Work and Energy Question Answers: NCERT Class 9 Science

Welcome to the Chapter 11 - Work and Energy, Class 9 Science - NCERT Solutions page. Here, we provide detailed question answers for Chapter 11 - Work and Energy.The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics and excel in their exams. By going through these Work and Energy question answers, you can strengthen your foundation and improve your performance in Class 9 Science. Whether you're revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

From very previous class we have learnt about work, that is the ability to do work. But these are many situation where work done is said to be zero. Satellites revolving around earth is the best example. In this chapter we will discuss about when the work done is positive, negative and zero and various forms of energy.

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Exercise 1

Q1	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?
Ans:	Force = 7 N Displacement = 8 m Work done = Force × Displacement W = F × S W = 7 × 8 = 56 Nm = 56 J Therefore, work done is 56 J.

Exercise 2

Q1	When do we say that work is done?
Ans:	A work is said to be done when a force is applied on a body and makes the body move then we said work is done.

Q2	Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Ans:	When a force is acting on a body in the direction of the applied force, then the work done on the body is given by the expression: Work done = Force × Displacement W = F × s

Q3	Define 1 J of work.
Ans:	1 J is defined as the amount of work done when a force of one newton is applied over a displacement of one meter.

Q4	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?
Ans:	Applied force, F = 140 N Displacement, d = 15 m Work done = force x displacement W = 140 × 15 = 2100 J Work done is 2100 J.

Exercise 3

Q1	What is the kinetic energy of an object?
Ans:	The energy of a body due to its motion is known as kinetic energy. It is a scalar quantity, i.e it does not depend on direction.

Q2	Write an expression for the kinetic energy of an object.
Ans:	Kinetic energy possessed by a body of mass m is moving with a uniform velocity v, is given by the expression, Its SI unit is Joule (J).

Q3	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?
Ans:	Mass of the object = m Velocity of the object = v = 5 m/s Kinetic energy, E_K = 25 J (i) If the velocity of an object is doubled, then v = 5 × 2 = 10 m/s Therefore, its kinetic energy becomes 4 times 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 because it is proportional to the square of the velocity. Hence, kinetic energy = 25 × 9 = 225 J.

Exercise 4

Q1	What is power?
Ans:	Power is the amount of work done per unit time. It is a scalar quantity. It is expressed in watt (W).

Q2	Define 1 watt of power.
Ans:	1 watt of power is defined as the power produced when 1 joule of work is done for 1 second

Q3	A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Ans:	Energy consumed by the lamp ( work done ) = 1000 J Time = 10 s = 100 W

Q4	Define average power.
Ans:	The average power is defined as the ratio of total work done and the total time taken.

Exercise 5

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 wind-mill 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.
Ans:	Work is done whenever the given two conditions are satisfied: (i) Force is applied. (ii) Displacement takes place. (iii) The angle between the force and displacement is not 90 degrees . (a) When Suma is swimming in a swimming pool, she applies force in forward direction and moves in the forward direction. So, displacement is in the forward direction. Both are in the same direction. Therefore, work is done. (b) When a donkey is carrying a load on his back, it is applied 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) When a wind-mill is lifting water from a well, it is applying a force in the upward direction and it is moving water in an upward direction. Hence, work is done by the wind-mill in lifting water from the well. (d) In this case, there is no force involved when a green plant is carrying photosynthesis. Therefore, the work done is zero. (e) When an engine is pulling a train, it is applying a force in the forward direction. This allows the train to move in the forward direction. Hence, there is a displacement in the train in the same direction. Therefore, work is done by the engine on the train. (f) When food grains are getting dried in the Sun, there is no force involved. Hence, the work done is zero during the process of food grains getting dried in the Sun. (g) When a sailboat is moving due to wind energy, force is applied by the boat 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.

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?
Ans:	There is no work done because the applied force is in the vertical direction but the displacement of the body is in the horizontal direction. Since the angle between force and displacement is 90 degrees.

Q3	A battery lights a bulb. Describe the energy changes involved in the process.
Ans:	When a battery lights a bulb , 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. Therefore, energy changes are shown as : Chemical Energy → Electrical Energy → Light Energy + Heat 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.
Ans:	Mass of the body = 20 kg Initial velocity, u = 5 m/s Final velocity, v = 2 m/s Kinetic energy is given by the expression, (i) Kinetic energy when the object was moving with a velocity 5 m/s (ii) Kinetic energy when the object was moving with a velocity 2 m/s Work done = change in kinetic energy. Therefore, work done = (E_k)₂ - (E_k)₅ = 40 − 250 = −210 J The negative sign indicates that the force is acting in the direction opposite to the motion of the object.

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.
Ans:	There is no work done because the force is gravitational force which is in downward direction but the displacement of the body is in horizontal direction. Since the angle between force and displacement is 90 degrees.

Q6	The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Ans:	No. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then it loses its potential energy. But as it falls, it gains some velocity. Due to increase in velocity, the body gains kinetic energy. During the process, total mechanical energy of the body remains conserved. Hence, the law of conservation of energy is not violated.

Q7	What are the various energy transformations that occur when you are riding a bicycle?
Ans:	While riding a bicycle, we use our muscular energy to make the cycle move and our body also warms up in the process. Therefore, the muscular energy gets converted into kinetic energy and heat energy. The energy transformation can be shown as: Muscular Energy → Kinetic Energy + Heat Energy

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?
Ans:	When we push a huge rock, we apply muscular force in order to move it. There is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is converted into heat energy, which causes our body to feel warm.

Q9	A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Ans:	1 unit of energy = 1 kilowatt hour (kWh) 1 kWh = 3.6 × 10₆ J Therefore, 250 units of energy = 250 × 3.6 × 10₆ = 9 × 10₈ 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.
Ans:	Vertical displacement = h = 5 m Mass of the object = m = 40 kg Acceleration due to gravity = g = 9.8 m/s Gravitational potential energy is given by the expression, W = mgh ∴ W = 40 × 9.8 × 5 = 1960 J. At half-way down, the potential energy of the object will be 1960 / 2 = 980 J.

Q11	What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Ans:	Work is done whenever the given conditions are satisfied: (i) Force is applied. (ii) Displacement takes place. (iii) The angle between the force and displacement is not 90 degrees. When a satellite moves around the Earth, then the direction of force of gravity on the satellite is perpendicular to its displacement. Therefore , the work done on the satellite by the Earth 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.
Ans:	Yes. There can be displacement of an object in the absence of any force acting on it, for a uniformly moving object. Suppose an object is moving with uniform velocity. Force will be zero when acceleration is zero. Hence, there can be a displacement without a 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.
Ans:	Work is done whenever the given conditions are satisfied: (i) Force is applied. (ii) Displacement takes place. (iii) The angle between the force and displacement is not 90 degrees. When a person holds a bundle of hay over his head, then the bundle of hay remains stationary. 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.

Q14	An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Ans:	Power of the heater, P = 1500 W = 1.5 kW Time taken, T = 10 h Energy consumed by an electric heater can be obtained with the help of the expression, Power = Energy consumed / time Therefore, Energy consumed = Power × Time = 1.5 × 10 = 15 kWh Hence, the energy consumed by the heater in 10 h 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?
Ans:	The law of conservation of energy states that energy can neither be created nor be destroyed. It can change 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.

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?
Ans:	Mass of the object = m Velocity = v Therefore ,Kinetic energy To bring the object to rest, amount of work is required to be done on the object.

Q17	Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Ans:	Mass of car, m = 1500 kg Velocity of car, v = 60 km/h Therefore, Kinetic energy Hence, 20.8 × 10₄ J of work is required to stop the car.

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.
Ans:	Work is done whenever the given conditions are satisfied: (i) Force is applied. (ii) Displacement takes place. (iii) The angle between the force and displacement is not 90 degrees . Case I In this case, the angle between the force and displacement is 90 degrees. Hence, work done by force on the block will be zero. Case II In this case, the angle between the force and displacement is 0 degree. Hence, work done by force on the block will be positive. Case III In this case, the angle between the force and displacement is 180 degrees. Hence, work done by force on the block will be 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?
Ans:	Yes, I agree with Soni that acceleration in an object could be zero even when several forces are acting on it. This happens when equal and opposite forces act on an object, they cancel out each other so that the net external force experienced by the object is zero. Therefore, the acceleration of the object is zero. Therefore, Soni is right.

Q20	Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Ans:	Power of one device, P = 500 W = 500 / 1000 = 0.50 kW Power of four devices = 500 x 4 = 2000 W = 2000 / 1000 = 2 kW Time taken, T = 10 h We know that Power = energy consumed / time Energy consumed = Power × Time = 2 × 10 = 20 kWh Therefore, the energy consumed by four equal rating devices in 10 h will be 20 kWh.

Q21	A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?
Ans:	When an object falls freely towards the ground, its potential energy decreases and kinetic energy increases. When a free falling object hits the ground, its kinetic energy becomes zero and its potential energy becomes maximum. The kinetic energy changes into heat and sound energy while the object comes to rest.