NCERT Solutions for Class 9 Science Chapter 10 - Work and Energy

Q1. 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:

(i) Work done by Suma is negative as the force and displacement are in the opposite direction.

(ii) Work done is zero as the gravity on the load is acting vertically downward, whereas its displacement is in a horizontal direction.

(iii) Work done is positive as both force and displacement are in an upward direction.

(iv) Work done is zero as there is no displacement involved.

(v) Work done is positive as force is acting in the direction of the motion.

(vi) Work done is zero as there is no displacement of the grains.

(vii) Work done by wind force is positive as it supports the motion of the boat.

Q2. 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:

We can see that there is no net displacement in the vertical direction. It has only horizontal displacement. Thus, work done by vertical forces on the stone will be zero.

The force of gravity is acting vertically downward; thus the work done by it is zero.

Q3. A battery lights a bulb. Describe the energy changes involved in the process.

Answer:

The chemical energy stored in a battery is transformed into the heat energy which glows the filament of the bulb. Further, the heat energy is converted into light energy.

Q4. Certain force acting on a $20 \; kg$ mass changes its velocity from $5\; m\; s ^{-1}$ to $2\; m\; s ^{-1}$ . Calculate the work done by the force.

Answer:

By the equations of motion, we can write :

$v^2\ =\ u^2\ +\ 2as$

$s\ =\ \frac{v^2\ -\ u^2}{2a}$

$=\ \frac{2^2\ -\ 5^2}{2a}\ =\ \frac{-21}{2a}\ m$

The work done is :

$W\ =\ F.s$

$=\ 20a\times \frac{-21}{2a}$

$=\ -210\ J$

Thus work done is - 210 J.

Q5. 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:

The displacement of the object is horizontal on the table. We know that the gravitational force is acting in a downward direction. There is no displacement vertically.

Thus the work done by the gravitational force is zero.

Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Answer:

No, there is no violation of the law of conservation of energy. At the maximum height, the energy is in the form of potential energy. When the object reaches the ground, its potential energy decreases, whereas its kinetic energy is increasing (as the velocity of the object is increasing). Thus, there is no loss of total energy (energy transformation may take place ).

Q7. What are the various energy transformations that occur when you are riding a bicycle?

Answer:

The muscular energy of a person is transformed in the form of mechanical energy, which helps to rotate the wheel of bicycle.

Q8. 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:

Since the rock doesn't move, the net displacement is zero. As a result, the work done by the force is zero.

The energy that we apply to the rock gets transformed in the form of heat.

Q9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Answer:

We know that 1 unit of energy is given by: 1 unit = 1 KWh.

Also, $1\ KWh\ =\ 3.6\times 10^6\ J$

Thus, 250 units in joules is given by :

$E\ =\ 250\times 3.6\times 10^6\ =\ 9\times 10^8\ J$

Q10. 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:

The potential energy of an object of mass m at a height h is given as = mgh

The potential energy of the given object is :

$P.E.\ =\ mgh\ =\ 40\times 10\times 5\ =\ 2000\ J$

The potential energy is being converted into the K.E..

Thus, at halfway kinetic energy of the object is :

$K.E.\ =\ \frac{P.E.}{2}\ =\ \frac{2000}{2}\ =\ 1000\ J$

Q11. 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 gravitational force is zero. This is because the satellite is moving in a circular orbit. Thus, the direction of displacement of the satellite is perpendicular to the force of gravity. Hence work done is zero.

Q12. 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:

This depends on the initial state of the object. If the object is in motion initially, then it will continue to be in this state as no external force is acting. But if the object is at rest initially, then the object can't move without an external force.

Q13. 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:

Since there was no displacement thus the work done by man is zero. This work should not be considered relatable to the term 'work' we use in daily life.

The weight against gravity led the man to get tired.

Q14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Answer:

The relation between energy and power is given by :

Energy = Power $\times$ Time

Thus, the energy used in 10 hours is :

$E\ =\ 1500\times 10\ =\ 15000\ Wh\ =\ 15\ KWh$

Hence, the energy used by the heater is 15 kWh.

Q15. 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 :

In the above figure, point B is the mean position about which the bob rotates.

When the bob is released from point C, it attains some velocity while moving down (up to B) and decelerates and stops at point A.

Thus, points A and C are the maximum height points. And the velocity of the bob at point B will be maximum.

The total energy at points A and C is only the potential energy, as their velocity at these points is zero. And at point B, as the height of the bob is zero thus the total energy is just the kinetic energy.

Thu,s in this manner, the conservation of energy takes place (by transforming into some other form).

It eventually comes to rest due to the air resistance. It decelerates the motion of the bob. (as it is a frictional force.)

There is no violation of the energy conservation law, as some amount of energy is converted in the form of heat.

Q16. 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:

At this moment, the energy of the object is :

$K.E.\ =\ \frac{1}{2}mv^2$

Thus, in bringing the object to rest, the work needed is: $=\ K.E.$

$W\ =\ \frac{1}{2}mv^2\ J$

Q17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer:

Firstly, convert the velocity into SI units.

$v\ =\ 60\times \frac{5}{18}\ =\ \frac{50}{3}\ m/s$

Thu,s the work done to stop the car is equal to the kinetic energy of the car.

$W\ =\ \frac{1}{2}mv^2\ =\ \frac{1}{2}\times 1500\times \left ( \frac{50}{3} \right )^2$

$=\ 208333.33\ J\ or\ 208.33\ KJ$

Q18. 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:

(i) In first case, the force acting on the object is perpendicular to the displacement of the body. Thus the work done by the force is zero.

(ii) In this, the force is in the direction of the displacement so the work done is positive.

(iii) In this case, the direction of force and displacement are opposite to each other. Thus the work done by the force is negative.

Q19. 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. If all the external forces acting on the body balance each other, then the net force acting on the object is zero. Thu,s the acceleration produced is zero even if forces are acting on the body. The necessary condition for the zero acceleration is that the net force acting on the body should be zero.

Q20. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.

Answer:

The energy consumed by 1 device is given by :

$E\ =\ 500\times 10\ =\ 5000\ Wh$

Thus the energy consumed by 4 devices is $=\ 4\times 5000\ =\ 20\ KWh$

Q21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer:

When an object is falling from a height, there is an increase in the kinetic energy of the object. Thus just before hitting the ground, the kinetic energy of the object is very high. But after hitting the groun,d its velocity comes to zero and thus the kinetic energy becomes zero. The kinetic energy is transformed into other forms of energy, such as heat, sound or some deformation on the ground.