Class IX Science

Chapter - 11 Work and Energy

NCERT Solutions for Class 9 Science

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In page question - Page 148
Q 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?

Class IX Science chapter 11 work and energy 1

Ans.

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
Where,
F = 7 N
S = 8 m
Therefore, work done, W = 7 × 8
= 56 Nm
= 56 J

In page question - Page 149
Q 1.

When do we say that work is done?

Ans.

Work is done whenever the given conditions are satisfied:

  • A force acts on the body.
  • There is a displacement of the body caused by the applied force along the direction of the applied force.
In page question - Page 149
Q 2.

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

Ans.

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

In page question - Page 149
Q 3.

Define 1 J of work.

Ans.

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. 80

In page question - Page 149
Q 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?

Ans.

Work done by the bullocks is given by the expression:
Work done = Force × Displacement
W = F × d
Where,
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 field.

In page question - Page 152
Q 1.

What is the kinetic energy of an object?

Ans.

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.

In page question - Page 152
Q 2.

Write an expression for the kinetic energy of an object.

Ans.

If a body mass m is moving with a velocity v, then its kinetic energy Ek is given by the expression,
Ek= 1⁄2 mv2.

Its SI unit is Joule (J).

In page question - Page 152
Q 3.

The kinetic energy of an object of mass, m moving with a velocity of 5 m/s 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?

Ans.

Expression for kinetic energy
Ek= 1⁄2 mv2
m = Mass of object
v = Velocity of the object in m/s
Given that kinetic energy, Ek=25J


(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m/s.

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.80

In page question - Page 156
Q 1.

What is power?

Ans.

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,
Power = Work/Time = Energy/Time
P=W/T
It is expressed in watt (W).

In page question - Page 156
Q 2.

Define 1 watt of power:

Ans.

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

In page question - Page 156
Q 3.

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

Ans.

Power is given by the expression,
Power=Work done /Time
Work done = Energy consumed by the lamp = 1000 J
Power=1000 / 10 = 100 Js−1
= 100 W

In page question - Page 156
Q 4.

Define average power.

Ans.

A body can do different amount of work in different time intervals. Hence, it is better to find average power. Average power is obtained by dividing the total amount of work done in the total time taken to do this work.

Average Power =Total work done / Total time taken

Chapter end exercise
Q 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’.

  • 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.
Ans.
  • Suma is swimming in a pond.
    Suma is doing work as she is able to move herself by applying force with the movement of her arms and legs in the water.
  • A donkey is carrying a load on its back.
    Donkey is not doing any work (in the sense of physics) as the weight he is carrying (the direction of force) and displacement are perpendicular to each other.
  • A wind-mill is lifting water from a well.
    Wind mill is lifting water from a well and doing work against the gravity.
    A green plant is carrying out photosynthesis.
    No force and displacement are present here, so work done is zero.
  • An engine is pulling a train.
    During the pulling a train, engine does the work against the friction, present between the railway track and wheels.
  • Food grains are getting dried in the sun.
    During the drying the grains, there is no force as well as displacement is present. So, no work is done.
  • A sailboat is moving due to wind energy.
    Work is done by the wind as it moves the sailboat towards the direction of
    the force (force of blowing air).
Chapter end exercise
Q 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.

When the object moves upwards, the work done by gravity is negative (as the direction of gravitational force is towards the Earth’s center) and when the object come downwards, there is a positive work done. So, the total work down is zero in throughout the motion.

Chapter end exercise
Q 3.

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

Ans.

Battery converts chemical energy into electrical energy. This electrical energy is further converted into light and heat energy.

Chapter end exercise
Q 4.

Certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. Calculate the work done by the force.

Ans.

Mass of the body = 20 kg
Initial velocity = 5 m/s
Final velocity = 2 m/s
We know that,
Work done = change in kinetic energy = 12mu2-12mv2

12mu2 - v2

12×20×52 - 22
= 10(25 − 4)
= 210 J

Chapter end exercise
Q 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.

The work down by the gravitational force acting on the body is zero because the direction of force is vertically downward and the displacement is horizontal i.e. force and displacement are perpendicular to each other.

Chapter end exercise
Q 6.

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

Ans.

The potential energy of freely falling object decreases and its kinetic energy increases (as its velocity increases) progressively. So, in this way the total mechanical energy (Kinetic energy + potential energy) remains constant. Thus, the law of conservation of energy is not violated.

Chapter end exercise
Q 7.

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

Ans.

The muscular energy of the cyclist is converted into kinetic (rotational) energy of wheels of cycle which is further converted into kinetic energy to run the bicycle.

Chapter end exercise
Q 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 the rock and fail to move it. Some of our energy is absorbed by the rock in the form of potential energy and the rest of our energy is goes to environment through our muscles and the surface between the rock and out hand.

Chapter end exercise
Q 9.

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

Ans.

We know that

1 unit = 3,600,000 J

So,   250 units = 250 × 3,600,000 J
= 900,000,000 J
= 9 × 108J
Hence, the energy consumed = 9 × 108 J

Chapter end exercise
Q 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.

We know that, potential energy = mgh
Where, m = 40 kg
g = 9.8 m/s2
h = 5 m
So, the potential energy = 40 × 9.8 × 5 J = 1960 J

According to law of conservation of energy, the total mechanical energy (Kinetic and potential energy) of an object remains constant.

Therefore, when the object is half-way down, its potential energy become half the original energy and remaining half converted into kinetic energy.

Hence, the kinetic energy = 1⁄2 (1960) J = 980 J

Chapter end exercise
Q 11.

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

Ans.

When a satellite moves around the Earth, the displacement in short interval is along the tangential direction and the force (gravitational force) is towards the centre of the Earth. Since, the force and displacement are perpendicular to each other, the work done by gravitational force is zero.

Chapter end exercise
Q 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.

Yes, it is true. There may be displacement in the absence of force.
We know that, F = ma,
In the absence of force, F = 0, then ma = 0

⇒ a = 0 [as m ≠ 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.

Chapter end exercise
Q 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.

The person holding a bundle of hay get tired because his muscular energy is converting into thermal energy. There is no displacement at all, so he had no work. As Work done = Force × displacement.

Chapter end exercise
Q 14.

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

Ans.

We know that, Energy = Power × time
Here, Power = 1500 W

Time = 10 hours = 10 × 60 × 60 seconds = 36000 seconds

Therefore, The energy used by heater = Power × time

= 1500 × 36000 J
= 54000000 J
= 5.4 × 107 J

Chapter end exercise
Q 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.

The law of conservation of energy states that the total energy of an isolated system remains constant and that energy can neither be created nor destroyed; rather, it can transform from one form to another.

When pendulum bob moves from its rest position (mean position) to another extreme position its kinetic energy will be converted into potential energy when it is raised at some height 'h'. Now when it come back to its mean position, then its potential energy will be converted into kinetic energy and this phenomenon will take place again and again.
 
The bob will eventually come to rest due to the frictional resistance offered by air on the surface of bob and pendulum loses its kinetic energy to overcome this friction and finally comes to rest.
 
The law of conservation of energy is not violated because the kinetic energy loss by pendulum to overcome the friction is gained by surrounding, So total energy of system will remain conserved.

 

Chapter end exercise
Q 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.

The object is in motion, so its energy = kinetic energy = 12mv2
The kinetic energy of the object, when it comes to rest = 0.
Work done on object = change in kinetic energy

 12mv2 - 0=12mv2

Chapter end exercise
Q 17.

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Ans.

Mass of the car m = 1500 kg
Velocity of the car v = 60 km/h

=60 × 100060 × 60 = 503m/s

The car is in motion, so its energy = kinetic energy = 12mv2
1215005032J

12150025009J

= 208333.3 J
The kinetic energy of the car, when it comes to rest = 0 J
Work done on object = change in kinetic energy
= 208333.3 − 0 J
= 208333.3 J

Hence, the work required to stop the car is 208333.3 J.

Chapter end exercise
Q 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.

Class IX Science chapter 11 work and energy 2

Ans.

In the first case, the force and displacement are perpendicular to each other, so work done is zero.

In the second case, the force and displacement are in the same direction, so the work done is positive.

In the third case, the force and displacement are in the opposite direction, so the work done is negative.

Chapter end exercise
Q 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.

If the resultant force acting on a body in different directions is zero, then the acceleration will be zero.
We know that, F = ma,
In the net force is zero, F = 0, then ma = 0

⇒ a = 0 [as m ≠ 0]

Chapter end exercise
Q 20.

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

Ans.

The power of four devices = 4 × 500 W = 2000 W

Time = 10 hours

Therefore, the energy consumed = power × time

= 2000 × 10 Wh
= 20000 Wh
= 20 kWh
= 20 units [1 unit = 1 kWh]

Chapter end exercise
Q 21.

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

Ans.

When a freely falling body eventually stops on reaching the ground, its kinetic energy gets converted into heat energy (as the body and ground become warm due to collision), sound energy and into potential energy (due to change of shape or deformation). 

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