# NCERT Class 9 Science Notes & Solutions – Chapter 11 – Work and Energy

## Exercises Questions & Answers - NCERT Chapter 11 - Work and Energy

##### # Questions & Answers ⇒ Work and Energy ⇒ Page No. 148 ⇒

1. A force of 7 N acts on an object. The displacement say 8 m in the direction of the force. Let us take the force acts on the object through the displacement. What is the work done in this case? Observe the following picture. Answer: Work = Force x Displacement = (7 x 8) J = 56 J.

###### Extra math question related to work ⇒

1. An object of mass 150 kg is lifted on top of a tower of 15 m height. Find the work done in lifting of the object.

Answer: Work = Force x Displacement ⇒ W = F x s.
Here s = displacement = 15 m. Now what is F ? Here F is the gravitational force acting on the object which is equal to the weight of the object. So-
F = m x g  [ ∵ m = mass of the object = 150 kg, g = acceleration to gravity = 9.8 ms-2 ]
= 150 x 9.8 N = 1470 N .
Here displacement is against the force as gravitational force attracts object towards the earth . So-
W = – ( F x s )    [ See Q3 ],
= – ( 1470 x 15 ) N = (-) 22,050 N.
Therefore, magnitude of work done is 22,050 N.

###### 2. A man carries a cement bag of mass 180 kg from ground floor to 2nd floor of height 10 m. Calculate the work done by him.

Answer: As work is done against gravitational force attracts. So –
W= – (F x s) = – (m x g x s) = – (180 x 9.8 x 10) N = (-) 17,640 N. Magnitude of work done is 17,640 N.

##### # Questions & Answers ⇒ Work and Energy ⇒ Page No. 149 ⇒
###### 1. When do we say that work is done?

Answer: If we apply a force on an object & it causes displacement of that object in the direction of the force, then we say that some work is done.

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

Answer: Basically work is defined and expressed mathematically as the product of the applied force and displacement of the object. If W = total work done, F = applied force and s = displacement of the object, then –
Work done = Applied force x Displacement along the direction of the  force
W = Fs

###### 3. Define 1 J work.

Answer: 1 Joule is the amount of work done by 1 Newton force when displacement is 1 meter.

###### 4. A pair of bullocks exerts a force of 140 N on plough. Length of the field ploughed is 15 m long. Calculate the work done in ploughing of the field.

Answer: Work = Force x Displacement = (140 x 15) J = 2100 J.

##### # Questions & Answers ⇒ Work and Energy ⇒ Page No. 152 ⇒
###### 1. What is the kinetic energy of an object?

Answer: It is the energy due to motion of the object.

###### 2. Write an expression for the kinetic energy of an object.

Answer: Let assume an object of mass m is in motion with speed v. Now kinetic energy Ek  ⇒

Ek = ½  x mv2 = ½  x mass x (speed)2 .

###### 3. The kinetic energy of an object of mass ‘m’ moving with a velocity of 5 ms-1 is 25 J. Find the kinetic energy when velocity is doubled. Also find the kinetic energy when its velocity is increased three times.

Answer: In this case, kinetic energy ⇒  W = ½ mv2,
⇒ 25 = ½ x m x 52  [∵ W = 25 J, v = 5 ms-1 ]
⇒ m = (25 x 2)/25 = 2 kg.

When velocity is double: velocity = 2 x  5 ms-1 = 10 ms-1 ⇒ W = ½ x 2 x 102 J = 100 J.

When velocity is three times , velocity = 3 x  5 ms-1 = 15 ms-1  ⇒ W = ½ x 2 x 152 J = 225 J.

###### 4. A bullet of mass 200 g is moving with velocity of 100 m/s. Find the kinetic energy of the bullet.

Answer: In this case, kinetic energy of the bullet = ½  x mv2
= ½  x 0.2 x 1002 J  [∵ mass = m =200 g = 0.2 kg, v = 100 m/s] = 1000 J.

##### # Questions & Answers ⇒ Work and Energy ⇒ Page No. 156 ⇒
###### 1. What is power?

Answer: Power is the work done in unit time. Basically it is the rate of work done per second.

Power (P) =  Work (W) / Time (T) = W/T

###### 2. Define 01 (one) watt of power.

Answer:  If one (01) joule work is done in one second, then it is called one (01) watt of power. So  ⇒ 1 watt = 1 Joule / 1 second = 1 J/s.

###### 3. A lamp consumes 1000 J of electrical energy in 10s. Calculate the power of the lamp.

Answer:  Power = Total Consumption/Time = 1000/10  J/s = 100 J/s = 100 W.

###### 4. Define average power.

Answer:  Average power arises in the case where different amounts of work is done in different intervals of time. It is basically the ratio of the total work done & the total time. Therefore ⇒

average power =  total work / total time.

##### Q15. What is Kilowatt hour (kW h)?

Answer: Do you know what reading taken from electric meter installed in your house? The people from electric office takes kW h reading from the electric meter. Basically it is the energy reading over a period of time. kW h is also the commercial unit of electricity consumption and 1 kWh = 1 unit. Now what do you mean by 1 kW h?

1 kW h = 1000 W h, is the total energy consumed in one hour at the rate of 1000 W i.e. energy consumption in one second is 1000 W (or 1000 joule per second  as 1 W = 1 Joule/ second). So total energy consumption in one hour (i.e. 3600 second) is: 1 kW h = 1000 W x 3600 s = 1000 (Joule/s) x 3600 s  = 36,00,000 Joule = 3.6 x 106 J.

###### Exercises Questions & Answers | Chapter 11 Work and Energy | Page Nos. 158 - 159

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’.
a. Suma is swimming in a pond.
Answer: As Suma is swimming, so she can move ahead. Therefore, Suma is doing work.
b. A donkey is carrying a load on its back.
Answer: In this case, there is no work. Because the weight the donkey is carrying & the displacement are perpendicular to each other.
c. A wind-mill is lifting water from a well.
Answer: As wind-mill lifts water from the well, so it does work against gravity.
d. A green plant is carrying out photosynthesis.
Answer: In this case, there is no force & displacement. So there is no work.
e. An engine is pulling a train.
Answer: By pulling a train, an engine overcomes the friction force between the railway track & the wheels of the train. So it does work.
f. Food grains are getting dried in the sun.
Answer: In this case, there is no force & displacement. So there is no work.
g. A sailboat is moving due to wind energy.
Answer: Wind causes the sailboat to move to its direction. So the work is done by the wind.

###### 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 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: Object thrown at a certain angle to the ground but initial and final points lie on the same horizontal line. So displacement with respect to the level of the initial point is 0 (zero). So, Work = Force x Displacement = Force x 0 =  0.

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

Answer:  In this case, battery converts chemical energy into electrical energy. When we connect a bulb to a battery, then the chemical energy of the battery gets converted into electrical energy. Further, the bulb converts that electrical energy into light & heat energy. The process is describes as follows ⇒

chemical energy ⇒ electrical energy ⇒ (light + heat) energy

###### 4. Certain force acting on 20 kg mass changes its velocity from 5 m s-1 to 2 m s-1 . Calculate the work done by the force.

Answer:  In this case, work done = ½ mv2  – ½  mu2  = ½  m (v2 – u2)
= ½ x 20 (22  – 52)   [∵ m = 20 kg, v = 2 m s-1, u = 5 m s-1]
= – 210 J   [(-) sign indicates reduction in velocity]. So magnitude of work done = 210 J.

###### 5. A mass of 10 kg is at a point A on a table. It moves to a point B. If the line joining A and B is horizontal, then what is the work done on the object by the gravitational force? Explain your answer.

Answer: In general, every points on the top of the table are equidistant from the earth. So points A and B are equidistant from the earth. So displacement with respect to the earth is zero. Hence, work done by the gravitational force is zero.

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

Answer:  Of course not. The free falling of an object does not violate the law of conservation of energy. Because when an object falls from a height, its potential energy changes into kinetic energy progressively. In this case, the decrease in potential energy is equal to the increase in kinetic energy. So the total energy always remains conserved. As a result, the free falling of an object does not violate the law of conservation of energy.

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

Answer: The energy conversion in this case is ⇒ Muscular Energy ⇒ (Kinetic +  Heat) Energy.

When we ride a bicycle, our muscular energy gets converted into rotational energy of the wheels of the bicycle. It further gets converted into kinetic energy which provide velocity to the bicycle. Also whole muscular energy does not convert into kinetic energy. Because some muscular energy coverts into heat energy which raises our body temperature.

###### 8. Does the transfer of energy take place when you push a huge rock with all your might & fail to move it? where is the energy you spend going?

Answer:  There is no transfer of energy in this case. When we fail to move the rock by pushing it, there is no transfer of muscular energy to the rock. But our muscular energy converts into heat energy. As a result, our body becomes hot.

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

Answer: Previously we learnt in Q15,  1 unit = 1 KW h = 1,000 W h = 1000 W x 3,600 s = 1000 J/s x 3,600 s = 3.6 x 106 J.
So, 250 units = 250 x 3.6 x 106 J = 9 x 108 J.

###### 10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is the potential energy? If the object falls, find its kinetic energy when it is half-way down.

Answer: Potential Energy due to gravity = mgh  [g = acceleration due to gravity]
= 40 x 10 x 5 J  [∵ m = 40 kg, g = 10 ms-2, h = 5 m]
= 2000 J.

When it starts falling, it falls with acceleration of 10 ms-2. In this case, we apply equation –
2gs = v2 – u2  [∵g = 10 ms-2, s = half-way = 2.55 m, u = 0 ms-1, v = ? ]
⇒ v2 = 2 x 10 x 2.5 = 50.
kinetic energy when it is half-way down = ½ m v2 = ½ x 40 x 50 J = 1000 J.

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

Answer: Satellite is moving round the earth continuously at a certain orbit i.e. a fixed height from the earth. So distance of the satellite from the earth does not vary that means displacement of the satellite with respect to the earth is zero. As there is no displacement, so there is no work by the force of gravity.

###### 12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends & teacher.

Answer: Of course it is possible. We know –
F = m x a [ where F = force, m = mass of the object, a = acceleration of the object]
⇒ 0 = m x a [ if the force is absent i.e. F = 0]
So, a = 0. Because mass never be zero (m ≠ 0).
When a = 0, there must be two situation. Either the object is in rest or the object is uniformly moving along a straight line. For a uniformly moving object, there is displacement in the absence of force.

###### 13. A person holds a bundle of hay over his head for 30 minutes & gets tired. Has he done some work or not? Justify your answer.

Answer:  There is no displacement of the bundle of hay over 30 minutes. So work done on the bundle by the person is zero. But the person gets tired. Because his muscular energy gets converted into heat energy.

###### 14. Rating of an electric heater is 1500 W. How much energy does it use in 10 hours?

Answer: Energy consumption in 10 hours = 1500 W x 10 h = 15000 W h = 15 KW h = 15 unit.

###### 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side & 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:  Energy can neither be destroyed nor created as per law of conservation of energy. Energy only can be converted from one form to another. In case of an oscillating pendulum, a pendulum moves from its mean position to extreme position on both side. When it moves from mean to extreme position on any side, it rises through a height above the mean level. At extreme points, the kinetic energy of the bob transfers completely into potential energy. Therefore, the bob only possesses potential energy & kinetic energy becomes zero.

When the bob moves from extreme points to mean position, its potential energy decreases progressively & accordingly, its kinetic energy increases. At the time when the bob reaches mean position, its potential energy becomes zero & it only possesses kinetic energy. As long as the pendulum oscillates, the said process repeats.

The bob comes to rest after some time due to resistance by air. The bob faces air resistance during its motion & it losses kinetic energy to overcome the friction. So it stops after some time. The loss in kinetic energy to overcome air friction is gained by the surroundings. Therefore, the total energy of the pendulum & the system remains constant. So this never violates the law of conservation of energy.

###### 16. An object of mass, m is moving with a constant velocity, v. Calculate the work needed to bring the object to rest?

Answer: When an object of mass m is in motion with speed v, its kinetic energy, Ek = ½  x mv2 = ½  x mass x (speed)2. When we bring it to rest by applying some force, its speed becomes zero. So its kinetic energy becomes zero. Therefore, the work done on the object to bring it to rest is equal to the change in the kinetic energy of the object. So, work done = change in kinetic energy = ½ mv2 – 0 = ½ mv2.

###### 17. Calculate the work required to do to stop a car of 1500 kg moving at a velocity of 60 km/h?

Answer: In this case, velocity = 60 km/h = 60000/3600 = 16.67 m/s.
Kinetic energy possessing by the car = ½ mv2 = ½ x 1500 x 16.672 J =  208416.675 J. To stop the car, we need to neutralize total kinetic energy that means to stop the car, we need to work equal to 208416.675 J.
On the other hand,  work done  = ½ mv2  – ½  mu2
= ½ m (v2 – u2)  [∵ u= 16.67 m/s, v = 0 as car stops at the end]
= ½ mv2 = ½ x 1500 x 16.672 J =  208416.675 J.

###### 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 & state whether the work done by the force is negative, positive or zero. Answer: Case – I ⇒ In this case, the force & the displacement is perpendicular to each other. So the work done will be zero.
Case – II ⇒ In this case, the force & the displacement are in the same direction. So the work done will be positive.
Case – III ⇒ In this case, the force & the displacement are in the opposite direction. So the work done will be negative.

###### 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: If multiple forces acting on an object cancel each other, then acceleration of that object could become zero. When all forces cancel each other, the net force becomes zero. Therefore, acceleration will be zero. We know that ⇒

F = m x a [ where F = force, m = mass of the object, a = acceleration of the object]
⇒ 0 = m x a [ when F = 0]
So, a = 0. Because mass never be zero (m ≠ 0).
For example, the net force is zero for a uniformly moving object. Hence, acceleration of it is zero & it moves with uniform velocity.

###### 20. Find the energy in KW h consumed in 10 hrs by four devices of power 500 W each.

Answer: Energy consumption = 4 x 500 W x 10 hrs = 20000 W h = 20 KW h.

###### 21. A free falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Answer: A free falling object posseses decreasing potential energy & increasing kinetic energy. Just before it touches the ground, it possesses only kinetic energy but no potential energy. When it touches the ground, the kinetic energy gets converted into heat & sound energy. Also a free falling object can deform the ground. Though it depends on the type of ground & the amount of the kinetic energy.

To Be Continued