# Work, Energy and Power

## Exercise 2(A)

#### Question 1

Define work. When is work said to be done by a force?

The definition of work states that when force is applied on a body and the body moves then work is said to be done.

The condition for work done is displacement of the body when force is applied.

#### Question 2

How is the work done by a force measured when

(i) force is in direction of displacement,

(ii) force is at an angle to the direction of displacement?

(i) When force is in direction of displacement then work done is given as —

Work done = Force x displacement of the point of application of force in the direction of force.

When,

W = work done

F = force applied

S = displacement of the body then we get,

W = F × S

(ii) When force acts at an angle to the direction of displacement then work done is given as —

Work done = Force x component of displacement in the direction of force.

When,

W = work done

F = force applied

S = displacement of the body then we get,

W = F × S Cos θ

#### Question 3

A force F acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force.

(a) Write the expression for the work done by the force.

(b) What should be the angle between force and displacement so that the work done is (i) zero, (ii) maximum?

(a) When force acts at an angle θ to the direction of displacement then work done is

W = F × S Cos θ

(b)

(i) For work done to be zero the angle between force and displacement should be equal to 90° as cos 90° = 0

Therefore, product will also be zero.

W = F × S Cos 90°
= 0

(ii) For work done to be maximum the angle between force and displacement should be equal to 0° as cos 0° = 1

Therefore, product will also be maximum.

W = F × S Cos 0°
= F × S

#### Question 4

A body is acted upon by a force. State two conditions when the work done is zero.

The two conditions for the work done to be zero are,

(i) When there is no displacement of the body on application of force

i.e. S = 0 and

(ii) When displacement is normal to the direction of force

i.e. θ = 90° as cos 90° = 0

#### Question 5

State the condition when the work done by a force is (i) positive, (ii) negative. Explain with the help of examples.

(i) When the displacement of the body is in the direction of force then the work done is said to be positive.

Therefore, W = F × S

Example - When a child pushes a toy table then the force applied by the child and the displacement of the table are in the same direction. So, work done is said to be positive.

(ii) When the displacement of the body is in opposite direction to the force applied then the work done is negative.

Therefore, W = – F × S

Example - When a ball is thrown upwards with a force, the balls moves to a height (h).

However, the displacement is opposite to the direction of force of gravity.

W = -mgh

#### Question 6

A body is moved in a direction opposite to the direction of force acting on it. State whether the work is done by the force or work is done against the force.

When a body is moved in a direction opposite to the direction of force then work is done against the force acting on the body.

When the displacement of the body due to application of force is in a direction opposite to the force then,

θ = 180°, and

cos 180° = -1

Therefore, Work done = (-) Force x displacement of the body.

#### Question 7

When a body moves in a circular path, how much work is done by the body? Give reason.

When a body moves in a circular path, no work is said to be done by the body as the force is directed towards the centre of circular path (the body is acted upon by the centripetal force).

At all points the displacement is along the tangent to the circular path, (i.e normal to the direction of the force).

#### Question 8

A satellite revolves around the earth in a circular orbit. What is the work done by the satellite? Give reason.

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 9

State whether work is done or not by writing yes or no, in the following cases?

(a) A man pushes a wall.

(b) A coolie stands with a box on his head for 15 min.

(c) A boy climbs up 20 stairs.

(a) No, work is not done by the man.

(b) No, work is not done by the coolie.

(c) Yes, work is done by the boy in climbing up the 20 stairs.

#### Question 10

A coolie X carrying a load on his head climbs up a slope and another coolie Y carrying the identical load on his head move the same distance on a frictionless horizontal platform. Who does more work? Explain the reason.

Work done is given by the formula:

W = F × S Cos θ

Coolie Y is moving the load on a frictionless horizontal platform. The force acting on the load is the force of gravity. In case of Coolie Y as the platform is horizontal so the angle between force and displacement is 90°.

Work done by Coolie Y = F × S Cos 90° = F x S x 0 = 0.

Therefore Coolie Y does zero work.

Coolie X is moving the load on a slope. Assuming the angle of slope as θ:

Work done by Coolie X = F × S Cos θ

Therefore, work done by coolie X is more than coolie Y.

#### Question 11

The work done by a fielder when he takes a catch in a cricket match, is negative. Explain.

The work done by a fielder when he takes a catch is said to be negative as the force applied by him is in the direction opposite to the displacement of the ball.

#### Question 12

Give an example when work done by the force of gravity acting on a body is zero even though the body gets displaced from its initial position.

When a coolie carrying a load moves on a horizontal platform, the force of gravity acting on him is normal to the displacement of the load i.e. the angle between force and displacement of load is 90°.

Work done by Coolie = F × S Cos 90° = F x S x 0 = 0.

Therefore, in this case work done by the force of gravity is zero even though the load gets displaced from its initial position.

#### Question 13

What are the S.I. and C.G.S. units of work? How are they related? Establish the relationship.

The S.I unit of work is Joule.

C.G.S unit of work is erg.

Relation between joule and erg:

1 joule = 1N × 1m

As we know, 1 N = 105 dyne and 1 m = 102 cm

Therefore,

1 joule = 105 dyne x 102 cm
⇒ 1 joule = 107 dyne cm
So, 1 joule = 107 erg.

#### Question 14

State and define the S.I. unit of work.

The S.I unit of work is Joule.

1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.

#### Question 15

Express joule in terms of erg.

1 joule = 1N × 1m

As we know, 1 N = 105 dyne and 1 m = 102 cm

Therefore,

1 joule = 105 dyne x 102 cm
⇒ 1 joule = 107 dyne cm
So, 1 joule = 107 erg.

#### Question 16

A body of mass m falls through a height h. Obtain an expression for the work done by the force of gravity.

When body of mass m falls through a height (h) either vertically or on an inclined plane work is said to be done.

We know, work done is

W = F x S

However, the force of gravity on the body is F = mg acting vertically downwards and the vertical displacement in the direction of force is S = h.

Applying the values of F and S we get,

Work done by the force of gravity is

W = F x S
= m x g x h

#### Question 17

A boy of mass m climbs up a stairs of vertical height h.

(a) What is the work done by the boy against the force of gravity?

(b) What would have been the work done if he uses a lift in climbing the same vertical height?

When a boy of mass m climb up stairs of vertical height h then

(a) Work done by the boy against the force of gravity is

W = FS = mgh

(b) The work done when the boy uses a lift in climbing the same vertical height

W = FS = mgh

#### Question 18

Define the term energy and state its S.I unit.

Energy is defined as the capacity to do work. The S.I unit of energy is Joule.

#### Question 19

What physical quantity does electron volt (eV) measure? How is it related to the S.I. unit of that quantity?

The energy of atomic particles is measured in electron volt (eV)

1 eV = charge on an electron × 1 volt
⇒ 1 eV = 1.6 x 10-19 coulomb x 1 volt
1 eV = 1.6 x 10-19 J

#### Question 20

Complete the following sentence

(a) 1 J = ..... calorie.

(b) 1 kWh = ..... J.

(a) 1 J = 0.24 calorie.

(b) 1 kWh = 3.6 x 106 J.

#### Question 21

Name the physical quantity which is measured in calorie. How is it related to the S.I unit of that quantity?

The physical quantity which is measured in calorie is heat energy.

1 calorie = 4.18 joule

#### Question 22

Define kilowatt hour. How is it related to joule?

kilowatt-hour is the energy spent (or work done) by a source of power 1 kW in 1 hour.

1 kWh = 3.6 × 106 joule.

#### Question 23

Define the term power. State its S.I. unit.

Power is defined as the rate of doing work. The S.I unit of power is watt (W)

#### Question 24

State two factors on which power spent by a source depends. Explain your answer with examples.

The two factors on which the power spent by a source depends are :

1. The amount of work done by the source
2. The time taken by the source to do the said work.

For example:

Let us suppose a man X takes 5 minute to lift a load to the roof of a house and a man Y takes 10 minutes to lift the same load to the roof of the same house.

The work done by both the persons remain the same, but the power spent by the man X is twice the power spent by the man Y because man X does the work faster than man Y.

#### Question 25

Differentiate between work and power.

WorkPower
Work done is the product of force and displacement produced due to the applied force.Power is the rate of doing work.
Work done is independent of time.Power spent depends on the time in which work is done.
S.I unit of work is joule (J)S.I unit of power is watt (W)

#### Question 26

Differentiate between energy and power.

EnergyPower
Energy is defined as the capacity of a body to do work.Power is the rate at which energy is supplied by a body.
Energy spent is independent of time.Power depends on the time in which energy is spent.
S.I unit of energy is joule (J).S.I unit of power is the watt (W).

#### Question 27

State and define the S.I unit of power.

The S.I unit of power is watt (W).
The power spent is said to be 1 watt, if 1 joule of work is done in 1 second.

#### Question 28

(a) Name the physical quantity measured in terms of horsepower.

(b) How is horsepower related to the S. I. unit of power?

(a) Power is the physical quantity measured in terms of horsepower.

(b) The relation between horsepower and S.I. unit of power is 1 horsepower = 746 watt.

#### Question 29

Differentiate between watt and watt-hour.

The unit of power is watt (W) while the unit of work is watt-hour (Wh), since power × time = work or energy.

#### Question 30

Name the quantity which is measured in

(a) kWh

(b) kW

(c) Wh

(d) eV

(a) Energy is measured in kWh

(b) Power is measured in kW

(c) Energy is measured in Wh

(d) Energy is measured in eV

#### Question 31

Is it possible that no transfer of energy takes place even when a force is applied to a body?

Yes there are cases when we will see no transfer of energy even on application of force.

Example - When a body moves in circular motion then force acts normal to the displacement of the body then there is no transfer of energy and work done is zero.

## Multiple Choice Type

#### Question 1

One horsepower is equal to

1. 1000 W
2. 500 W
3. 764 W
4. 746 W ✓

#### Question 2

kWh is the unit of:

1. Power
2. Force
3. Energy ✓
4. None of these

## Numericals

#### Question 1

A body, when acted upon by a force of 10 kgf, gets displaced by 0.5 m. Calculate the work done by the force, when the displacement is

(i) in the direction of force,

(ii) at an angle of 60 degree with the force, and

(iii) normal to the force where g = 10Nkg-1

Given,

Force = 10 kgf and g = 10N per kg

So,

Force = 10 x 10 = 100 N

Displacement = S = 0.5 m

As we know,

Work done = force × displacement in the direction of force

(i) When displacement is in the direction of force

$W = F × S \\[0.5em] W = 100 \times 0.5 \\[0.5em] W = 50 J \\[0.5em]$

Therefore, work done is 50J.

(ii) When displacement is at an angle of 60° with the force

$W = F × S \space cos \space \theta \\[0.5em] W = 100 \times 0.5 \space Cos \space 60 \\[0.5em] W = 100 \times 0.5 \times 0.5 \\[0.5em] W = 25 J$

Therefore, work done is 25J.

(iii) When displacement is normal to the force

$W = F × S \space cos \space \theta \\[0.5em] W = 100 \times 0.5 \space Cos \space 90 \degree\\[0.5em] W = 100 \times 0.5 \times 0 \\[0.5em] W = 0$

Therefore, work done is 0.

#### Question 2

A boy of mass 40kg climbs up the stairs and reaches the roof at a height 8m in 5s. Calculate:

(i) The force of gravity acting on the boy,

(ii)The work done by him against gravity,

(iii)The power spent by the boy.

Take g = 10 ms-2

Given,

m = 40 kg

h = 8 m

t = 5 s

(i) Force of gravity acting on the boy

$F = mg \\[0.5em] F = 40 \times 10 = 400N \\[0.5em]$

(ii) Work done against gravity

$W = F × S \\[0.5em] W = 400 \times 8 = 3200J \\[0.5em]$

(iii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] W = \dfrac{3200}{5} = 640 W$

#### Question 3

A man spends 6.4 kJ energy in displacing a body by 64 m in the direction in which he applies force, in 2.5 s.

Calculate:

(i) the force applied and

(ii) the power spent (in H.P) by the man.

Given,

Work done = 6.4 kJ

S = 64 m

t = 2.5 s

(i) W = F x S

$\text{Force} = \dfrac{\text{work done}}{\text{distance}} \\[0.5em] F = \dfrac{6.4 \times 10^3}{64} \\[0.5em] \Rightarrow F = 100 N \\[0.5em]$

(ii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] P = \dfrac{6.4 \times 10^3}{2.5} \\[0.5em] \Rightarrow P = 2560 W \\[0.5em]$

We know,

$\text{1 H.P} = \text{746 W} \\[0.5em] \text{1 W}= \dfrac{1}{746} \text{ H.P} \\[0.5em] \therefore \text {2560 W} = \dfrac {2560}{746} \\[0.5em] \Rightarrow \text {2560 W} = \text{3.43 H.P}$

#### Question 4

A weight lifter lifted a load of 200 kgf to a height of 2.5 m in 5 s.

Calculate:

(i) the work done, and

(ii) the power developed by him.

Take g = 10Nkg-1

Given,

mass = 200kgf

height = 2.5m

time = 5s

Force = mg
= 200 x 10 = 2000N

So, we get force is equal to 2000N

(i) Work done = F x h

$W = 2000 \times 2.5 = 5000 J\\[0.5em]$

Therefore, work done is 5000 J.

(ii)

$\text{Power developed} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] P = \dfrac{5000}{5} = 1000W$

Therefore, power developed by the boy 1000 W.

#### Question 5

A machine raises a load of 750N through a height of 16m in 5s. Calculate:

(i) the energy spent by the machine.

(ii) the power of the machine if it is 100% efficient.

Given,

Force = 750N

height or distance moved = 16 m

time = 5 s

(i) Energy spent or the work done = F x S

$W = 750 \times 16 = 12000 J \\[0.5em]$

Therefore, energy spent is equal to 12000 J

(ii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] W = \dfrac{12000}{5} = 2400 W \\[0.5em]$

Therefore, power spent is equal to 2400 W.

#### Question 6

An electric heater of power 3 kW is used for 10 h. How much energy does it consume? Express your answer in

(i) kWh

(ii) joule.

Given,

Power of the electric heater = 3kW

Time for which the electric heater is used = 10 hours

We know,

Energy consumed = power × time

(i) Energy consumed in kWh

$\text{Energy} = 3 \times 10 \\[0.5em] \Rightarrow \text{Energy} = 30 kWh \\[0.5em]$

(ii) Energy consumed in joule.

$1 kWh = 3.6 \times 10^6 J \\[0.5em] 30 kWh = 30 \times 3.6 \times 10^6 \\[0.5em] \Rightarrow 30 kWh = 1.08 \times 10^8 J \\[0.5em]$

#### Question 7

A water pump raises 50 litres of water through a height of 25m in 5s. Calculate the power of the pump used (assuming 100% efficiency).

(Take g = 10 N kg-1 and density of water = 1000 kg m-3).

Given,

Volume of water raised = 50 L and
50 L = 50 × 10-3 m3

height = 25m

time spent = 5s

As we know,

Mass of water = Volume of water × density of water

$Mass =\dfrac{50\times 1000}{1000} = 50 Kg$

Work done in raising 50 kg water to height of 25 m is

W = mgh

Power = work done / time taken

Substituting the values we get

$\text{Power} = \dfrac{50 \times 10 \times 25}{5} = 2500W\\[0.5em]$

Therefore, power of the pump is 2500W.

#### Question 8

A pump is used to lift 500kg of water from a depth of 80m in 10s.

Calculate:

(a) The work done by the pump,

(b) The power at which the pump works, and

(c) The power rating of the pump if its efficiency is 40%. (Take g = 10m s-2).

Given,

Mass of water = 500kg

Height to which the water has to be raised = 80 m

Time = 10 s

(a) Work done by the pump = mgh

Substituting the values we get,

$W = 500 \times 10 \times 80 \\[0.5em] \Rightarrow W = 4 \times 10^5 J \\[0.5em]$

(b) Power of the pump = work done / time taken

Power = mgh / t

Substituting the values we get,

$\text{Power} = \dfrac{500 \times 10 \times 80 }{10} = 4000 W\\[0.5em] \Rightarrow \text{Power} = 40 kW \\[0.5em]$

(c)

$\text{Efficiency} = \dfrac{\text{useful power}}{\text{power input}}$

Given,

Efficiency = 40 %

$0.4 = \dfrac{40}{\text{input power}} \\[0.5em] \text{input power} = \dfrac{40}{0.4} \\[0.5em] = 100 kW \\[0.5em]$

∴ Power rating of pump = 100 kW.

#### Question 9

An ox can apply a maximum force of 1000N. It is taking part in a cart race and is able to pull the cart at a constant speed of 30ms-1 while making its best effort. Calculate the power developed by the ox.

Given,

Force = 1000 N

Velocity = 30 m/s

Power = force × velocity

Power = 1000 x 30 = 30,000 W

Therefore, power developed by the ox is 30 kWh

#### Question 10

The power of a motor is 40 kW. At what speed can the motor raise a load of 20,000 N?

Given,

Power = 40 kW

Force = 20,000 N

Power = force × velocity

Hence,

$\text{velocity} = \dfrac{\text{power}}{\text{force}} \\[0.5em] \Rightarrow \text{velocity} = \dfrac {40,000}{20,000} \\[0.5em] \Rightarrow \text{velocity} = 2 \space ms^{-1} \\[0.5em]$

#### Question 11

Rajan exerts a force of 150 N in pulling a cart at a constant speed of 10ms-1. Calculate the power exerted.

Power exerted by Rajan = force x velocity

Power = 150 x 10 = 1500 W

#### Question 12

A boy weighing 350 N climbs up 30 steps, each 20 cm high in 1 minute.

Calculate:

(i) the work done, and

(ii) the power spent.

We know,
work done = Force × distance moved in direction of force
i.e. W = F × S

Given,

Force = 350 N

Distance covered in 30 steps (S)

S = 30 x 20 cm = 600cm
⇒ S = 6m

Substituting the values of F and S

$W = 350 \times 6 \\[0.5em] = 2100J \\[0.5em]$

∴ Work done by boy = 2100J

(ii)

$\text{Power developed} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] = \dfrac{2100}{60} \\[0.5em] = 35 W \\[0.5em]$

∴ Power spent by boy = 35W

#### Question 13

It takes 20 s for a person A of mass 50 kg to climb up the stairs, while another person B of same mass does the same in 15 s.

Compare the

(i) work done and

(ii) power developed by the persons A and B.

(i) We know that work done is a product of force and displacement and is independent of time. Hence both persons A and B will do the same amount of work.

Therefore, ratio of work done by A and B = 1:1

(ii) The power developed by the persons A and B :

We know that Power = work done / time

Work done is same by A and B

Time A took = 20 s

Time B took = 15 s

Therefore, more power is spent by B as B does the work at a faster rate.

Substituting the values in formula we get :

Power = 1 / time

${\text{Power spent by A}} : {\text{Power spent by B}} = \dfrac{15}{20} \\[0.5em] \Rightarrow {\text {Power spent by A}} : {\text {Power spent by B}} = 3 : 4 \\[0.5em]$

#### Question 14

A boy of weighing 40 kgf climbs up a stair of 30 steps, each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes.

Compare :

(i) the work done by them, and

(ii) the power developed by them. Take g = 10 N kg-1.

(i) We know that work done is a product of force and displacement

W = F x S = mgh

Given,

m1 = 40 kgf
m2 = 30 kgf
t1 = 4 min = 4 x 60 =240 sec
t2 = 3 min = 3 x 60 = 180 sec
h1 = h2 = 30 x 20 = 600 cm = 6 m

$\dfrac{\text {Work done by boy}}{\text {Work done by girl}} = \dfrac{m_1gh_1}{m_2gh_2} \\[0.5em] = \dfrac{m_1}{m_2} = \dfrac{40}{30} \\[0.5em] = \dfrac{4}{3} \\[0.5em]$

Therefore,

Work done by boy : Work done by girl = 4 : 3

(ii) The power developed —

We know that,

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] \dfrac{\text{Power developed by boy}}{\text{Power developed by girl}} \\[1em] = \dfrac{\dfrac{\text{work done by boy}}{\text{time taken by boy}}}{\dfrac{\text{work done by girl}}{\text{time taken by girl}}} \\[1em] = \dfrac{\dfrac{40 \times 10 \times 6}{240}}{\dfrac{30 \times 10 \times 6}{180}} \\[1em] = \dfrac{\dfrac{2400}{240}}{\dfrac{1800}{180}} \\[1em] = \dfrac{10}{10} \\[1em] = \dfrac{1}{1}$

Therefore,

Power developed by boy : Power developed by girl = 1:1

#### Question 15

A man raises a box of mass 50kg to a height of 2m in 20s, while another man raises the same box to the same height in 50s.

(a) Compare: (i) the work done, (ii) the power developed by them.

(b) Calculate the (i) the work done, (ii) the power developed by each man.

Take g = 10N kg-1.

(a)

(i) Work done = force x displacement

Hence, work done is same for both men as both carry 50 kg weight to a height of 2m.

Therefore,

Work done by first man : Work done by second man = 1 : 1

(ii)

Let the work done by both men be W.

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[1em] \dfrac{\text{Power developed by Man A}}{\text{Power developed by Man B}} \\[1em] = \dfrac{\dfrac{\text{work done by Man A}}{\text{time taken by Man A}}}{\dfrac{\text{work done by Man B}}{\text{time taken by Man B}}} \\[1em] = \dfrac{\dfrac{W}{20}}{\dfrac{W}{50}} \\[1em] = \dfrac{5}{2}$

Power developed by Man A : Power developed by Man B = 5:2

(b)

(i) As we know,

W = F x S = mgh

$W = 50 \times 10 \times 2 = 1000 J \\[0.5em] \Rightarrow W = 1000 J \\[0.5em]$

(ii) Power developed by Man A

$P = \dfrac{1000}{20} \\[0.5em] \Rightarrow P = 50 W\\[0.5em]$

Power developed by Man B

$P = \dfrac{1000}{50} \\[0.5em] \Rightarrow P = 20 W\\[0.5em]$

#### Question 16

A boy takes 3 minutes to lift a 20 litre water bucket from a 20 m deep well, while his father does it in 2 minutes.

(a) Compare:

(i) the work, and (ii) power developed by them.

(b) How much work each does?

Take density of water = 103kg m -3 and g = 9.8 N kg-1

(a)

(i) Work done = force x displacement

Work done is same for boy and father as both lift 20 litre water from 20 m deep well.

Work done by boy : Work done by father = 1:1

(ii) Comparing Power developed:

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] {\text {Power of boy}} : {\text {Power of father }} = \dfrac{\dfrac{W}{3}}{\dfrac{W}{2}} \\[0.5em] \Rightarrow {\text {Power developed by boy}} : {\text {Power developed by father}} = \dfrac{2}{3} \\[0.5em]$

Power developed by boy : Power developed by father = 2:3

(b) Work done = mgh

Given,

density of water = 103kg m -3

To convert, density of water to mass per litre

We know,

1 litre = 1000 cm3
and
1 m3 = 106 cm3

So,

$10^3kgm^-3 = \dfrac{1000kg}{1m^3} \\[0.5em] = \dfrac{1000kg}{10^6 cm^3} \\[0.5em] = \dfrac{1kg}{1000cm^3} \\[0.5em] = \dfrac{1kg}{1 litre}$

Hence,

20L of water = 20 Kg by mass

height = 20m

Now, substituting the values in the formula for work done we get

W = 20 x 9.8 x 20
= 3.92 kJ

## Exercise 2(B)

#### Question 1

What are the two forms of mechanical energy?

The two forms of mechanical energy are

1. Kinetic energy
2. Potential energy

#### Question 2

Name the form of energy which a wound-up watch spring possesses.

Wound up watch spring possesses an elastic potential energy.

#### Question 3

Name the type of energy (kinetic energy K or potential energy U) possessed in the following cases:

(a) A moving cricket ball

(b) A compressed spring

(c) A moving bus

(d) A stretched wire

(e) An arrow shot out of a bow.

(f) A piece of stone placed on the roof.

a) A moving cricket ball posses Kinetic energy — K.

(b) A compressed spring posses Potential energy — U.

(c) A moving bus posses Kinetic energy — K.

(d) A stretched wire posses Potential energy — U.

(e) An arrow shot out of a bow posses Kinetic energy — K.

(f) A piece of stone placed on the roof posses Potential energy — U.

#### Question 4

Define the term potential energy of a body. Name its two forms and give one example of each.

The energy possessed by a body at rest due to its position or size and shape is called the potential energy.

Different forms of potential energy are as follows:

(i) Gravitational potential energy — It is the energy possessed by a body due to the force of attraction of the earth.

Example — A stone at a height has gravitational potential energy.

(ii) Elastic potential energy — It is the energy possessed by a body due to the deformed state because of change in its shape and size.

Example — When a spring is compressed it has has elastic potential energy.

#### Question 5

Name the form of energy which a body may possess even when it is not in motion. Give an example to support your answer.

A body possesses potential energy even when it is not in motion.

For example — a stone raised at a height has gravitational potential energy.

#### Question 6

What is meant by gravitational potential energy? Derive an expression for it for a body placed at a height above the ground.

The potential energy possessed by a body due to the force of attraction of the earth on it, is called the gravitational potential energy.

When a body is placed at a height above the ground then the amount of work done in lifting the body to that height against the force of gravity is called the gravitational potential energy.

When we lift a body of mass m from the ground to a height h.Then, force of gravity (mg) acts on the body acting vertically downward which must be equal to the least upward force F required to lift the body.

Work done = force of gravity (mg) × displacement (h)

⇒ W = mgh

So, the work done on the body is stored in the form of its gravitational potential energy when it is at a height h.

Hence,

Gravitational potential energy U = mgh.

#### Question 7

Write an expression for the potential energy of a body of mass m placed at a height h above the earth’s surface. State the assumptions made, if any.

The work done in lifting a body to a height h is

W = force of gravity (mg) × displacement (h) = mgh

Therefore, this work is stored in the body in the form of its gravitational potential energy.

Hence, Gravitational potential energy U = mgh

#### Question 8

What do you understand by the kinetic energy of a body?

Kinetic Energy of a body is the energy possessed by a body in motion.

#### Question 9a

A body of mass m is moving with a velocity v. Write the expression for its kinetic energy.

(a) The expression for a body of mass m moving with a velocity v is given by

${\text {Kinetic energy} = \dfrac{1}{2} \times m \times velocity^2} \\[0.5em] {\text {Kinetic energy} = \dfrac{1}{2} \times m \times v^2} \\[0.5em]$

#### Question 9b

Show that the quantity 2K/v2 has the unit of mass, where K is the kinetic energy of the body.

2K / v2 = Joules / (ms-1)2 = kg

So, Kg is the unit of mass

#### Question 10

State the work–energy theorem.

According to the work theorem,

The work done by a force in the same direction on a moving body is equal to the increase in its kinetic energy.

#### Question 11

A body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity increases from u to v. How much work is being done by the force?

Mass = m

Change in velocity from u to v, so acceleration acts (a)

Force = f

Work done by the force = force × displacement

From equation of motion we know,

$v^2 = u^2 + 2aS \\[0.5em] \therefore S = \dfrac{v^2 -u^2}{2a}$

Substituting it in the formula for work done:

$W = m \times a \times (\dfrac{v^2-u^2}{2a}) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = \dfrac{1}{2} m (v^2 - u^2) \\[0.5em] W = KE_f - KE_i$

#### Question 12

A light mass and a heavy mass have equal momentum. Which will have more kinetic energy?

Kinetic energy (KE) and momentum (p) are related as

$KE = \dfrac{p^2}{2m} \\[0.5em]$

As both masses have the same momentum p. The kinetic energy k is inversely proportional to the mass of the body.

Hence, a body of light mass has more kinetic energy as smaller the mass, larger is its kinetic energy.

#### Question 13

Two bodies A and B of masses m and M (M ≫ m) have same kinetic energy. Which body will have more momentum?

Kinetic energy and momentum are related as

$p = \sqrt{2mk}$

Since,

the kinetic energy of both bodies are same, momentum is directly proportional to the square root of mass.

As we can see that mass of body B is greater than that a body A.

Therefore, body B will have more momentum than body A.

#### Question 14

Name the three forms of kinetic energy and give one example of each.

The three forms of kinetic energy are:

(i) Vibrational kinetic energy

Example — A wire clamped at both the ends vibrates when struck.

(ii) Rotational kinetic energy

Example — A spinning top.

(iii) Translational kinetic energy

Example — A car moving in a straight path.

#### Question 15

State two differences between the potential energy and the kinetic energy.

Potential energyKinetic energy
It is the energy possessed by a body due to its changed position or change in shape or size.It is the energy possessed by a body due to its state of motion.
It does not depend on the speed of the body.It depends on the speed of the body.

#### Question 16

Complete the following sentences

(a) The kinetic energy of a body is the energy by virtue of its ......

(b) The potential energy of a body is the energy by virtue of its ......

(a) The kinetic energy of a body is the energy by virtue of its motion.

(b) The potential energy of a body is the energy by virtue of its position.

#### Question 17

When an arrow is shot from a bow, it has kinetic energy in it. Explain briefly from where does it get its kinetic energy?

When the string of a bow is pulled, the work done is stored in the form of elastic potential energy.

When the string is released to shoot the arrow, the potential energy of the bow changes into the kinetic energy.

The kinetic energy is the same potential energy stored when the string was pulled.

#### Question 18

A ball is placed on a compressed spring. What form of energy does the spring possess? On releasing the spring, the ball flies away. Give a reason.

When the ball is placed on a compressed spring, the compressed spring has elastic potential energy.

The potential energy of the spring changes into kinetic energy when it is released and as a result it flies. #### Question 19

A pebble is thrown up. It goes to a height and then comes back on the ground. State the different changes in the form of energy during its motion.

On being thrown upwards, the kinetic energy of the pebble changes to potential energy.

Its kinetic energy is completely converted into potential energy when the pebble reaches its maximum height.

The potential energy is then converted into kinetic energy while coming down and the potential energy is completely converted to kinetic energy when the pebble reaches the ground.

#### Question 20

In what way does the temperature of water at the bottom of a waterfall differ from the temperature at the top? Explain the reason.

When the water falls from the height, the potential energy stored in water at a height changes into the kinetic energy.

On reaching the ground, a part of the kinetic energy of water changes into the heat energy which increases the temperature of the water.

#### Question 21

Name the form of energy in which potential energy can change.

The potential energy can change only in the form of kinetic energy.

#### Question 22

Name the form of mechanical energy, which is put to use.

Kinetic energy is put to use in the form of mechanical energy.

#### Question 23

Name six different forms of energy?

The six different forms of energy are

(i) Heat energy

(ii) Solar energy

(iii) Nuclear energy

(iv) Hydro energy

(v) Light energy

(vi) Mechanical energy

#### Question 24

Energy can exist in several forms and may change from one form to another. For each of the following, state the energy changes that occur in:

(a) the unwinding of a watch spring

(b) a loaded truck when started and set in motion

(c) a car going uphill

(d) photosynthesis in green leaves

(e) charging of a battery

(f) respiration

(g) burning of a match stick

(h) explosion of crackers

(a) Potential energy of a watch spring converts into kinetic energy

(b) Chemical energy of diesel or petrol converts into mechanical energy

(c) Kinetic energy converts into potential energy

(d) Light energy converts into chemical energy

(e) Electrical energy converts into chemical energy

(f) Chemical energy converts into heat energy

(g) Chemical energy converts into heat and light energy

(h) Chemical energy converts into heat, light and sound energy

#### Question 25

State the energy changes in the following cases while in use:

(a) Loudspeaker

(b) A steam engine

(c) Microphone

(d) Washing machine

(e) A glowing electric bulb

(f) Burning coal

(g) A solar cell

(h) Biogas burner

(i) An electric cell in a circuit

(j) A petrol engine of a running car

(k) An electric iron

(l) A ceiling fan

(m) An electromagnet.

(a) Electrical energy changes into sound energy

(b) Heat energy changes into mechanical energy

(c) Sound energy changes into electrical energy

(d) Electrical energy changes into mechanical energy

(e) Electrical energy changes into light energy

(f) Chemical energy changes into heat energy

(g) Light energy changes into electrical energy

(h) Chemical energy changes into heat energy

(i) Chemical energy changes into electrical energy

(j) Chemical energy changes into mechanical energy

(k) Electrical energy changes into heat energy

(l) Electrical energy changes into mechanical energy

(m) Electrical energy changes into magnetic energy

#### Question 26

Name the process used for producing electricity from nuclear energy.

Nuclear fission is the process used for producing electricity from nuclear energy.

#### Question 27

Is it practically possible to convert a form of energy completely into the other useful form? Explain your answer.

It is not possible to completely convert one form of energy into other as a part of the energy is dissipated in the form of heat which is lost to the surroundings.

#### Question 28

While transforming energy from one from to another desired form, the entire energy does not change into the desired form. A part of energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and becomes useless. This conversion of energy to the undesirable form is called dissipation of energy.

Since, this part of energy is not available to us for any productive purpose so we call this part of energy as degraded energy.

#### Question 29

What do you mean by degradation of energy? Explain it by taking one example of your daily life.

While transforming energy from one from to another desired form, the entire energy does not change into the desired form. A part of energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and becomes useless. This conversion of energy to the undesirable form is called dissipation of energy or degradation of energy.

Since, this part of energy is not available to us for any productive purpose so we call this part of energy as degraded energy.

Example — When a vehicle is run by using the chemical energy of its fuel a major part is wasted as heat and sound. Only a part of it is changed into useful mechanical energy.

#### Question 30

Complete the following sentence

The conversion of part of energy into an undesirable form is called ......

The conversion of part of energy into an undesirable form is called degradation of energy

## Multiple Choice Type

#### Question 1

A body at a height possesses:

1. Kinetic energy
2. Potential energy ✓
3. Solar energy
4. Heat energy

#### Question 2

In an electric cell while in use, the change in energy is from:

1. Electrical to mechanical
2. Electrical to chemical
3. Chemical to mechanical
4. Chemical to electrical ✓

## Numericals

#### Question 1

Two bodies of equal masses are placed at heights h and 2h. Find the ratio of their gravitational potential energies.

Given,

Mass of body A and body B = m each

Height HA = h

Height HB = 2h

We know, Gravitational potential energy = mgh

Substituting the values we get,

$\text {Ratio of gravitational P.E.} = \dfrac{mgh}{mg2h} \\[0.5em] \Rightarrow \text {Ratio of gravitational P.E.} = \dfrac{1}{2} \\[0.5em]$

Hence, ratio of gravitational potential energies of the two bodies = 1 : 2

#### Question 2

Find the gravitational potential energy of 1 kg mass kept at a height of 5m above the ground. Calculate its kinetic energy when it falls and hits the ground. Take g = 10ms-2.

Given,

Mass, m = 1 kg

Height, h = 5 m

Gravitational potential energy = mgh
= 1 × 10 × 5
= 50 J

When the mass hits the ground, its entire Potential energy is converted to Kinetic energy.

Hence, Kinetic energy of mass on hitting the ground = 50 J.

#### Question 3

A box of mass 150 kgf has gravitational potential energy stored in it equal to 14700 J. Find the height of the box above the ground.

(Take g = 9.8 N kg-1)

Given,

Mass of box = 150 Kgf

Gravitational potential energy = 14700 J

We know,

Gravitational potential energy = mgh

Therefore,

$h = \dfrac{\text{Gravitational P.E.}}{\text{mass} \times \text{gravity}}$

Substituting the values in formula we get,

$\text{height} = \dfrac{14700}{150 \times 9.8} \\[0.5em] \Rightarrow \text{height} = \dfrac{14700}{1470} \\[0.5em] \Rightarrow \text{height} = 10m$

#### Question 4

A body of mass 5 kg falls from a height of 10 m to 4 m.

Calculate:

(i) the loss in potential energy of the body,

(ii) the total energy possessed by the body at any instant? (Take g = 10 ms-2).

Given,

Mass = 5Kg

$\text {Fall in height} = 10 - 4 \\[0.5em] \Rightarrow \text {Fall in height} = 6m \\[0.5em]$

(i) Loss in potential energy = mg(h1 - h2)

Substituting the vales in formula we get,

$\text{Loss in potential energy} = 5 \times 10 \times 6 \\[0.5em] \Rightarrow \text{Loss in potential energy} = 300J \\[0.5em]$

Total energy = P.E + K.E

So, at height 10 m, K.E = 0

Therefore total energy = P.E + K.E

Total energy = mgh + 0

$5 \times 10 \times 10 = 500J \\[0.5em]$

#### Question 5

Calculate the height through which a body of mass 0.5 kg is lifted if the energy spent in doing so is 1.0 J. Take g = 10ms-2.

Given,

Mass = 0.5 kg

Energy = 1 J

We know that,

$\text{Gravitational P.E.} = \text{mgh} \\[0.75em] \therefore h = \dfrac{\text{Gravitational P.E.}}{\text{mg}} \\[0.5em]$

Substituting the values we get,

$h = \dfrac{1}{0.5 \times 10} \\[0.5em] \Rightarrow h = 0.2m \\[0.5em]$

#### Question 6

A boy weighing 25 kgf climbs up from the first floor at a height of 3 m above the ground to the third floor at a height of 9 m above the ground. What will be the increase in his gravitational potential energy?

(Take g = 10 N kg-1).

Mass = 25 Kgf

Increase in gravitational potential energy = mg(h2 – h1)

$\text {Increase in gravitational P.E.} = 25 \times 10 \times 6 \\[0.5em] \Rightarrow \text {Increase in gravitational P.E.} = 1500J \\[0.5em]$

#### Question 7

A vessel containing 50 kg of water is placed at a height 15m above the ground. Assuming the gravitational potential energy at the ground to be zero, what will be the gravitational potential energy of water in the vessel?

(g = 10ms-2).

Given,

Mass = 50 kg

Height = 15 m

Gravitational potential energy = mgh

Substituting the values we get,

$\text {Gravitational P.E.} = 05 \times 10 \times 15 \\[0.5em] \Rightarrow \text {Increase in gravitational P.E.} = 7500J \\[0.5em]$

#### Question 8

A man of mass 50 kg climbs up a ladder of height 10m.

Calculate:

(i) the work done by the man,

(ii) the increase in his potential energy.

(g= 9.8m s-2).

Given,

Mass = 50 kg

h1 = 0m

h2 = 10m

(i) Work done by man = mgh2

Substituting the values we get,

$W = 50 \times 9.8 \times 10 \\[0.5em] \Rightarrow W = 4900 J \\[0.5em]$

(ii) Increase in his potential energy = Mg (h2 – h1)

$\text{Increase in gravitational P.E.} = 50 \times 9.8 \times 10 \\[0.5em] \Rightarrow \text {Increase in gravitational P.E.}= 4900 J \\[0.5em]$

#### Question 9

A block A, whose weight is 100N, is pulled up a slope of length 5m by means of a constant force F (=100N) as illustrated. (a) What is the work done by the force F in moving the block A, 5m along the slope?

(b) What is the increase in potential energy of the block A?

(c) Account for the difference in the work done by the force and the increase in potential energy of the block.

Given,

F = 100 N

h = 3m

displacement = 5m

We know,

(i) Work done = Force × displacement in the direction of the force

$W = 100 \times 5 \\[0.5em] \Rightarrow W = 500 J \\[0.5em]$

(ii) The potential energy gained by the block U = mgh

$\text{P.E.} = 100 \times 3 \\[0.5em] \Rightarrow \text { P.E. gained by the block}= 300 J \\[0.5em]$

(iii) The difference (200 J) energy is used in doing work against the force of friction between the block and the slope and it will appear as heat energy.

#### Question 10

Find the kinetic energy of a body of mass 1kg moving with a uniform velocity of 10m s-1.

Given,

m = 1 kg

Velocity, v = 10 m / s

Kinetic energy = 1 / 2 × mass × (velocity)2

Substituting the values we get,

$KE = \dfrac{1}{2} \times 1 \times 10^2 \\[0.5em] \Rightarrow KE = 50 J \\[0.5em]$

#### Question 11

If the speed of a car is halved, how does its kinetic energy change?

Let speed of car be v.

Kinectic Energy at speed v = $\dfrac{1}{2}$ x m x v2 = $\dfrac{\text{mv}^2}{2}$

Kinectic Energy at speed $\dfrac{\text{v}}{2}$ = $\dfrac{1}{2}$ x m x $(\dfrac{\text{v}}{2})^2$ = $\dfrac{\text{mv}^2}{8}$ = $\dfrac{1}{4}\Big(\dfrac{\text{mv}^2}{2}\Big)$

Hence, when the speed of the car is halved the kinetic energy becomes one-fourth.

#### Question 12

Two bodies of equal masses are moving with uniform velocities v and 2v. Find the ratio of their kinetic energies.

Given,

Velocity of first body v1 = v

Velocity of second body, v2 = 2v

As the mass of the two bodies are same and as kinetic energy is directly proportional to the square of the velocity (K α v2)

Hence, ratio of their kinetic energies is

$\dfrac{K_1}{K_2} = \dfrac{(v_1)^2}{(v_2)^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{(2v)^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{4v^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{1}{4} \\[0.5em]$

#### Question 13

Two bodies have masses in the ratio 5 : 1 and kinetic energies in the ration 125 : 9. Calculate the ratio of their velocities.

We know that,

$\text{K.E.} = \dfrac{1}{2} \times m \times v^2 \\[0.5em]$

Let the two bodies be A and B.

Let mass of body A be mA and body B be mB. Let velocity of body A be vA and body B be vB.

Given,

Ratio of masses:

$\dfrac {m_A}{m_B} = \dfrac {5}{1} \\[0.5em]$

Ratio of K.E.:

$\dfrac {KE_A}{KE_B} = \dfrac {125}{9} \\[1em] \dfrac {KE_A}{KE_B} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{m_A}{m_B} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{5}{1} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{125}{9} \times \dfrac{1}{5} \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{25}{9} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \dfrac{5}{3} \\[0.5em]$

∴ Ratio of their velocities = 5:3

#### Question 14

A car is running at a speed of 15 km h-1 while another similar car is moving at a speed of 45 km h-1. Find the ratio of their kinetic energies.

We know that,

$KE =\dfrac{1}{2} \times m \times v^2 \\[0.5em]$

Hence, KE α v2

So we get,

$\dfrac{KE_1}{KE_2} = \dfrac{v_1^2}{v_2^2} \\[0.5em] \dfrac{KE_1}{KE_2} = (\dfrac{15}{45})^2 \\[0.5em] \dfrac{KE_1}{KE_2} = (\dfrac{1}{3})^2 \\[0.5em] \Rightarrow \dfrac{KE_1}{KE_2} = (\dfrac{1}{9}) \\[0.5em]$

#### Question 15

A ball of mass 0.5 kg slows down from a speed of 5ms-1 to that of 3ms-1. Calculate the change in kinetic energy of the ball.

Given,

Mass = 0.5 kg

Initial velocity = 5 m / s

Final velocity of the ball = 3 m / s

Kinetic energy = 1 / 2 × mass × (velocity)2

Substituting the values in the equation we get,

$\text{Initial KE} = \dfrac{1}{2} \times 0.5 \times 5^2 \\[0.5em] \text{Initial KE} = \dfrac{1}{2} \times 0.5 \times 25\\[0.5em] \Rightarrow \text{Initial KE} = 6.25 J$

Final kinetic energy,

$\text{Final KE} = \dfrac{1}{2} \times 0.5 \times 3^2 \\[0.5em] \text{Final KE} = \dfrac{1}{2} \times 0.5 \times 9 \\[0.5em] \Rightarrow \text{Final KE} = 2.25 J$

So, change in the kinetic energy of the ball = 2.25 J – 6.25 J = – 4J

Hence, there is a decrease in the kinetic energy of the ball by 4J.

#### Question 16

A canon ball of mass 500g is fired with a speed of 15m/s-1.

Find:

(i) its kinetic energy and

(ii) its momentum.

Given,

mass = 500 g = 0.5 kg

Speed = v = 15 m / s

(a) We know,

Kinetic energy = 1 / 2 × mass × (velocity)2

Substituting the values in the equation we get,

$KE = \dfrac{1}{2} \times 0.5 \times 15^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.5 \times 225 \\[0.5em] \Rightarrow KE = 56.25J \\[0.5em]$

(b) We know,

Momentum = mass × velocity

Substituting the values in the equation we get,

Momentum = 0.5 x 15
= 7.5 kgm/s

#### Question 17

A body of mass 10 kg is moving with a velocity 20 m s-1.

If the mass of the body is doubled and its velocity is halved,

find:

(i) the initial kinetic energy, and

(ii) the final kinetic energy.

Given,

Initial Mass = 10 kg and

Final mass = 2 × 10 kg = 20kg

Initial Velocity = 20 m / s

Changed velocity v2 = 20 / 2 = 10 m / s

(i) Initial kinetic energy = 1 / 2 × mass × (velocity)2

$\text {Initial Kinetic Energy} = \dfrac{1}{2} \times 10 \times (20)^2 \\[0.5em] \Rightarrow \text{Initial kinetic energy} = 2000J \\[0.5em]$

(ii) Final kinetic energy

$\text {Final Kinetic Energy} = \dfrac{1}{2} \times 20 \times (10)^2 \\[0.5em] \Rightarrow \text{Final kinetic energy} = 1000J \\[0.5em]$

#### Question 18

A truck weighing 1000 kgf changes its speed from 36 km/h to 72 km/h in 2 minutes.

Calculate:

(i) the work done by the engine and

(ii) its power.

(g =10 m/ s-2)

Given,

Mass = 1000kg

$\text {Initial velocity (u) } = 36 km / h \\[0.5em] = \dfrac {36 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Initial velocity (u) } = 10ms^{-1} \\[0.5em]$

$\text {Final velocity (v) } = 72 km / h \\[0.5em] = \dfrac {72 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Final velocity (v) } = 20ms^{-1} \\[0.5em]$

From, equation of motion we get

$S = ( v^2 - u^2) / time \\[0.5em]$ (i) Work done = force x displacement

$W = \dfrac{1000 \times (20^2 -10^2)}{2} \\[0.5em] W = \dfrac{1000 \times (400 -100)}{2} \\[0.5em] W = 500 \times 300 \\[0.5em] \Rightarrow W = 150000J = 1.5 × 10^5J \\[0.5em]$

(ii) Power = work done / time taken

$Power = \dfrac{1.5 \times 10^5}{120} \\[0.5em] \Rightarrow Power = 1.25 × 10^3 W$

#### Question 19

A body of mass 60 kg has the momentum of 3000 kgms-1.

Calculate:

(i) the kinetic energy and

(ii) the speed of the body.

Given,

Mass = 60 kg

Momentum = 3000 kgm / s

(a)

$\text{Kinetic Energy} = \dfrac{\text{momentum}^2}{2 \times \text{mass}} \\[0.5em] \Rightarrow \text{KE} = \dfrac{3000^2}{2 \times 60} \\[0.5em] \Rightarrow \text{KE} = \dfrac{3000 \times 3000}{120} \\[0.5em] \Rightarrow \text{KE} = 75000\text{ J} = 7.5 × 10^4\text{ J}$

(b)

$\text{Momentum} = \text{Mass} \times \text{Velocity} \\[0.5em] \Rightarrow \text{Velocity} = \dfrac{\text{Momentum}}{\text{Mass}} \\[0.5em] \Rightarrow \text{Velocity} = \dfrac{3000}{60} \\[0.5em] \Rightarrow \text{Velocity} = 50\text{ ms}^{-1}$

#### Question 20

How much work is needed to be done on a ball of mass 50g to give it a momentum of 5 kg m s-1?

Given,

mass = 50gm = 0.05kg

momentum = 500 gcm / s = 0.005 kgm / s

KE = momentum2 / 2 x mass

$\text{Work done} = KE = \dfrac {momentum^2}{2 \times mass} \\[0.5em] \text{Work done} = KE = \dfrac{0.005^2} {2 \times 0.05} \\[0.5em] \Rightarrow KE =250 J \\[0.5em]$

#### Question 21

How much energy is gained by a box of mass 20 kg when a man

(a) carrying the box waits for 5 minutes for a bus?

(b) runs carrying the box with a speed of 3 ms-1 to catch the bus?

(c) raises the box by 0.5 m in order to place it inside the bus?

(g=10 ms-2)

Given,

Mass = 20 kg

(a) As there is no displacement of the man, so work done is Zero.

(b) We know,

Work done = Kinetic energy of man = 1 / 2 × mass × (velocity)2

$\text{Work done} = \text{Kinetic energy of man} \\[0.5em] = \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \\[0.5em] = \dfrac{1}{2} \times 20 \times 3^2 \\[0.5em] = \dfrac{1}{2} \times 20 \times 9 \\[0.5em] = 90 J \\[0.5em]$

∴ Energy Gained = 90 J

(c) We know,

Potential energy = mgh

$\text{PE} = 20 \times 10 \times 0.5 \\[0.5em] \Rightarrow U = 100 J \\[0.5em]$

#### Question 22

A bullet of mass 50g is moving with a velocity of 500ms-1. It penetrates 10 cm into a still target and comes to rest.

Calculate:

(a) the kinetic energy possessed by the bullet, and

(b) the average retarding force offered by the target.

Given,

Mass = 50 g = 0.05 kg

Velocity = 500 m / s

Distance (S) = 10 cm = 0.1 m

(a) The kinetic energy possessed by the bullet = $\dfrac{1}{2}$ × mass × (velocity)2

Substituting the values in equation we get,

$KE = \dfrac {1}{2} \times 0.05 \times 500^2 \\[0.5em] KE = \dfrac {1}{2} \times 0.05 \times 250000 \\[0.5em] \Rightarrow KE = 6250J \\[0.5em]$

(b)

$\text{Retarding Force} = \dfrac{\text{Work done by bullet}}{\text{distance}} \\[0.5em] = \dfrac{6250}{0.1} \\[0.5em] = 62500N \\[0.5em]$

∴ Average retarding force offered by target = 62500N

#### Question 23

A spring is kept compressed by a small trolley of mass 0.5 kg lying on a smooth horizontal surface as shown. When the trolley is released, it is found to move at a speed of v = 2 ms-1. What potential energy did the spring possess when compressed? Given,

Mass = 0.5 kg

Velocity = 2 m / s

Potential Energy possessed by the spring is equal to the kinetic energy of moving trolley.

Kinetic energy of trolley

$= \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \\[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \\[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \\[0.5em] = 1 \text{ J} \\[0.5em]$

∴ Potential energy = Kinetic energy = 1.0 J

## Exercise 2(C)

#### Question 1

State the Principle of conservation of energy.

The principle of conservation of energy states that energy can neither be created nor can be destroyed. It only changes from one form to other.

#### Question 2

What do you understand by the conservation of mechanical energy? State the condition under which the mechanical energy is conserved.

Whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant. This is principle of conservation of mechanical energy.

i.e K + U = constant provided there are no frictional forces.

The condition for mechanical energy to be conserved is that there should be no frictional forces. This condition is only applicable in vacuum.

#### Question 3

Name two examples in which the mechanical energy of a system remains constant.

1. Simple pendulum and

2. Motion of a freely falling body.

#### Question 4

A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?

When a body is thrown vertically upwards, its kinetic energy changes into potential energy and its velocity becomes zero.

#### Question 5

A body falls freely under gravity from rest. Name the kind of energy it will possess

(a) At the point from where it falls.

(b) While falling

(c) On reaching the ground.

(a) The energy possessed at the point from where it falls is potential energy.

(b) Potential energy and kinetic energy, both are possessed at the time of fall.

(c) The energy possessed by the body on reaching the ground is kinetic.

#### Question 6

Show that the sum of kinetic energy and potential energy (i.e., total mechanical energy) is always conserved in the case of a freely falling body under gravity (with air resistance neglected) from a height h by finding it when

(i) the body is at the top,

(ii) the body has fallen a distance x,

(iii) the body has reached the ground. We know that,

Kinetic energy + potential energy = constant

So, when a body falls from a height h under free fall

(i) At the position A — height h

Initial velocity = 0

Kinetic energy K = 0

Potential energy U = mgh

As, Total energy = KE + PE

Total energy = 0 + mgh

Total energy = mgh       (1)

(ii) At the position B — when it has fallen a distance x.

Then, velocity at B = v1

Then u = 0, s = x, a = g

From equation of motion:
v2 = u2 + 2aS
v12 = 0 + 2gx = 2gx

$\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_1^2 \\[0.5em] = \dfrac{1}{2} \times m \times 2gx \\[0.5em] = mgx$

∴ Potential energy U = mg (h – x)

Hence, total energy = K + U = mgx + mg (h – x) = mgh

Total energy = mgh       (2)

(iii) At position C (on the ground) —

Let the velocity acquired by the body on reaching the ground be v2.

Then u = 0, s = h, a = g

We know,
v2= u2 + 2aS
v22 = 0 + 2gh
v22 = 2gh

$\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_2^2 \\[0.5em] = \dfrac{1}{2} \times m \times 2gh \\[0.5em] = mgh$

And potential energy U = 0 (at the ground when h = 0)

So, total energy = K + U = mgh + 0

Total energy = mgh       (3)

Thus, from equation (1), (2) and (3) we note that the total mechanical energy i.e. the sum of kinetic energy and potential energy always remain constant at each point of motion and it is equal to the initial potential energy at height h.

#### Question 7

A pendulum is oscillating on either side of its rest position. Explain the energy changes that take place in the oscillating pendulum. How does the mechanical energy remain constant in it? Draw necessary diagram. 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 repeat again and again.

So, when the bob in its state of to and from 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.

#### Question 8

A pendulum with a bob of mass m is oscillating on either side from its resting position A between the extremes B and C at a vertical height h above A. What is the kinetic energy K and potential energy U when the pendulum is at positions

(i) A, (ii) B and (iii) C?

(i) For position A,

Pendulum has the maximum kinetic energy and potential energy is zero at its resting position.

So, K = mgh and U = 0

(ii) For position B,

The kinetic energy starts decreasing and the potential energy starts increasing.

So, K = 0 and U = mgh

(iii) At position C,

Kinetic energy K = 0 and potential energy U = mgh.

#### Question 9

Name the type of energy possessed by the bob of a simple pendulum when it is at

(a) the extreme position,

(b) the mean position, and

(c) between the mean and extreme positions.

(a) When the bob is at extreme position the energy possessed is potential energy.

(b) When the bob is at mean position the energy possessed is kinetic energy.

(c) When the bob is in between mean position and extreme position the energy possessed by the bob are both kinetic energy and potential energy.

## Multiple Choice Type

#### Question 1

A ball of mass m is thrown vertically up with an initial velocity so as to reach a height h. The correct statement is:

1. Potential energy of the ball at the ground is mgh.
2. Kinetic energy to the ball at the ground is zero.
3. Kinetic energy of the ball at the highest point is mgh.
4. The potential energy of the ball at the highest point is mgh. ✓

#### Question 2

A pendulum is oscillating on either side of its rest position. The correct statement is :

1. It has only the kinetic energy at its each position.
2. It has the maximum kinetic energy at its extreme position.
3. It has the maximum potential energy at its mean position.
4. The sum of its kinetic and potential energy remains constant throughout the motion. ✓

## Numericals

#### Question 1

A ball of mass 0.20 kg is thrown vertically upwards with an initial velocity of 20ms-1. Calculate the maximum potential energy it gains as it goes up.

Given,

Mass = 0.20 kg

Initial velocity = 20ms-1

We know,

$KE = \dfrac{1}{2} \times m \times v^2$

Maximum potential energy at the maximum height = initial kinetic energy

Substituting the values in the equation

$KE = \dfrac{1}{2} \times 0.20 \times 20^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.20 \times 400$

∴ Maximum potential energy at the maximum height = Initial kinetic energy = 40 J

#### Question 2

A stone of mass 500 g is thrown vertically upwards with a velocity of 15ms-1.

Calculate:

(a) the potential energy at the greatest height,

(b) the kinetic energy on reaching the ground

(c) the total energy at its half waypoint.

Mass = 500g

Velocity = 15ms-1

(a) We know,

Potential energy at maximum height = initial kinetic energy

mgh = $\dfrac{1}{2}$ mv2

Substituting the values in the equation we get,

$KE = \dfrac{1}{2} \times 0.500 \times 15^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.500\times 225 \\[0.5em] = 56.25J$

∴ Potential energy at maximum height = Initial kinetic energy = 56.25J

(b) As we know that,

Kinetic energy on reaching the ground = potential energy at the greatest height

= 56.25J

(c) Total energy at its halfway point = $\dfrac{1}{2}$ (K + U) = 56.25J

#### Question 3

A metal ball of mass 2kg is allowed to fall freely from rest from a height of 5m above the ground.

1. Taking g = 10ms-1, calculate:
1. the potential energy possessed by the ball when it is initially at rest.
2. the kinetic energy of the ball just before it hits the ground?
2. What happens to the mechanical energy after the ball hits the ground and comes to rest?

Given,

Mass = 2 kg

Height = 5m

(a)

(i) the potential energy possessed by the ball when it is initially at rest = mgh = 2 x 10 x 5 = 100J

(ii) As we know that,

The kinetic energy of the ball just before hitting the ground = initial potential energy = mgh = 2 x 10 x 5 = 100J

(b) Mechanical energy of the ball gets converted into heat and sound energy after the ball hits the ground and comes to rest.

#### Question 4

The diagram given below shows a ski jump. A skier weighing 60kgf stands at A at the top of ski jump. He moves from A and takes off for his jump at B. (a) Calculate the change in the gravitational potential energy of the skier between A and B.

(b) If 75% of the energy in part (a) becomes the kinetic energy at B, calculate the speed at which the skier arrives at B.

(Take g = 10 ms-2).

Given,

Mass = 60 kg

(a)

$\text {Loss in potential energy} = \text{mg (h1 – h2)} \\[0.5em] = 60 \times 10\times (75-15) \\[0.5em] = 60 \times10 \times 60 \\[0.5em] = 3.6 \times 10^4 J \\[0.5em]$

(b) When kinetic energy at B is 75% of (3.6 × 104)

$\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \\[0.5em] = 27000J \\[0.5em] = 2.7 \times 10^4 J \\[0.5em]$

Since,

Kinetic energy = $\dfrac{1}{2}$ mv2

Substituting the values in equation we get,

$27000 = \dfrac{1}{2} \times 60 \times v^2 \\[0.5em] \Rightarrow 27000 = 1 \times 30 \times v^2 \\[0.5em] \Rightarrow v^2 = \dfrac{27000}{30} \\[0.5em] \Rightarrow v^2 = 900 \\[0.5em] \Rightarrow v = \sqrt{900} \\[0.5em] \Rightarrow v = 30 \\[0.5em]$

∴ The speed at which the skier arrives at B = 30ms-1

#### Question 5

A hydroelectric power station takes its water from a lake whose water level is 50m above the turbine. Assuming an overall efficiency of 40%, calculate the mass of water which must flow through the turbine each second to produce power output of 1MW. (Take g = 10 m s-2).

Given,

Efficiency = 40%

Work done = 40% of potential energy

$\text{Work done} = 0.4 \times mgh \\[0.5em] = 0.4 \times m \times 10 \times 50 \\[0.5em] = 0.4 \times m \times 500 \\[0.5em] = 200 \times m$

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] 1MW = 200 \times \text {mass of water flowing per second} \\[0.5em] \Rightarrow 1 \times 10^6 = 200 \times m \\[0.5em] \Rightarrow \text m = \dfrac{1 \times 10^6 }{200} \\[0.5em] = 5000$

Hence,

Mass of water which must flow through the turbine = 5000Kg

#### Question 6

The bob of a simple pendulum is imparted a velocity of 5 m s-1 when it is at its mean position. To what maximum vertical height will it rise on reaching at its extreme position if 60% of its energy is lost in overcoming the friction of air?

(Take g = 10 m s-2).

Given,

Energy lost = 0.6 x KE

KE = $\dfrac{1}{2}$ x mass x velocity2

Substituting the values we get,

$KE = \dfrac{1}{2} \times mass \times 5^2 \\[0.5em]$

Amount of energy lost,

$\text{Energy lost} = \dfrac{1}{2} \times mass \times 5^2 \times 0.6 \\[0.5em]$

Amount of energy available,

$\text{Energy available} = \dfrac{1}{2} \times mass \times 5^2 \times 0.4\\[0.5em]$

Applying the rule for the conservation of energy we get,

Kinetic energy available = potential energy

$\dfrac{1}{2} \times mass \times 5^2 \times 0.4 = mass \times gravity \times height\\[0.5em] \Rightarrow \dfrac{1}{2} \times 5 \times 5 \times 0.4 = 10 \times height\\[0.5em] \Rightarrow \dfrac{1}{2} \times 5 \times 2 = 10 \times height\\[0.5em] \Rightarrow 5 = 10 \times height\\[0.5em] \Rightarrow height = 0.5m \\[0.5em]$

∴ Maximum vertical height reached = 0.5m