Tag: Chapter 10 – Work and Energy

Exercises – Chapter 10: Work and Energy

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Question 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’.
(i) Suma is swimming in a pond.
(ii) A donkey is carrying a load on its back.
(iii) A wind-mill is lifting water from a well.
(iv) A green plant is carrying out photosynthesis.
(v) An engine is pulling a train.
(vi) Food grains are getting dried in the sun.
(vii) A sailboat is moving due to wind energy.

Answer:

(i) Yes, work is done (force causes displacement).
(ii) No, work is not done (no displacement in the direction of force).
(iii) Yes, work is done (water is lifted against gravity).
(iv) No, work is not done (no mechanical displacement).
(v) Yes, work is done (train is displaced by force).
(vi) No, work is not done (no displacement due to force).
(vii) Yes, work is done (wind causes displacement of boat).

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Question 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?

Answer:
The work done by the force of gravity is zero, because the vertical displacement of the object is zero.

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Question 3

A battery lights a bulb. Describe the energy changes involved in the process.

Answer:
Chemical energy of the battery is converted into electrical energy, which is further converted into light energy and heat energy in the bulb.

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Question 4

A 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:

$$W=\frac{1}{2}m(v^2-u^2)$$

$$W=\frac{1}{2}\times 20\times (2^2-5^2)$$

$$W=10\times (4-25)=-210J$$

Work done = –210 J

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Question 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.

Answer:
The work done by the gravitational force is zero, because there is no vertical displacement of the object.

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Question 6

The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:
No, this does not violate the law of conservation of energy.
As the object falls, its potential energy decreases and an equal amount of kinetic energy increases. The total energy remains constant.

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Question 7

What are the various energy transformations that occur when you are riding a bicycle?

Answer:
Chemical energy of food → muscular energy → mechanical energy → kinetic energy of the bicycle.

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Question 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?

Answer:
No work is done on the rock as there is no displacement.
The energy spent is converted into heat energy in the muscles and surroundings.

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Question 9

A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer:

1 unit = 1 kWh = $3.6\times {10}^{6}J$

$$\text{Energy}=250\times 3.6\times {10}^6$$

$$=9.0\times {10}^{8}J$$

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Question 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.

Answer:

Potential energy at height 5 m:

$$E_p=mgh=40\times 10\times 5=2000J$$

At half-way down (height = 2.5 m):

Potential energy remaining:

$$E_p=40\times 10\times 2.5=1000J$$

Kinetic energy at that point:

$$E_k=2000-1000=1000J$$

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Question 11

What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Answer:
The work done by the force of gravity is zero.
The force of gravity acts towards the centre of the earth, while the displacement of the satellite is along the circular path. Since the force is perpendicular to the displacement, no work is done.

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Question 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.

Answer:
Yes, an object can have displacement without any force acting on it if it is moving with uniform velocity.
In such a case, no net force is required to maintain motion.

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Question 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.

Answer:
No work is done on the bundle of hay because there is no displacement of the bundle in the direction of the force.
The person gets tired due to energy spent by the muscles.

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Question 14

An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer:

$$E=P\times t$$

$$E=1500\times (10\times 3600)$$

$$E=5.4\times {10}^{7}J$$

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Question 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?

Answer:
When the pendulum bob is raised, it has potential energy.
As it swings down, potential energy is converted into kinetic energy.
During oscillation, energy keeps changing between potential and kinetic forms.

The bob eventually comes to rest due to air resistance and friction, which convert mechanical energy into heat energy.
This does not violate the law of conservation of energy because energy is only transformed, not destroyed.

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Question 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?

Answer:
Initial kinetic energy $=\frac{1}{2}m{v}^2$
Final kinetic energy $=0$

Work done $=$ change in kinetic energy

$$W=0-\frac{1}{2}m{v}^2=-\frac{1}{2}m{v}^2$$

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Question 17

Calculate the work required to be done to stop a car of mass 1500 kg moving at a velocity of 60 km h⁻¹.

Answer:

$$u=60{\text{km h}}^{-1}=\frac{50}{3}{\text{m s}}^{-1},v=0$$

$$W=\frac{1}{2}m(v^2-u^2)$$

$$W=\frac{1}{2}\times 1500(0-{\left(\frac{50}{3}\right)}^2)$$

$$W=-\frac{1}{2}\times 1500\times \frac{2500}{9}=-2.08\times {10}^{5}J(\text{approximately})$$

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Question 18

In each of the following cases a force $F$ acts on an object of mass $m$. The direction of displacement is from west to east. State whether the work done by the force is positive, negative or zero.

Answer:

• Work done is positive if the force acts from west to east.

• Work done is negative if the force acts from east to west.

• Work done is zero if the force acts perpendicular to the direction of displacement.

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Question 19

Soni says that the acceleration of an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Answer:
Yes, the acceleration can be zero if the net force acting on the object is zero, even though several forces may be acting on it.

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Question 20

Find the energy in joules consumed in 10 hours by four devices of power 500 W each.

Answer:

Total power $=4\times 500=2000W$

$$E=P\times t=2000\times (10\times 3600)$$

$$E=7.2\times {10}^{7}J$$

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Question 21

A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:
The kinetic energy of the object is converted into heat energy, sound energy, and energy used in deforming the object and the ground.

Chapter 10: Work and Energy

Page 115 – NCERT Class 9 Science

Questions with Answers

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Question 1

A force of 7 N acts on an object. The displacement is 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?

Answer

$$\text{Work done}=\text{Force}\times \text{Displacement}$$

$$W=7\times 8=56J$$

Work done = 56 joules

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Page 116 – NCERT Class 9 Science

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Question 1

When do we say that work is done?

Answer:
Work is said to be done when a force acts on an object and the object gets displaced in the direction of the force.

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Question 2

Write an expression for the work done when a force is acting on an object in the direction of its displacement.

Answer:

$$W=F\times s$$

where $F$ is force and $s$ is displacement.

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Question 3

Define 1 J of work.

Answer:
1 joule is the work done when a force of 1 newton displaces an object by 1 metre in the direction of the force.

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Question 4

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?

Answer:

Given Force = 140 N, Distance covered = 15 m

$$W=F\times s=140\times 15=2100J$$

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Page 119

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Question 1

What is the kinetic energy of an object?

Answer:
The energy possessed by an object due to its motion is called kinetic energy.

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Question 2

Write an expression for the kinetic energy of an object.

Answer:
The kinetic energy of an object of mass $m$ moving with velocity $v$ is given by:

$$E_k=\frac{1}{2}m{v}^2$$

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Question 3

The kinetic energy of an object of mass $m$ moving with a velocity of $5m{s}^{-1}$ is $25J$. 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:

Since kinetic energy is proportional to the square of velocity:

• When velocity is doubled:

$$E_k=4\times 25=100J$$

• When velocity is tripled:

$$E_k=9\times 25=225J$$

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Page 123

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Question 1

What is power?

Answer:
Power is defined as the rate of doing work or the rate of transfer of energy.

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Question 2

Define 1 watt of power.

Answer:
1 watt is the power when 1 joule of work is done in 1 second.

$$1W=1J{s}^{-1}$$

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Question 3

A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Answer:

$$P=\frac{W}{t}=\frac{1000}{10}=100W$$

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Question 4

Define average power.

Answer:
Average power is defined as the total work done divided by the total time taken.

$$\text{Average power}=\frac{\text{Total work done}}{\text{Total time taken}}$$

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