant but the height becomes
zero since the initial and final points of the path of the object lie on the
same horizontal line.
Ques 3 :- A battery lights a bulb. Describe the energy
changes involved in the process.
Ans: - When
a battery lights a bulb, first energy transformation takes place in the battery
and then in the bulb.
In the battery, the chemical energy gets converted into electrical energy.
Battery :- Chemical energy ----> Electrical energy
In the bulb, the electrical energy from the bulb gets converted into heat energy and then into light energy
Bulb :- Electrical energy ----> Heat energy ----> Light energy
So, Energy Changes are
Chemical
energy ----> Electrical energy ----> Heat energy ---> Light energy
Ques 4 :- Certain force
acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done
by the force.
Ans: - Given Mass of the body = 20 kg
Initial velocity (u) = 5 ms-1
Final velocity (v) = 2 ms-1
The initial kinetic energy
Ei = 1/2 mu2 ===> 1/2 × 20 × (5)2
= 10 × 25 ===> 250 J
Final kinetic energy
Ef = 1/2 mv2 ===> 1/2 × 20 × (2)2
= 10 × 4 ===> 40 J
∴ Work done = Change in kinetic energy
∴ Work done = Ef – Ei
= Work done = 40 J – 250 J
= Work done = - 210 J
The negative sign indicates
that the force is acting in the direction opposite to the motion of the object.
OR
Given Mass of the body = 20 kg
Initial velocity = 5 ms-1
Final velocity = 2 ms-1
∴ Work done = change in kinetic energy
∴ Work done = 1/2 mv2 - 1/2 mu2
Work done = 1/2 m ( v2 - u2 )
= 1/2 x 20 ( 22 - 52 ) ===> 10 (4 - 25)
= 10 x (-21) = - 210 J
= Work done = - 210 J
The negative sign indicates
that the force is acting in the direction opposite to the motion of the object.
Ques 5 :- 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.
Ans :- Work done by gravity depends only on the vertical displacement of the body.
work done by gravity is given
by the expression,
W = m x g x h
Where Vertical displacement (h)
= 0
∴ W = m x g × 0 = 0
Therefore the work done on
the body by gravity is zero.
Ques 6: - The
potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Ans :- 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 (potential energy + kinetic energy) of the body remains conserved. Therefore, the law of conservation of energy is not violated.
Ques 7: - What are the various energy transformations that occur when you are riding a bicycle?
Ans :- 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
During the transformation the total energy remains conserved.
Ques 8 :- 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?
Ans :- When we push a huge rock there is no transfer of muscular energy to the stationary rock as there is no displacement takes place in the direction of applied force. Also there is no loss of energy since muscular energy is transferred into heat energy which causes our body to become hot.
Ques 9 :- A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Ans :- Given Energy consumed in one month = 250 unit
1 unit = 1 kWh
∴ 250 units = 250 kWh
= 250 k x W × h
==> 250 × 1000 W × 3600 s
==> 25 x 36 x 10 × 1000 × 100 J
( J = W X s )
= 900 x 10 × 1000 × 100 J
==> 9 x 100 x 10 × 1000 × 100 J
= 9 × 108 J.
Ques 10 :- An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Ans: - Given Mass (m) = 40 kg
==> Acceleration due to gravity (g)= 10 ms-2
Height (h)= 5m
==> Potential energy= m × g× h
P.E = 40 × 10 × 5 = 2000 J
==> Potential energy = 2000 J (joules)
At a height of 5 metres, the
object has a potential energy of 2000 J.
Now , when the object is
half-way down
Height (h)=5/2 m = 2.5 m.
==> Height (h) =2.5m. P.E at Halfway down= m× g × h
P.E= 40 × 10 × 2.5= 1000 J [h= 2.5 m]
Potential Energy halfway down= 1000 joules.
According to the law of conservation of energy:
Total potential energy at this point = potential energy halfway down + kinetic energy halfway down
2000 = 1000 + K.E halfway down
==> K.E at halfway down = 2000- 1000 = 1000 J
Kinetic energy at halfway down= 1000 joules.
Ques 11: - What is the work
done by the force of gravity on a satellite moving round the earth? Justify
your answer.
Ans: - The work done on a satellite moving round the earth is zero.
--- When the satellite moves in a circular path, then the centripetal force acts towards the centre along the radius, and the direction of motion is tangential to the circle. The two are thus perpendicular to each other So,
Work done = F x S x cos0
W = FS cos 900 = 0
So in the case of uniform circular motion the work done is zero.
Ques 12: - 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.
Ans: - We know that,
Force = mass x acceleration
==> F = m x a
In the absence of force, F = 0,
then ma = 0
If a = 0, the object is either at rest or in a state of uniform motion in a straight line.
In case the object is moving in a straight line, there must be displacement.
So, in the absence of force there may be displacement in the object.
Ques 13 :- 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.
Ans :- Work is done when
--- Force is applied
--- Displacement takes place
--- The angle between the force and displacement is not 900 .
When a person holds a bundle of hay over his head, he applies force to hold it but there is no displacement.
Work done = Force (F) x displacement (s)
Work done = Force (F) x 0
==> Work done = 0
Since the displacement is zero, no work is done.
Ques 14 :- An electric heater is rated 1500 W. How much energy does it use in 10 hours ?
Ans :- Given, Power of the heater = 1500 W
==> 1500 /1000 ==> 1.5 kw
Time taken = 10 hours
Energy consumed by an electric heater can be obtained with the help of the expression,
==> Power = Energy consumed / Time taken
==> Energy consumed = Power x Time taken
Energy consumed = 1.5 kw x 10 h = 15 kWh
Energy consumed = 15 kWh
Therefore, the energy consumed by the heater in 10 hours is 15 kWh.
Ques 15 :- 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?
Ans :- In a pendulum as it swings energy alternates between kinetic energy (KE) and potential energy (PE):
-- At extreme positions (A or B) :- The pendulum rises to a height h,
where its KE is zero, and it has maximum PE.
--
At
the mean position (P) :- The pendulum is at its lowest point with
maximum speed.
Here, PE is zero and it has maximum KE.
-- This energy exchange continues during oscillation.
-- However due to air resistance, the pendulum loses KE which is converted to heat in the surroundings. Eventually the pendulum stops.
-- The law of conservation of energy holds because the total energy
(pendulum + surroundings) remains constant, even though the pendulum's energy decreases.
Ques 16 :- 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 ?
Ans :- Given
, Mass of the object = m
==> Velocity of the object = v
∴ Initial Kinetic Energy (Ei) = ½ mu2
Since the object finally comes to rest v = 0
∴ Final Kinetic Energy (Ef) = ½ mv2
Final Kinetic Energy (Ef) = 0
Hence, Work Done = Change in Kinetic Energy
Work Done = Ef - Ei
Work Done = ½ mv2 - 0
Work Done = ½ mv2
∴ Work done on the object in order to bring the object to rest = ½ mv2
Ques 17 :- Calculate the work
required to be done to stop a car of 1500 kg moving at a velocity of 60 km h-1
.
Ans :- Given , Mass of the car m = 1500 kg
==> Velocity of the car v = 60 km h-1
Velocity of the car v = 60 x 1000 / 60 x 60 = 100 / 60 = 50 / 3
Velocity of the car v = 50 / 3 ms-1
The car is in motion, so its energy = kinetic energy = 1/2 mv2
= 208333.3
The kinetic energy of the car, when it comes to rest = 0 J
Work done on object = change in kinetic energy
Work done on object = 208333.3 - 0 J
Work done on object = 208333.3 J
Hence the work required to stop the car is 208333.3 J.
Ques 18 :- 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.
Ans :- Work is done when
--- Force is applied
--- Displacement takes place
--- The angle between the force and displacement is not 900 .
Case l : - F = Force , D = Displacement
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.
Case II :-
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.
Case III :-
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.
Ques 19 :- 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?
Ans :- 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. Hence, the acceleration of the object is zero. Hence Soni is right.
We know that , Force = Mass x Acceleration
If Net Force = 0 Then, Acceleration = 0
Ques 20: - Find the energy in
kW h consumed in 10 hours by four devices of power 500 W each.
Ans: - Given, Power rating of the
device (P) = 500 W = 500 /1000 kW = 0.50 kW
Time for which the device runs
(T) = 10 h
Energy consumed by an electric
device can be obtained by the expression
Power = Energy consumed / Time
taken
Energy consumed = Power ×
Time
Energy consumed = 0.50 kw × 10 h = 5 kWh
Thus, the energy consumed by four equal rating devices in 10 h will be = 4 × 5 kWh = 20 kWh
Ques 21 :- A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy ?
Ans :- When a freely falling body falls, its potential energy is converted into kinetic energy.
As it hits the ground, Velocity (v) = 0
K.E = ½ mv2 ==> K.E = 1/2 x m x 0
K.E = 0
Its kinetic energy is converted into :-
--- Heat Energy :- Heat is generated in the object and the ground
--- Sound Energy :- Sound is produced when object falls into the ground
--- Transferred to the ground/earth :- It can also deform the ground and transfer some energy to the ground .
0 Comments
Please do not enter any spam link in the comment box.