## Intext Questions 1

#### Question 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?

**Answer**

Given,

Force (f) = 7 N

Displacement (S) = 8 m

As,

Work done = Force × Displacement

Substituting we get,

W = 7 × 8 = 56 Nm or 56 J

Hence, **work done = 56 J**

## Intext Questions 2

#### Question 1

When do we say that work is done?

**Answer**

Work is said to be done when force applied on an object shows the displacement in that object. It is equal to the product of force and displacement.

Work done = force x displacement

or

W = F x S

#### Question 2

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

**Answer**

Work done = Force × Displacement

#### Question 3

Define 1 J of work.

**Answer**

1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

#### 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 (F) = 140 N

Displacement (S) = 15 m

Work done = Force × Displacement

Substituting we get,

W = 140 x 15 = 2100 J

Hence, **2100 J of work is done in ploughing the length of the field.**

## Intext Questions 3

#### Question 1

What is the kinetic energy of an object?

**Answer**

The kinetic energy is the energy possessed by an object due to its motion. It increases when the speed increases.

#### Question 2

Write an expression for the kinetic energy of an object.

**Answer**

Kinetic Energy (K_{E}) = $\dfrac{1}{2}$mv^{2}. Its SI unit is Joule (J).

#### Question 3

The kinetic energy of an object of mass m, moving with a velocity of 5 ms^{-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**

Given,

Kinetic Energy (K_{E}) = 25 J

Velocity (v) = 5 ms^{-1}

We know,

Kinetic Energy (K_{E}) = $\dfrac{1}{2}$mv^{2}.

Substituting we get,

25 = $\dfrac{1}{2}$ x m x 5^{2}

25 x 2 = 25 x m

m = $\dfrac{50}{25}$

m = 2 kg

When velocity is doubled :

v' = 10 ms^{-1}

m = 2 kg

Substituting we get,

K_{E} = $\dfrac{1}{2}$ x 2 x 10^{2}

K.E. = 100 J

When velocity is increased three times, then

v'' = 15 ms^{-1}

m = 2 kg

Substituting we get,

K_{E} = $\dfrac{1}{2}$ x 2 x 15^{2}

K.E. = 225 J

Hence, **Kinetic Energy becomes 100 J when velocity is doubled and it becomes 225 J when velocity is increased three times.**

## Intext Questions 5

#### Question 1

What is power?

**Answer**

Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by:

Power = $\dfrac{\text{Work done}}{\text{Time}}$

It is expressed in watt (W).

#### Question 2

Define 1 watt of power.

**Answer**

1 watt is the power of an agent, which does work at the rate of 1 joule per second.

#### Question 3

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

**Answer**

Given,

Time = 10 s

Work done = Energy consumed by the lamp = 1000 J

We know,

Power = $\dfrac{\text{Work done}}{\text{Time}}$

Substituting we get,

Power = $\dfrac{1000}{10}$ = 100 Js^{-1} or 100 W

Hence, **the power of the lamp is 100 W**

#### Question 4

Define average power.

**Answer**

Average power is defined as the ratio of total energy consumed to the total time taken by the body.

## Exercises

#### 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'.

(a) Suma is swimming in a pond.

(b) A donkey is carrying a load on its back.

(c) A wind-mill is lifting water from a well.

(d) A green plant is carrying out photosynthesis.

(e) An engine is pulling a train.

(f) Food grains are getting dried in the sun.

(g) A sailboat is moving due to wind energy.

**Answer**

Work is said to be done when force applied on an object shows the displacement in that object.

(a) While swimming, Suma applies a force to push the water backwards. She swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, **the work is done**.

(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 windmill works against gravity to elevate water. The windmill lift water by applying a force in an upward direction, and thus the water is moving in the same upward direction itself. Hence, **work is done**.

(d) No force is required when a green plant is carrying out photosynthesis and there is no displacement of plant. Hence, **no work is done**.

(e) When an engine is pulling a train, it is applying a force in the forward direction and the train is moving. As, displacement and force are in the same direction. Hence, **work is done**.

(f) As there is no force applied when grains are dried and there is no displacement as well. Hence, **no work is done**.

(g) 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. As, displacement and force are in the same direction. Hence, **work is done**.

#### 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**

Given,

Work done by gravity depends on the vertical displacement of the body.

So, work done by gravity is :

W = m g h

where,

Vertical displacement, h = 0 [∵ initial and the final points of the path of the object lie on the same horizontal line.]

∴ W = m × g × 0 = 0

Hence, **work done by the force of gravity on the object = 0.**

#### Question 3

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

**Answer**

A battery converts chemical energy into electrical energy. When the bulb receives this electrical energy, it converts it into light and heat energy. Hence, energy changes involved in the process are :

Chemical Energy ⟶ Electrical Energy ⟶ Light Energy + Heat Energy.

#### Question 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.

**Answer**

Given,

Initial velocity u = 5 ms^{-1}

Mass of the body = 20 kg

Final velocity v = 2 ms^{-1}

Initial kinetic energy

E_{i} = $\dfrac{1}{2}$mu^{2}

Substituting we get,

E_{i} = $\dfrac{1}{2}$ x 20 x 5^{2}

= 10 × 25

= 250 J

Final kinetic energy

E_{f} = $\dfrac{1}{2}$mv^{2}

= $\dfrac{1}{2}$ x 20 x 2^{2}

= 10 × 4

= 40 J

As, Work done = Change in kinetic energy = E_{f} – E_{i}

Work done = 40 J - 250 J

Work done = -210 J

Hence, **work done by the force = -210 J.**

#### 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**

Given,

Mass (m) = 10 kg

Work done by gravity depends on the vertical displacement of the body. It is independent of the horizontal path.

So, work done by gravity is = m g h

Where,

Vertical displacement, h = 0 [∵ the line joining A and B is horizontal. ]

∴ W = m × g × 0 = 0

Hence, **work done on object by gravity is zero.**

#### Question 6

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

**Answer**

No. When the body falls from a height, 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. Hence, throughout the process, the total mechanical energy of the body remains conserved.

Hence, the law of conservation of energy is not violated.

#### Question 7

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

**Answer**

The rider's muscular energy is converted to heat energy and the bicycle's kinetic energy while riding a bicycle. Hence, energy transformations are :

Muscular energy ⟶ Kinetic energy + Heat energy

#### 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**

When we push a huge rock and fail to move it, our muscular energy is not transferred to the rock as kinetic energy as there is no displacement of the rock. However, as per the law of conservation of energy, our muscular energy is transformed into heat energy that heats up our body and makes us sweat.

#### Question 9

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

**Answer**

Given,

Energy (E) = 250 units

1 kWh = 3.6 x 10^{6} J

1 unit of energy = 1 kWh

So, 250 units of energy = 250 × 3.6 × 10^{6} = 9 × 10^{8} J.

Hence, **250 units of energy = 9 × 10 ^{8} J.**

#### 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**

Given,

Mass (m) = 40 kg

Acceleration due to gravity = 10 ms^{-2}

Height (h) = 5 m

Potential energy = ?

We know,

Potential energy = m g h

Substituting we get,

P.E. = 40 × 10 × 5 = 2000 J

Height when halfway down : = $\dfrac{5}{2}$ = 2.5 m

Potential energy when halfway = ?

Substituting we get,

P.E. = 40 × 10 × 2.5 = 1000 J

According to the law of conservation of energy:

Total potential energy = potential energy halfway down + kinetic energy halfway down

2000 = 1000 + K.E. halfway down

K.E. at halfway down = 2000 - 1000 = 1000 J

Hence, **Potential energy = 2000 J and kinetic energy at halfway down = 1000 joules**.

#### Question 11

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

**Answer**

When a satellite revolves around the earth in a circular orbit the work done is **zero as force of gravity acting on satellite is perpendicular to its displacement.**

#### 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, a uniformly moving object can experience displacement even when no force is acting on it. According to Newton's first law of motion, an object in motion will continue in its straight-line motion unless acted upon by an external force. Therefore, displacement of an object can occur in the absence of any force acting on it.

#### 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**

Work is said to be done when force applied on an object shows the displacement in that object. In this case, as there is no displacement of the hay bundle, hence no work is done.

#### Question 14

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

**Answer**

Given,

Power of the heater = 1500 W

Converting W to kW

1000 W = 1 kW

So, 1500 W = $\dfrac{1500}{1000}$ = 1.5 kW

Time taken = 10 h

We know,

Power = $\dfrac{\text{Energy consumed}}{\text{Time taken}}$

Or

Energy consumed = Power x Time taken

Substituting we get,

Energy consumed = 1.5 x 10 = 15 kWh

Converting kWh to J

1 kWh = 3.6 x 10^{6} J

So, 15 kWh = 3.6 x 10^{6} x 15 = 5.4 x 10^{7} J

Hence, **the energy consumed = 5.4 x 10 ^{7} J**

#### 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**

The law of conservation of energy states that energy can only be converted from one form to another; it can neither be created or destroyed. The total energy before and after the transformation remains the same.

Refer the figure of an oscillating pendulum bob shown below:

The kinetic energy decreases and the potential energy becomes maximum at B.

After a moment the the to and fro movement starts again.

So, from B to A, again the potential energy changes into kinetic energy and this process repeats again and again.

So, when the bob is in its state of to and fro movement it has potential energy at the extreme position B or C and kinetic energy at resting position A.

It has both the kinetic energy and potential energy at an intermediate position. However, the sum of kinetic and potential energy remain same at every point of movement.

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 lost by the pendulum to overcome the friction is gained by surroundings. Hence, total energy of the system will remain conserved.

#### 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**

The kinetic energy of an object of mass m, moving with a velocity, v, is given by the expression,

Kinetic energy = $\dfrac{1}{2}$mv^{2}

In order to bring it to rest, its velocity has to be reduced to zero.

An external force has to absorb energy from the object, i.e., do negative work on it, equal to its kinetic energy = $-\dfrac{1}{2}$mv^{2}

#### Question 17

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 kmh^{-1}.

**Answer**

Given,

Mass (m) = 1500 kg

Initial velocity (v) = 60 kmh^{-1}

Final velocity = 0

Converting kmh^{-1} to ms^{-1} : multiply by $\dfrac{5}{18}$

60 x $\dfrac{5}{18}$ = $\dfrac{50}{3}$ ms^{-1}

Work required to stop the moving car = Kinetic energy of car (K.E.)

K.E. = $\dfrac{1}{2}$mv^{2}

= $\dfrac{1}{2}$ x 1500 x $(\dfrac{50}{3})$^{2}

= $\dfrac{1}{2}$ x 1500 x $\dfrac{2500}{9}$

= 208333.3 J

Hence, **work done = 208333.3 J.**

#### Question 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.

**Answer**

**Case I**

The direction of force acting on the block is perpendicular to the displacement. Hence, work done by force on the block will be **zero**.

**Case II**

The direction of force acting on the block is in the direction of displacement. Hence, work done by force on the block will be **positive**.

**Case III**

The direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block will be **negative.**

#### Question 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?

**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.

Consider the below object:

F = ma

F = 0 [∵ equal and opposite forces cancel out each other]

∴ 0 = ma

⇒ a = 0

∴ V_{final} = V_{initial}

Hence, if the object was initially moving, it will continue to move with uniform velocity without acceleration.

Therefore, I agree with Soni's statement.

#### Question 20

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

**Answer**

Given,

Power rating (P) = 500 W

Converting W into kW

1000 W = 1 kW

500 W = $\dfrac{500}{1000}$ = 0.5 kW

Time (T) = 10 h

We know,

Power = $\dfrac{\text{Energy consumed}}{\text{Time taken}}$

Or

Energy consumed by each device = Power x Time taken

Substituting we get,

Energy consumed by each device = 0.5 x 10 = 5 kWh

Energy consumed by four devices = 4 x 5 = 20 kWh

Converting kWh to J

1 kWh = 3.6 x 10^{6} J

So, 20 kWh = 3.6 x 10^{6} x 20 = 7.2 x 10^{7} J

Hence, **the energy consumed = 7.2 x 10 ^{7} J**

#### Question 21

A freely falling object eventually stops on reaching the ground. What happens 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 becomes kinetic energy. When the object hits the ground, all its kinetic energy gets converted into heat energy and sound energy.