These NCERT Solutions for Class 9 Science Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

## Work and Energy NCERT Solutions for Class 9 Science Chapter 11

### Class 9 Science Chapter 11 Work and Energy InText Questions and Answers

Question 1.

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Answer:

When a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:

Work done = Force x Displacement

W = F × S

Where,

F = 7N

S = 8m

Therefore, work done, W = 7 × 8

= 56 Nm

= 56 J

Question 2.

When do we say that work is done?

Answer:

Work is done whenever the given conditions are satisfied:

- A force acts On the body.
- There is a displacement of the body caused by the applied force along the direction of the applied force.

Question 3.

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

Answer:

When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:

Work done = Force × Displacement

W = F × s

Question 4.

Define 1 J of work.

Answer:

1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.

Question 5.

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:

Work done by the bullocks is given by the expression:

Work done = Force × Displacement

W = F × d

Where,

Applied force, F = 140 N

Displacement, d = 15m

W = 140 × 15 = 2100 J

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

Question 6.

What is the kinetic energy of an object?

Answer:

Kinetic energy is the energy possessed by a body by the virtue of its motion. Every moving object possesses kinetic energy. A body uses kinetic energy to do work. Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run windmills, etc.

Question 7.

Write an expression for the kinetic energy of an object.

Answer:

If a body of mass ‘m’ is moving with a velocity v, then its kinetic energy E_{k} is given by the expression,

E_{k} = \(\frac {1}{2}\)mv^{2}

Its SI unit is Joule (J j.

Question 7.

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

Expression for kinetic energy is

E_{K} = \(\frac {1}{2}\)mv^{2}

m = Mass of the object

v = Velocity of the object = 5ms^{-1}

Given that kinetic energy, E_{K} = 25 J

(i) If the velocity of an object is doubled, then v = 5 × 2 = 10 ms^{-1}.

Therefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 4 = 100 J.

(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225 J.

Question 8.

What is power?

Answer:

Power is the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by the expression,

Power = \(\frac{\text { Work }}{\text { Time }}=\frac{\text { Energy }}{\text { Time }}\)

P = \(\frac{\mathrm{W}}{\mathrm{T}}\)

It is expressed in watt (W).

Question 9.

Define 1 watt of power:

Answer:

A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e.,

1W = \(\frac{1 \mathrm{~J}}{1 \mathrm{~s}}\)

Question 10.

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

Answer:

Power is given by the expression,

Work done = \(\frac{\text { Workdone }}{\text { Time }}\)

Work done = Energy consumed by the lamp = 1000 J

Time = 10 s

Power = \(\frac{1000}{10}\) = 100J S^{-1} = 100 W

Question 11.

Define average power.

Answer:

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

Average Power = \(\frac{\text { Total work done }}{\text { Total time taken }}\)

### Class 9 Science Chapter 11 Work and Energy Textbook Questions and Answers

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’.

- Suma is swimming in a pond.
- A donkey is carrying a load on its back.
- A windmill is lifting water from a well.
- A green plant is carrying out photosynthesis.
- An engine is pulling a train.
- Food grains are getting dried in the sun.
- A sailboat is moving due to wind energy.

Answer:

Work is done whenever the given two conditions are satisfied:

(i) A force acts on the body.

(ii) There is a displacement of the body by the application of force in or opposite to the direction of force.

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

(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 the gravitational force to lift water. Hence, work is done by the windmill in lifting water from the well.

(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.

(e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.

(f) Foodgrains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun.

(g) Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.

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:

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions/heights of the object, which is zero.

Work done by gravity is given by the expression,

W = mgh

Where,

h = Vertical displacement = 0

W = mg × 0 = 0 J

Therefore, the work done by gravity on the given object is zero joule.

Question 3.

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

Answer:

When a bulb is connected to a battery, then the chemical energy of the battery is transferred into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:

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:

Kinetic energy is given by the expression,

(Ek)v = \(\frac {1}{2}\)mv2

Where,

Ek = Kinetic energy of the object moving with a velocity, v

m = Mass of the object

(i) Kinetic energy when the object was moving with a velocity 5 m s-1

(Ek)5 = \(\frac {1}{2}\) × 20 × (5)2 = 250J

ii. Kinetic energy when the object was moving with a velocity 2 ms-1

(Ek)2= \(\frac {1}{2}\) × 20 × (2)2 = 40J

Work done by force is equal to the change in kinetic energy.

Therefore, work done by force = (Ek)2 – (Ek)5

= 40 – 250 = -210 J

The negative sign indicates that the force is acting in the direction opposite to the motion of the object.

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:

Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,

W = mgh

Where,

Vertical displacement, h = 0

∴ W = mg × 0 = 0

Hence, the work done by gravity on the body 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. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then 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. During the process, total mechanical energy of the body remains conserved. Therefore, 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:

While riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle. Heat energy heats the rider’s body. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:

Muscular Energy → Kinetic Energy + Heat Energy

During the transformation, the total energy remains conserved.

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, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy, which causes our body to become hot.

Question 9.

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

Answer:

1 unit of energy is equal to 1 kilowatt hour (kWh).

1 unit = 1 kWh

1 kWh = 3.6 × 10^{6} J

Therefore, 250 units of energy = 250 × 3.6 × 10^{6} = 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:

Gravitational potential energy is given by the expression,

W = mgh Where,

h = Vertical displacement = 5 m

m = Mass of the object = 40 kg

g = Acceleration due to gravity = 9.8 ms^{-2}

∴ W = 40 × 5 × 9.8 = 1960 J.

At half-way down, the potential energy of the object will be \(\frac{1960}{2}\) = 980 J.

At this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.

Question 11.

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

Answer:

Work is done whenever the given two conditions are satisfied:

- A force acts on the body.
- There is a displacement of the body by the application of force in or opposite to the direction of force.

If the direction of force is perpendicular to displacement, then the work done is zero.

When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Hence, the work done on the satellite by the Earth is zero.

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. For a uniformly moving object.

Suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.

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 done whenever the given two conditions are satisfied:

- A force acts on the body.
- There is a displacement of the body by the application of force in or opposite to the direction of force.

When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.

Question 14.

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

Answer:

Energy consumed by an electric heater can be obtained with the help of the expression,

P = \(\frac{W}{T}\) Where,

Power rating of the heater, P = 1500 W = 1.5 kW

Time for which the heater has operated, T = 10 h

Work done = Energy consumed by the heater

Therefore, energy consumed = Power × Time

= 1.5 × 10 = 15 kWh

Hence, the energy consumed by the heater in 10 h is 15 kWh.

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 be neither created nor destroyed. It can only be converted from one form to another.

Consider the case of an oscillating pendulum.

When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates.

The bob does not oscillate forever. It comes to rest because air resistance resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time.

The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain conserved.

Question 16.

An object of mass, m is moving with a constant velocity, v. How much work should be done oh the object in order to bring the object to rest?

Answer:

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

E_{k} = \(\frac {1}{2}\)mv^{2}

To bring the object to rest, \(\frac {1}{2}\)mv^{2} amount of work is required to be done on the object.

Question 17.

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

Answer:

Kinetic energy, Ek = \(\frac {1}{2}\)mv^{2}

Where,

Mass of car, m = 1500 kg

Velocity of car, v = 60 km/h = 60 × \(\frac {5}{18}\)ms-1

∴ Ex = \(\frac {1}{2}\) × 1500 × \(\left(60 \times \frac{5}{18}\right)^{2}\) =20.8 × 104J

Hence, 20.8 × 104 J of work is required to stop the car.

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:

Work is done whenever the given two conditions are satisfied:

- A force acts on the body.
- There is a displacement of the body by the application of force in or opposite to the direction of force.

Case I:

In this case, the direction of force acting on the block is perpendicular to the displacement.

Therefore, work done by force on the block will be zero.

Case II:

In this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.

In this case, the direction of force acting on the block is in the direction of displacement.

Therefore, work done by force on the block will be positive.

Case III:

In this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, 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. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

Question 20.

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

Answer:

Energy consumed by an electric device

can be obtained with the help of the expression for power,

P = \(\frac{W}{T}\)

Where,

Power rating of the device, P = 500 W = 0.50 kW

Time for which the device runs, T = 10 h

Work done = Energy consumed by the device

Therefore, energy consumed = Power × Time

= 0.50 × 10 = 5 kWh

Hence, the energy consumed by four equal rating devices in 10 h will be 4 × 5 kWh = 20 kWh = 20 Units.

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 gets converted into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. It can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.

### Class 9 Science Chapter 11 Work and Energy Additional Important Questions and Answers

Multiple Choice Questions

Choose the correct option:

Question 1.

When a body falls freely towards the earth, then its total energy

(a) increases

(b) decreases

(c) remains constant

(d) first increases and then decreases

Answer:

(c) remains constant

Question 2.

A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car

(a) does not change

(b) becomes twice to that of initial

(c) becomes 4 times that of initial

(d) becomes 16 times that of initial

Answer:

(a) does not change

Question 3.

In case of negative work the angle between the force and displacement is

(a) 00

(b) 450

(c) 900

(d) 1800

Answer:

(d) 1800

Question 4.

An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same

(a) acceleration

(b) momenta

(c) potential energy

(d) kinetic energy

Answer:

(a) acceleration

Question 5.

A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 ms^{-2})

(a) 6 × 103 J

(b) 6J

(c) 0.6 J

(d) zero

Answer:

(d) zero

Question 6.

Which one of the following is not the unit of energy?

(a) joule

(b) newton metre

(c) kilowatt

(d) kilowatt hour

Answer:

(c) kilowatt

Question 7.

The work done on an object does not depend upon the

(a) displacement

(b) force applied

(c) angle between force and displacement

(d) initial velocity of the object

Answer:

(d) initial velocity of the object

Question 8.

Water stored in a dam possesses

(a) no energy

(b) electrical energy

(c) kinetic energy

(d) potential energy

Answer:

(d) potential energy

Question 9.

A body is falling from a height h. After it has fallen a height h/2, it will possess

(a) only potential energy

(b) only kinetic energy

(c) half potential and half kinetic energy

(d) more kinetic and less potential energy

Answer:

(c) half potential and half kinetic energy

Very Short Answer Type Questions

Question 1.

A boy climbs up the staircase to reach to second, floor from first floor. Has he done any work?

Answer:

Yes, he has done work against the gravitational force by climbing up.

Question 2.

In the following activities, has work been done or not? Why?

(i) A girl pulls a trolly and trolley moves.

(ii) A book is lifted through a height ‘h’.

Answer:

(i) When a girl pulls a trolley, work is done because trolley experiences a displacement from its mean position.

(ii) When a book is lifted through a height ‘h’ because book has gained displacement in its height from the ground level.

Question 3.

Is work a scalar or vector quantity?

Answer:

Although work done = Force × Displacement, yet it is taken as scalar quantity.

Question 4.

What is the work done in lifting an object of mass ‘m’ through a height ‘h’ from the ground?

Answer:

Force acting on body, F = m × g

Displacement = h.

Therefore, workdone = Force × Displacement

= mg × h = mgh

Question 5.

What is the angle between the line of force and the direction of displacement for maximum work?

Answer:

With W = F.s. cos θ, θ = 0 for the maximum work.

Question 6.

Define energy. What is the S.I. unit?

Answer:

Energy is the capacity to do work. Its S.I. unit is joule.

Question 7.

Which physical quantity has the unit Js^{-1} ?

Answer:

Power.

Question 8.

Which of the two out of work, power and energy have same unit?

Answer:

Work and energy.

Question 9.

What are the two forms of mechanical energy?

Answer:

The two forms of mechanical energy are kinetic energy and potential energy.

Question 10.

State the relation in kinetic energy of a body in motion with respect to its (i) mass and (ii) velocity.

Answer:

- Kinetic energy of a body is directly proportional to its mass.
- Kinetic energy of a body is directly proportional to the square of its velocity.

Question 11.

State the relation in potential energy of a body at rest with respect to its mass and height.

Answer:

Potential energy of a body is directly proportional to both the mass and height.

Question 12.

What is the relationship between Watt and Horsepower?

Answer:

1 Horsepower = 746 Watt.

Question 13.

The work done by the heart for each beat is 1J. Calculate the power of the heart if it beats 72 times in a minute.

Answer:

Total work done in minute = 1J × 72 = 72J

Time = 1 minute = 1 × 60 = 6 seconds

Power = \(\frac{\text { Work done }}{\text { Time }}\) = \(\frac {72}{60}\) =1.2 Watt

Question 14.

What type of energy transformation takes place in (i) dynamo and (ii) Electric motor?

Answer:

(i) Dymamo—Mechanical energy into electrical energy,

(ii) Electric motor—Electrical energy into mechanical energy

Question 15.

Which type of energy is possessed by stretched rubber band?

Answer:

Elastic potential energy

Question 16.

A rubber ball and a leather cricket ball are moving with same velocity. Which will have greater kinetic energy? Why ?

Answer:

The leather cricket ball will have greater kinetic energy because of its large mass as compared to rubber ball.

Question 17.

Which will require more force to be pushed forward, a truck in rest or a slowly rolling truck?

Answer:

A truck in state of rest would require more force because of greater static friction than rolling friction.

Question 18.

What happens to kinetic energy of a body in motion when it comes to rest?

Answer:

Kinetic energy of body-in motion gets transformed into its potential energy when brought to rest while some is lost in overcoming friction as heat.

Short Answer Type Questions

Question 1.

A student when preparing for examinations draws diagram, read hooks and discuss problems with his friends. In the process does he do any work?

Answer:

No work is done by the student in reading or discussing problems with his friends as long as he remains sitting at one position to avoid any displacement.

Question 2.

What is work, a scalar or vector quantity? why?

Answer:

Work is a scalar quantity because on the application of force, the work done is in the direction of force. Hence, there is no change in the direction for work to be called a vector quantity.

Question 3.

Derive an expression for the work done on an object when the force is acting on the body in a direction making an angle ‘θ’ with the direction of displacement of the object.

Answer:

Let a force F acts on a body of mass ‘m’ by making an angle ‘θ’ with the direction of the displacement of the object.

The component of the force in direction of the displacement = F cos θ

Let displacement of body = s

Work done on the object = (F cos θ) × s = F × s cos θ

Question 4.

What is the workdone against force of gravity if a person carries a 30 kg load on his head?

Answer:

Work done against force of gravity W = F. s. cos θ

With θ = 90° and cos θ = 0

W = 0 × s = 0

Hence, total work done will be zero.

Question 5.

What do you mean by negative work and positive work? Give example.

Answer:

When a force acts on a body in a direction opposite to the direction of displacement

Work done = – F cos θ × s

The negative sign indicates that work done is negative.

When a force acts on a body in a direction of the displacement of body, the work done is positive.

When a bullock pulls a cart, the work done is called positive while when a player tries to stop a moving ball, the work done is called negative.

Question 6.

If a ball tied to a string is whirled in a circular orbit with centre being the hand holding the free end of string, then what is the work done on the ball?

Answer:

The force acting on the ball is along the radius of the circular path i.e. the direction of the force is perpendicular to the direction of motion of the ball.

Therefore, work done,

W = F. cos θ × s

= O × s = 0

Hence, work done on the ball is zero.

Question 7.

Take a toy car, wind it using key. Place the car on the ground and answer the following questions:

(i) Did the car move? Why?

(ii) Does the energy acquired by the car depends upon the number of times you turn the key?

(iii) From where did it get energy?

Answer:

(i) The car would move because the potential energy acquired by the car, due to widing of the spring of car got transformed into its kinetic energy.

(ii) The potential energy gained by the car depends upon the number of turns of the widing of the key with increasing number of turns, the greater potential energy is acquired by the car. This acquired potential energy later changes into car’s kinetic energy.

(iii) Car got the energy to move freon the spring inside it.

Question 8.

Two bodies having equal masses are kept at a height of 20m and 30m respectively. What is the ratio of their potential energy?

Answer:

Ratio of potential energies = mgh_{1} : mgh_{2}

mg × 20 : mg × 30

= 20 : 30

= 2 : 3.

Question 9.

State the type of energy changes taking place in the following:

(i) A body is thrown up in air.

(ii) In a green leaf during photosynthesis.

(iii) In an oscillation of a pendulum.

Answer:

(i) When a body is thrown in upward direction, the muscular energy which is chemical potential energy is transformed into the kinetic energy of the sail. As the ball keeps rising, kinetic energy continues to change into potential energy. When at the maximum height, the body has no kinetic energy but the maximum potential energy.

(ii) In a leaf during photosynthesis, solar energy changes into chemical potential energy.

(iii) An oscilating pendulum has the maximum potential energy at its peak position where it temporarily come to state of rest while at its mean position, it has maximum kinetic energy.

Question 10.

Suppose a hammer which falls purely on a nail placed on a 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?

Answer:

Mass of the hammer = 1 kg

Height of the hammer, h = 1m

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

Potential energy at the highest point = mgh = 1 × 10 × 1 = 10 J

When the hammer falls, its whole potential energy would be converted into kinetic energy.

Kinetic energy just before hitting the nail 10 J.

Question 11.

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

Answer:

Mass of the man, m = 50 kg

Number of the stairs = 30

Height of one stair = 20 cm = 0.20 m

Total height of stairs = 30 × 20 = 6 m

Workdone in climbing the stairs=m × g × h

=50 × 10 × 6 = 3000 J

Tune taken by man to climb up 30 stairs =30

Power used inclimbing stairs = \(\frac {3000}{30}\) = 100 w

Question 12.

A porter lifts a luggage of 15 kg from the ground and put it on his head 2 m above the ground. Calculate the work done by him on the luggage Tig = 10 ms^{-2})

Answer:

Mass of luggage, m = 15 kg

Height or displacement, h = 2m

Work done by porter = m.g × h

= 15 × 10 × 2 = 300 J

Question 13.

A force of 10 N acts on an object in the direction of displacement. If the displacement of the object is 2 m. Calculate the work done by the force.

Answer:

Force applied, F = 10 N.

Displacement, s = 2 m

Work done = F × S = 10 × 2 = 20 N.

Question 14.

An object of mass 15 kg is moving with a constant velocity of 4 ms^{-1}. What is the kinetic energy possessed by the body?

Answer:

Mass of the object, m = 15 kg.

Velocity of the object, v = 4 ms^{-1}

Kinetic energy possessed by the body = ½ mv^{2}.

= ½ × 15 × (4)^{2} = ½ × 15 × 16= 120 J

Question 15.

What is the work done to increase the velocity of a car from 30 km h^{-1} to 60 km h^{-1} if the mass of the car is 1500 kg ?

Answer:

Mass of the car, m = 1500 kg

Work done = change in kinetic energy

= ½ m (v^{2} – u^{2})

Question 16.

Find the energy present in any object of mass 10 kg when it is at a height of 6m above the ground. (Take g = 10 ms^{-2}).

Answer:

Mass of the object, m = 10 kg

Height, h = 6 m

g = 10 ms^{-2}

Potential energy possessed by the object = mg × h

= 10 × 10 × 6 = 600 J

Question 17.

An object of mass 12 kg is at a certain height above the ground. If the gravitational potential energy of the object is 480 J. Find the height at which the object is with respect to the ground, (Give g =10 ms^{-1}).

Answer:

Mass of the object, m = 12 kg

Potential energy, P.E. = 480 J

Let height of the object from ground = h

Then P.E. = m × g × h

h = \(\frac{\mathrm{P.E}}{m \times g}=\frac{480}{12 \times 10}=4 m\)

Question 18.

A woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10s. Calculate the power used by her. (Take g = 10 ms^{-2}).

Answer:

Mass of the bucket, m = 5 kg

Height, h = 10 m

g = 10 ms^{-2}

Energy consumed in pulling the bucket = m × g × h

= 5 × 10 × 10 = 500 J

Time taken = 10s

Question 19.

Two girls A and B each of weight 400 N climb up a rope to a height of 8m. Girl A takes 20 seconds while girl B takes 50 seconds to accomplish their task. What is the power spent by each girl?

Answer:

(i) Weight of the girl ‘A’ = 400 N.

Displacement (height) h = 8m

Work done by girl A = mg × h

=400 × 8 = 3200 J

Time taken by girl A, t = 20 seconds

Power spent by girl A,

= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 160 W

(ii) Weight of the girl B, = 400 N

Displacement (height), h = 8 m

= 400 × 8 = 3200 J

Time taken, t = 50 seconds

Power spent by girl B

= \(\frac{\text { Workdone }}{\text { Time taken }}=\frac{3200}{20}\) = 64 W

Question 20.

A boy of 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step is 15 cm. Calculate the power of the boy.

Answer:

Mass of the boy, m = 50 kg

Height of each step = 15 cm = 0.15 m

Number of steps in stair case = 45

Total height of stair case = 45 × 0.15 = 6.75 m

Time taken to climb, t = 9 seconds

Question 21.

A boy pulls a toy car with a force of SON through a string which makes an angle of 30° with the horizontal so as to move the toy 2 m horizontally. Calculate the work done by the boy on the toy car.

Answer:

Force applied by the boy = F = SON

Angle of direction of force with the direction of displacement, θ = 30°

Displacement, S = 2m

Work done = F × s cos θ

Work done = 50 × 2 cos 30°

= 50 × 2 × \(\frac{\sqrt{3}}{2}\) = 50\(\sqrt{3}\)J

= 50 × 1.732 = 86.5 J

Question 22.

Calculate the velocity of an object of mass 100 g moving with a kinetic energy of 200 J.

Answer:

Mass of the object = 100g = 0.1 kg

Kinetic Energy (K.E.) = 20 J

Let velocity of the object = v

K.E. = ½ mu^{2}

20 = ½ × 0.1 × v^{2}

or v^{2} = 400

v = 20 ms^{-1}

Question 23.

Calculate the power of a pump which can lift 200 kg of water to store it in a tank at height of 19m in 25 seconds. (Take g = 10 ms^{-2}).

Answer:

Mass of water, m =200 kg

Height, h = 19 m

Time taken to lift the water, t = 25 s

Power of the pump = \(\frac{\text { Workdone }}{\text { Time taken }}\)

= \(\frac{m \times g \times h}{t}=\frac{200 \times 10 \times 19}{25}\)

= 1520 W

Question 24.

Calculate the power of a pump in horse power (H.P.) which can lift 600 kg of water into the water tank at a height of 40m in 10 minutes.

Answer:

Mass of water m =600 kg

Height, h =40 m

g = 10ms^{-2}

Time taken by pump, t = 10 min = 600 s

Power of the pump = \(\frac{\text { Work done }}{\text { Time taken }}=\frac{m g}{600}\)

= \(\frac{600 \times 10 \times 40}{600}\) = 400 W

or = \(\frac {400}{746}\)H.P = 0.52H.P. (∵ 1H.P. = 746W)

Question 25.

An object of mass 2 kg is thrown vertically upward with an initial velocity of 20m/ s. What will-be the potential energy at the highest point (Take g= 10 ms^{-2}).

Answer:

Given u =20 m/s

g = -10 ms^{-2}

v = 0

Let maximum height of the object = h

We know that

v^{2} – u^{2} = 2gh

h = \(\frac{v^{2}-u^{2}}{2 g}=\frac{0-(20)^{2}}{2 \times(-10)}=\frac{-400}{-20}\) = 20 m

Potential energy of the object = mgh.

= 2 × 10 × 20 = 400 J

Long Answer Type Questions

Question 1.

Define kinetic energy and potential energy ? Give some examples in each case.

Answer:

Kinetic energy: The energy possessed by an object by virtue of its motion is known as its kinetic energy.

Kinetic energy (E_{k}) = ½ mv^{2}.

Kinetic energy is directly proportional to it is mass and its is directly proportional to the square of its velocity.

Example:

- The energy possessed by a moving wind can move the blade of a wind-mill due to its kinetic energy.
- The flowing water also have kinetic energy which can make the water-mill to work.
- The speeding car, moving bullet, rotating wheel all have kinetic energy and can do work.

Potential Energy: The energy posessed by a an object by virute of its position or change in configuration is known as its potential energy.

Potential energy =m × g × h

where m = mass of the object, h = height, g = acceleration due to gravity.

Example:

- When a body is lifted to a certain height against the force of gravitation, the work done on the body is stored in the body in the form of potential energy.
- The water stored at a height in a dam possesses potential energy
- The energy possessed b a stretched rubber or stretched spring or compressed spring is also an example of potential energy.

Question 2.

What is gravitational potential energy and elastic potential energy? Give examples.

Answer:

(i) Gravitational potential energy : When a body is lifted to a certain height then die work done in raising the object from the ground to that point against gravity is stored in the body and is called gravitational potential energy.

The graviational potential energy of the object = m.g.h

The work done by the gravity depends upon the difference in height of initial position and final position and not on the path along which object is moved.

Example: Any object lifted from the ground level or any other reference level possess potential energy.

(ii) Elastic potential energy : The energy possesed by an object due to change in its shape is known as elastic potential energy It is mainly associated with the compression or extension of an object.

Example :

- A stretched spring or rubber possesses elastic potential energy.
- A compressed spring also possesses elastic kinetic energy
- Derive an expression for the kinetic energy of an object of mass ‘m’ moving with the velocity V.

Or

Derive the expression of kinetic energy (Ek = ½ mv^{2}, where m is the mass of the object and v is the constant velocity with which the body is moving.

Answer:

The kinetic energy of an object is the energy possessed by the object by virtue of its motion. The kinetic energ of a body moving with a vertain velocity is equal to the work done on it to make it acquire that velocity from rest.

Consider an object of mass ‘tri at rest. Letitbe displaced through a distance ‘S’ when a constant force ‘F acts on its in the direction of displacement. The work done is given by

W = F × S ……..(i)

The work done on the object will cause a change in its velocity. Let velocity of the object changed from zero (position of rest) to v. Let ‘a’ be the acceleration produced in the body, then

According to Newtons second law of motion F = m × a …….(ii)

But we know that

v^{2} – u^{2} = 2as

or s = \(\frac{v^{2}-u^{2}}{2 a}\)

Object starts moving from rest, therefore u = 0

or s = \(\frac{v^{2}}{2 a}\) …….(iii)

Now substituting the value of ‘F’ and ‘s’ from equation (ii) and (iii) in equation (i) we get

W = m × a\(\left(\frac{v^{2}}{2 a}\right)\)

= ½ mv^{2}

Now, the work done is stored in the form of kinetic energy of the object.

Hence, kinetic energy (E_{k}) = ½ mv^{2}

Question 4.

What is the law of conservation of energy ? Prove it when an object falls freely from a certain height under the force of gravity.

Answer:

Law of conservation of energy: Energy can neither be created nor destroyed but it can be changed from’one form into another form. The total energy after and before the transformation always remains constant i.e. total energy of a closed system remains constant.

Let an object of mass ‘m’ is dropped freely from ascertain height ‘h’ from the ground.

(i) The potential energy of the body at point A = mgh

Kinetic energy at A = 0 (Because velocity is zero at A).

Total energy at point A (highest point)

= mgh + 0 mgh

(ii) After a certain interval of time it reaches at B. It moves through a distance V due to force of gravitation in downward direction.

The potential energy at B = m.g. (h – x)

Let the velocity of the body at B = v

We know that v^{2} – u^{2} = 2 gs

v^{2} = 2g(x)(u = 0)

Kinetic energ at B = ½ mv^{2}

= ½ m(2gx)

= mgx

Total energy at B = Potential energ + kinetic energy

= m.g. (h – x) + mgx

= m.g.h. – mgx + mgx

= mgh.

(iii) As the fall continues, the potential energy goes on changing into kinetic energy. The potential energu would decrease while kinetic energy would increase, when the object is about the reach the ground then, h – 0.

The potential energy at C = mg × h = mg × 0 = 0.

Let v be the velocity of the object just before reaching the ground.

u = 0, s = 0

v^{2} – u^{2} = 2gh

or v^{2} – 0 = 2gh

Kinetic energy of the object = ½ mv^{-1}

= ½ m (2gh) – mgh

Total energy = mgh + 0 = mgh.

The total energy i.e. the sum of potential energy and kinetic energy at all the three point is same. Similarly, the total energy at any point will be the same. This proves the law of conservation of energy.