Haryana State Board HBSE 9th Class Science Solutions Chapter 11 Work and Energy Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Science Solutions Chapter 11 Work and Energy

**HBSE 9th Class Science Work and Energy Intext Questions and Answers**

Questions from Sub-Section 11.1

Question 1.

A force of 7N acts on an object. The displacement is, say 8 m in the direction of the force (Fig. 11.1). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Solution:

Here,

Force (F) = 7N

Displacement (s) = 8m

Work done (W) =?

We know that, Work done (W) = Force (F) × Displacement (s) = 7 × 8N.m = S6 Joule

Questions from Sub-Section 11.1

Question 1.

When do we say that work is done?

Answer:

When an object is displaced by applying force, it is said to be work done.

Thus, work done (W) = Force (F) × Displacement (s)

Question 2.

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

Answer:

When a force is acting on an object in the direction of its displacement. The equation of work will be as follows:

Work done (W) = Force (F) X Displacement (s)

Thus, work done ¡s positive when the force is ¡n the direction of displacement.

Question 3.

Define 1 J of work.

Answer:

Work is said to be done 1 joule when a force of 1 N acting on an object and it is displaced through 1 m in the direction of force.

Thus, 1 J = 1 N x lm

Question 4.

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

Solution:

Here,

Force (F) = 140 N

Displacement (s) = 15 m

Work done (W) = F × S = 140 × 15 = 21001

Questions from Sub-Section 11.2

Question 1.

What is the kinetic energy of an object?

Answer:

The energy possessed by an object due to its motion is called kinetic energy. A body of mass m moving with velocity v has kinetic energy (E) equal to \(\frac {1}{2}\)mv^{2}.

Question 2.

Write an expression for the kinetic energy of an object.

Answer:

The kinetic energy of a body of mass m moving with velocity v is; K.E = \(\frac {1}{2}\) v^{2}

Question 3.

The kinetic energy of an object of mass m moving with a velocity of 5 ms^{-1} is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times ?

Answer:

The kinetic energy of an object of mass, m moving with a velocity of 5 ms^{-1} is 25 J. If its velocity is doubled (10 ms^{-1}), its kinetic energy will increase four (2^{-1}) times i.e. 100 J. But if its velocity is tripled (15 ms^{-1}), its kinetic energy will increase nine (3^{2}) times i.e. 225 J.

Questions from Sub-Section 11.3

Question 1.

What is power ?

Answer:

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

Unit of power is watt (W).

Question 2.

Define 1 watt of power.

Answer:

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

Question 3.

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

Solution:

Here,

Work done (W) = 1000J Time(t) = 10 s Power (P) = ?

We know that,

Work done(W) = 1000 J

Time(t) = 10 s

Power(P) = ?

We know that,

P = \(\frac {W}{t}\) = \(\frac {1000}{10}\) Js^{-1}= 100 W(Watt)

Question 4.

Define average power.

Answer:

The ratio of total energy used to total time given is called average power. Thus,

**HBSE 9th Class Science 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’.

(i) Suma is swimming in a pond.

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

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

(iv) A green plant is carrying out photosynthesis.

(v) A engine is pulling a train.

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

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

Answer:

(i) There is work when suma is swimming in a pond, because she is displacing in the direction of force.

(it) No work is done because its displacement is zero.

(iii) Work is done in lifting the water because it displaces in the direction of force.

(iv) No work is done because displacement is zero.

(v) Work is done in pulling the train because the train is displacing in the direction of engine.

(vi) No work is done because the displacement is zero.

(vii) Work is done by the wind energy in producing motion in sailboat because sail boat is displacing in the direction of force.

Question 2.

An object was thrown at a certain angle to the ground moves in a curved path and falls back to. the ground. The initial and the final point 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:

The work done by the force of gravity on the object is zero because the displacement of the object is zero.

Question 3.

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

Answer:

When a battery lights a bulb, chemical energy is converted in electric energy.

Question 4.

The 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.

Solution:

Here,

Mass of the body (m) = 20 kg

Initial velocity of the body (u) = 5 ms^{-1}

Final velocity of the body (v) = 2 ms^{-1}

Initial kinetic energy of the body (E_{1}) = \(\frac {1}{2}\) mu^{2} = \(\frac {1}{2}\) × 20 × (5)^{2} J = 250J

Final kinetic energy of the body (E_{2}) = \(\frac {1}{2}\)m(v)^{2} = \(\frac {1}{2}\) × 20 × (2)^{2} J = 40 J

Thus, Work done = Change in the Kinetic Enegry = E_{1} – E_{2}

= 250 J – 40J = 210 J

Question 5

A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal. What is the work done on the object by the gravitational force ? Explain your answer.

Solution:

Here,

Mass of the object (m) = 10 kg

Force of gravity (g) = 9.8 ms^{-2}

Height (h) = 0

Work done by force of gravity (w) = mgh = 10 × 9.8 × 0 = 0

Thus, the work done by force of gravity is zero because displacement is horizontal.

Question 6.

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

Answer:

The potential energy of a freely falling object decreases progressively but this does not violate the law of conservation of energy because this energy is transferred in kinetic energy. The total potential energy nearest to earth changes in kinetic energy and on reaching the earth, it converts in potential energy.

Question 7.

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

Answer:

When we ride a bicycle, our muscular energy changes in kinetic energy and we get muscular energy from the food after converting in chemical energy.

Question 8.

Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going?

Answer:

When we push a huge rock with all our might and fail to move it, the transformation of energy in this stage wastes against friction and the work done is considered as zero.

Question 9.

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

Solution:

Here,

Energy consumed = 250 units = 250 kWh = 250 × 1000 × 3600s = 900000000 watt second = 9 × 108J

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 halfway down, (g = 10 ms^{-2} )

Solution:

Here,

The mass of the object (m) = 40 kg

Height of the object from the earth (h) = 5 m

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

Potential energy of the object (w) = mgh

40 × 5 × 10 J = 200J

If the object is allowed to fall freely, the half of the potential energy of the object in the halfway will convert in kinetic energy. Therefore the kinetic energy of the object when it is half-way down = \(\frac {2000}{2}\) = 1000J

Question 11.

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

Answer:

The work done by the force of gravity on a satellite moving around the earth is zero because the displacement of both 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:

Ho, there cannot be displacement of an object in the absense of any force acting on it because the displacement is always due to unbalanced 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:

The work done by the person is zero when he hold a bundle of hay over his head for 30 minutes because the displacement is zero.

Question 14.

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

Solution:

Here,

The power of the electric heater (P) = 1500W

Time (t) = 10 hours

Energy (w) = P × t = 15000 Wh = \(\frac {15000}{1000}\) kWh = 15 k Wh

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.gy ?

Solution:

When we take the bob or pendulum from its middle position to one side (let left side) upto some height, the work done by it is conserved in the form of potential energy. When the bob is allow to swing, then the total potential energy of the bob changes into kinetic energy on reaching left-to-right and middle position. This kinetic energy takes the bob upto that height at which the kinetic energy is changed into potential energy. Due to this potential energy, the bob oscillates again from right to middle. This continuous and bob or pendulum moves from right-left.

After sometime bob comes in rest because frictional force of air works on it and the energy is wasted to work against it. This is not the violation of the law of conservation of energy.

Question 16.

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

Answer:

An object of mass is moving with a constant velocity v, its kinetic energy is \(\frac {1}{2}\) mv^{2}. The work is equal to kinetic energy should be done on the object in order to bring the object to rest.

Question 17.

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

Solution:

Here,

Mass of the car (m) = 1500 kg

Velocity of the car (v) = 60 km/h

= \(\frac{60 \times 1000}{3600}\) ms^{-1} = \(\frac{50}{3}\) ms^{-1}

Thus, kinetic energy of the car = \(\frac {1}{2}\) mv^{2} = \(\frac {1}{2}\) = 1500 × \(\frac {1}{2}\) × \(\frac {1}{2}\)J = 208333.3 J

∴ The Work Required to be done to stop the car = 208333.3 J

Question 18.

In each of the following, a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:

The work done in the first stage is zero, in the second stage is positive and in the third stage is 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:

Yes, we agree with Soni because when several forces are acting on an object, the total effect of the force becomes zero, Thus, F = 0, from this, ma = 0, but m cannot be zero, Therefore, the acceleration can be zero.

Question 20.

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

Solution:

Here,

The total power of four devices (P) = 500 W × 4 = 2000W

Time (t) = 10 hours

The energy consumed = Power × Time

= 2000 W × 10 h

\(\frac {20000}{1000}\) k Wh = 20kWh

Question 21.

A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy ?

Answer:

Its kinetic energy is transferred in potential energy.