# Question-1

**When do we say that work is done?**

**Solution:**

Work is done whenever a force acts on a body and the body moves in the direction of the force.

# Question-2

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

**Solution:**

Work done = Force x Displacement

# Question-3

**Define 1 J of work**

**Solution:**

1 J of work is said to be done, when a force of one Newton displaces a body through a distance of 1 metre in its own direction.

# Question-4

**A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?**

**Solution:**

Here, Force (F) = 140 N, Displacement (S) = 15m

Work done (W) = FS = 140 N x 15 m =2100 J

# Question-5

**A force of 7 N acts on an object. The displacement is 8 m in the direction of the force. Let us assume that the force acts on the object through the displacement. What is the work done in this case?**

**Solution:**

Work done = Force x displacement

= 7 N x 8 m = 56 Nm = 56 J

# Question-6

**What is the kinetic energy of an object?**

**Solution:**

The energy possessed by an object by virtue of its motion is called its kinetic

energy.

# Question-7

**Write an expression for the kinetic energy of an object.**

**Solution:**

The Kinetic energy possessed by an object of mass m, moving with a uniform velocity v is E

_{k }= ½ mv

^{2}

# Question-8

**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?**Solution:**

Initial Kinetic energy = 25 J

Now, E

_{k }∝ v

^{2}

When velocity is doubled, Kinetic energy becomes

E

_{k }= (2)

^{2}x Initial K.E. = 4 x 25 = 100 J.

When velocity is tripled, Kinetic energy becomes

E

_{k}= (3)

^{2}x Initial K.E. = 9 x 25 = 225 J.

# Question-9

**What happens to the kinetic energy of an object when its velocity is doubled?**

**Solution:**

The Kinetic energy becomes four times its initial value when the velocity of the object is doubled because E

_{k }∝

_{ }v

^{2 }

# Question-10

**What is power?**

**Solution:**

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

# Question-11

**Define 1 watt of power.**

**Solution:**

The power of an agent is one watt if it does work at the rate of 1 joule per second.

# Question-12

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

**Solution:**

P = = 100 Js

^{-1 }=

^{ }100 W.

# Question-13

**Define average power.**

**Solution:**

The average power of an agent may be defined as the total energy consumed by it in the total time taken.

Average power =

# Question-14

**Suppose that a hammer, which falls freely on a nail placed on a piece of wood, has a mass of 1 kg. If it falls from a height of 1m, how much kinetic energy will it have just before hitting the nail? (Take value of g = 10 m/s**

^{2})**.**

**Solution:**

Mass of the hammer, m = 1 kg

Height, h = 1 m

We know that,

v

^{2}– u

^{2}= 2gs

v

^{2}– 0 = 2 × 10 × 1

v

^{2}= 20

Kinetic energy = × m × v

^{2}

= × 1 × 20 = 10 joule.

# Question-15

**Suppose that a car is moving with uniform velocities: 18 km/h, 36 km/h, 54 km/h and 72 km/h at some intervals. Calculate the kinetic energy of the boy sitting in the car at these velocities. Plot a graph between the kinetic energy and the velocity. What is the nature of the curve?**

**Solution:**

Mass of the boy = 40 kg

18 km/h = = 5 m/s

Kinetic energy = mv

^{2}= × 40 × (5)

^{2}

= 500 joule

36 km/h = = 10 m/s

Kinetic energy = mv

^{2}= × 40 × (10)

^{2}

= 2000 joule

54 km/h = = 15 m/s

Kinetic energy = mv

^{2}

= × 40 × (15)

^{2}

= 4500 joule

72 km/h = = 20 m/s

Kinetic energy = mv

^{2}= × 40 × (20)

^{2}

= 8000 joules

Graph between the kinetic energy and the velocity:

# Question-16

**A man whose mass is 50 kg, climbs up 30 steps in 30 seconds. If each step is 20 cm high calculate the power used in climbing the stairs.**

**Solution:**

Mass of the man = 50 kg

Number of steps = 30

Height of each step = 20 cm = 0.2 cm

Work done = mgh

= 50 × 10 × 6

= 3000 Joule

Power =

= 100 J/s or 100 watt.

# Question-17

**A mass of 10 kg is dropped from a height of 50 cm, find**

(a) its kinetic energy and

(b) velocity

Just as it reaches the ground, does the velocity depend on the mass of the particle? Explain.

(a) its kinetic energy and

(b) velocity

Just as it reaches the ground, does the velocity depend on the mass of the particle? Explain.

**Solution:**

Mass of the body, m = 10 kg

Height, h = 50 cm = 0.5 m

Initial velocity, u = 0

We know that,

v

^{2}– u

^{2}= 2gs

v

^{2}– (0)

^{2}= 2 × 10 × 0.5

v

^{2}= 10

v = m/s

Kinetic Energy = mv

^{2}

= × 10 × 10 = 50 Joule.

# Question-18

**When you compress a coil spring, you do work on it. The elastic potential energy:**

(i) increases

(ii) decreases

(iii) disappears

(iv) remains unchanged.

(i) increases

(ii) decreases

(iii) disappears

(iv) remains unchanged.

**Solution:**

(i) Increases.

# Question-19

**If you apply 1 J of work to lift a book of 0.5 kg, how high will it rise?**

**Solution:**

Work done = 1 J

Mass of the book m = 0.5 kg

Acceleration due to gravity g = 10 m/s

^{2}

We know that

Work done = m × g × h

1 = 0.5 × 10 × h

∴ h = = 0.2 m.

# Question-20

**If an electric bulb of 100 watt is lighted for 2 hours, how much electric energy would be consumed?**

**Solution:**

Electric energy consumed = 100 watt × 2 hours

= 200 watt hour

= kilowatt hour

= 0.2 unit.

# Question-21

**Distinguish between work, energy and power. State the S.I. units for each of these quantities.**

**Solution:**

Work | Energy | Power |

Work is done where a force F acting a body produces a displacement in it. | The capacity of a body to do work is called its energy. | The rate of doing work is called power. |

The S.I. unit of work is joule. | The S.I. unit of energy is joule (J). | The S.I. unit of power is watt (W). |

# Question-22

**A woman pulls a bucket of water of total mass 5 kg from a well, which is 10 m deep in 10 sec. Calculate the power used by her.**

**Solution:**

Mass of the bucket of water, m = 5 kg

Depth of the well, h = 10 m

Acceleration due to gravity, g = 10 m/s

^{2}

Work done by the woman = mgh

= 5 × 10 × 10

= 500 Joule

Time taken t = 10 s

Power =

= = 50 J/s or 50 watt.

# Question-23

**About how many kg of boiled potatoes would you have to eat to supply energy for half-hour of swimming? Assume that your body utilizes only 20% of the total energy stored.**

**Solution:**

Energy used up in swimming in 1 minute = 25.6 kJ.

Energy used up in swimming in 30 minutes = 25.6 × 30 = 768 kJ

Energy given by 1 kg cooked potatoes = 3.7 × 10

^{6}J

Our body utilizes only 20% of total energy stored.

∴ = 0.74 × 10

^{3}kJ

740 kJ energy is obtained by 1 kg potatoes

∴ The energy obtained by 768 kJ = kg

= 1.4 kg.

# Question-24

**If a stone of mass m falls a vertical distance d, the decrease in gravitational potential energy is**

(i) mgh (ii) md

(i) mgh (ii) md

^{2}/2, (iii) mgd (iv) md/g.**Solution:**

Mass of the stone = m

Vertical Distance, i.e height h = d

Potential energy = mgh

= mgd

Hence (iii) is correct.

# Question-25

**No work is done when:**

(i) a nail is plugged into a wooded board.

(ii) a box is pushed along a horizontal floor.

(iii) there is no component of force parallel to the direction of motion.

(iv) there is no component of force normal to the direction of force.

(i) a nail is plugged into a wooded board.

(ii) a box is pushed along a horizontal floor.

(iii) there is no component of force parallel to the direction of motion.

(iv) there is no component of force normal to the direction of force.

**Solution:**

(iii) There is no component of force parallel to the direction of motion.

# Question-26

**A rocket of 3 × 10**

^{6}kg mass takes off from a launching pad and acquires a vertical velocity of 1 km/h at an altitude of 25 km. Calculate (a) the potential energy and (b) kinetic energy. (take the value of g = 10 m/s^{2})**.**

**Solution:**

Mass of the rocket, m = 3 × 10

^{6}kg

Velocity = 1 km/s = 1000 m/s

Altitude, h = 25 km = 25 × 1000 m

= 25000 m

(a) Potential energy = mgh

= 3 × 10

^{6}× 10 × 25000

= 75 × 10

^{10}Joule.

(b) Kinetic energy = mv

^{2}

= × 3 × 10

^{6}× (1000)

^{2}

= 1.5 × 10

^{12}Joule

# Question-27

**A horse exerts a pull on a cart of 300 N so that the horse-cart moves with a uniform speed of 18 km/h on a level road. Calculate the power developed by the horse in watt and also find its equivalent in horse power.**

**Solution:**

Force that the horse exerts F = mg = 300 N

Velocity v = 18 km/h = = 5 m/s

Work done = mgh

= 300 × 5 = 1500 watt

746 watt = 1 horse power

1500 watt = = 2 horse power.

# Question-28

**Potential energy of your body is minimum when:**

(a) you are standing

(b) you are sitting on a chair

(c) you are sitting on the ground

(d) you lie down on the ground

(a) you are standing

(b) you are sitting on a chair

(c) you are sitting on the ground

(d) you lie down on the ground

**.**

**Solution:**

(d) you lie down on the ground.

# Question-29

**A girl sits and stands repeatedly for 5 minutes. Draw a graph to show the variation of potential energy of her body with time.**

**Solution:**

# Question-30

**A boy pulls a toy with a force of 50 N through a string, which makes an angle of 30°**

**with the horizontal so as to move the toy by a distance of 1m horizontally. If the string were inclined making an angle of 45° with the horizontal, how much pull would he apply along the string in order to move it the same distance of 1m?**

**Solution:**

Force F = 50 N

Angle θ = 30°

Distance moved d = 1m

W = F cos θ × 1

= 50 × cos 30° × 1

= 50 × 0.86 × 1

= 43 J

W = F cos θ × 1

43 = F cos θ

∴ F =

= = 60.8 N