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#### Page No 28:

#### Question 1:

^{o}with respect to the direction of the applied force.

#### Answer:

a.

Potential Energy |
Kinetic Energy |

It is possessed by a body by virtue of its configuration and position. | It is possessed by a body by virtue of its motion. |

eg: A book kept on a shelf has potential energy. | eg: A moving car posses kinetic energy. |

b. Let the initial velocity of the object be

*u*. Let an external force be applied on it so that it gets displaced by distance

*s*and its velocity becomes

*v*. In this scenario, the kinetic energy of the moving body is equal to the work that was required to change its velocity from

*u*to

*v*.

Thus, we have the velocity−position relation as:

*v*^{2} = *u*^{2} + 2*as*

or

Where, *a* is the acceleration of the body during the change in its velocity

Now, the work done on the body by the external force is given by:

*W** *= *F* × *s*

*F* = *ma *…(ii)

From equations (i) and (ii), we obtain:

If the body was initially at rest (i.e., *u* = 0), then:

Since kinetic energy is equal to the work done on the body to change its velocity from 0 to *v*, we obtain:

c. Let the object be at point A i.e. at height

*h*above the surface of Earth as shown in the figure below.

__At point A__

The object is stationary i.e. its initial velocity,

*u*= 0.

Kinetic energy, $K=\frac{1}{2}m{v}^{2}=\frac{1}{2}m{u}^{2}=0$

Potential energy,

*U = mgh*.....(i)

__At point B__

Let the velocity of the object be

*v*

_{B}and the object has fallen through distance

*x*.

Using third equation of motion, we have

${v}_{B}^{2}=0+2gx\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{B}^{2}=2gx$

Kinetic energy, $K=\frac{1}{2}m{v}_{B}^{2}=mgx$

Potential energy,

*U = mg(h-x)*......(ii)

__At point C__

Let the velocity of the object be

*v*

_{C}and the object has fallen through a distance

*h*.

Using third equation of motion, we have

${v}_{C}^{2}=0+2gh\phantom{\rule{0ex}{0ex}}\Rightarrow {v}_{B}^{2}=2gh$

Kinetic energy, $K=\frac{1}{2}m{v}_{C}^{2}=mgh$

*......(iii)*

Potential energy,

*U =*0

From (i) and (iii), we see that the potential energy of the object at point A has transformed to its kinetic energy at point C. Thus, it can be concluded that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

d. Work done,

*W = F$\times $S*cos $\theta $

Here, $\theta $ = 30

^{o}

Thus,

*W = F$\times $S*cos 30 = $\frac{\sqrt{3}}{2}\times F\times S$

e. Momentum of an object,

*P = mv*

If

*P*= 0

$\Rightarrow v=0$

This is because mass of the object can never be 0.

Now, kinetic energy of the object,

*K*= $\frac{1}{2}\times m\times {v}^{2}$

Since,

*v*= 0

$\Rightarrow K=0$

Hence, when the object has zero momentum, its kinetic energy is also zero.

f. Work done on an object is given as

*W = F$\times $S*cos $\theta $

In circular motion, the direction of force acting on the object is radially inward and the direction of motion of the object is tangential to the circular path at every instant of time. Thus, the angle $\theta $ between the force vector and displacement vector is always 90

^{o }i.e. $\theta $ = 90

^{o}. Hence,

*W = F$\times $S*cos 90 = 0

Hence, the work done on an object moving in uniform circular motion zero.

#### Page No 28:

#### Question 2:

(ii) concentrated

(iii) transformed from one type to another

(iv) destroyed

b. Joule is the unit of ...

(ii) work

(iii) power

(iv) energy

c. Which of the forces involved in dragging a heavy object on a smooth, horizontal surface, have the same magnitude?

(ii) gravitational force

(iii) reaction force in vertical direction

(iv) force of friction

d. Power is a measure of the .......

(ii) amount of energy required to perform the work

(iii) The slowness with which work is performed

(iv) length of time

e. While dragging or lifting an object, negative work is done by

(ii) gravitational force

#### Answer:

a. For work to be performed, energy must be transferred from one place to another.

b. Joule is the unit of work and energy.

c. The gravitational force and the reaction force in vertical direction have same magnitude. Friction is not action as the horizontal surface is smooth.

d. Power is a measure of the rapidity with which work is done

e. While dragging or lifting an object, the negative work is done by frictional and gravitational force, respectively.

#### Page No 28:

#### Question 3:

a. The potential energy of your body is least when you are .....

(ii) sitting on the ground

(iii) sleeping on the ground

(iv) standing on the ground

b. The total energy of an object falling freely towards the ground ...

(ii) remains unchanged

(iv) increases in the beginning and then decreases

c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy ....

(iii) will be 4 times its original energy

(iv) will be 16 times its original energy.

d. The work done on an object does not depend on ....

(ii) applied force

(iii) initial velocity of the object

(iv) the angle between force and displacement.

#### Answer:

a. The potential energy of your body is least when you are __sleeping on the ground__.

b. The total energy of an object falling freely towards the ground __remains unchanged__.

c. If we increase the velocity of a car moving on a flat surface to four times its original speed, its potential energy __will not change__.

d. The work done on an object does not depend on __initial velocity of the object__.

#### Page No 29:

#### Question 4:

1. Take two aluminium channels of different lengths.

Questions

1. At the moment of releasing the balls, which energy do the balls have?

#### Answer:

1. At the moment of releasing the balls, they posses potential energy.

2. The potential energy of the balls converts to kinetic energy as they roll down.

3.** Note: **The question is wrong. It should be "Why do the balls cover same distance** after **rolling down"?

The balls will posses same potential energy because the channels are of same height as well as the weight of the balls are same. So, the balls on reaching the bottom of their respective channels will posses same kinetic energy (following the law of conservation of energy). Thus, the balls will have same velocity at the bottom of the channel. Because of the same velocity, they will cover equal distance on the ground. Thus, both the balls covers equal distance after rolling down the incline because they possessed same potential energy at the top of the channels.

4. The total energy of the ball will eventually be in the form of kinetic energy.

5. Law of conservation of energy is demonstrated using this activity. At the top of the channels, the total mechanical energy of the balls is in the form of potential energy. Now, as the balls roll down, they come into motion. This motion shows that the balls now posses kinetic energy too. As law of conservation of energy states that energy cannot be created nor destroyed, this means this kinetic energy came into existence because of decrease in potential energy of the ball (decrease in potential energy is due to decrease in height). Also, when the ball reaches the bottom, the total mechanical energy of the balls is in the form of kinetic energy. This shows that all the potential energy of the balls got converted to kinetic energy at the bottom.

#### Page No 29:

#### Question 5:

a. An electric pump has 2 kW power. How much water will the pump lift every minute to a height of 10 m?

b. If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?

c. If the energy of a ball falling from a height of 10 metres is reduced by 40%, how high will it rebound?

d. The velocity of a car increase from 54 km/hr to 72 km/hr. How much is the work done if the mass of the car is 1500 kg?

e. Ravi applied a force of 10 N and moved a book 30 cm in the direction of the force. How much was the work done by Ravi?

#### Answer:

a. Power, *P *= 2 kW

$\mathrm{Power},P=\frac{\mathrm{Work}\mathrm{done}\left(W\right)}{\mathrm{Time}\mathrm{taken}\left(T\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow W=P\times T$

For every minute i.e. 60 s, work done by the pump is

$W=2\times 1000\times 60=120000\mathrm{J}$

Now, this work done is stored in water as its potential energy. Thus,

$mgh=120000\mathrm{J}\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{120000}{9.8\times 10}=1224.5\mathrm{kg}$

Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.

b. Total electricity consumed in the month of April is

$E=P\times T\times (\mathrm{Total}\mathrm{days}\mathrm{in}\mathrm{April}\mathrm{month})\phantom{\rule{0ex}{0ex}}E=1200\times 30\times 60\times 30=64.8\times {10}^{6}\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know},\phantom{\rule{0ex}{0ex}}1\mathrm{unit}=3.6\times {10}^{6}\mathrm{J}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}E=18\mathrm{units}\phantom{\rule{0ex}{0ex}}$

c. Energy of the ball at the height of 10 m = *mgh* = 10*mg*

Let the ball rebounds to a height *h*' where the energy reduces by 40%. Thus,

Energy at height *h*' = *mgh*' = 60% of energy at height of 10 m = $\frac{60}{100}\times 10mg=6mg$

or, *mgh*' = 6*mg* = 6 m

d. Here, *v* = 72 km/h = 20 m/s, *u* = 54 km/h = 15 m/s, *m* = 1500 kg

Work done by the car = Change in kinetic energy of the car

i.e.

$W=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}\phantom{\rule{0ex}{0ex}}W=\frac{1}{2}\times m\times {v}^{2}-\frac{1}{2}\times m\times {u}^{2}=\frac{1}{2}\times m({v}^{2}-{u}^{2})\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 1500({20}^{2}-{15}^{2})=131250\mathrm{J}$

e. Work done, *W = F$\times $S*$\times $cos $\theta $

Here, $\theta $ = 0^{o}

*F *= 10 N

*S* = 30 cm = 0.03 m

$\therefore W=10\times .03\times 1=3\mathrm{J}$

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