Chapter 10 – Work and Energy

NCERT Answers for Class 9 Science Chapter 10 Work and Energy .delves into fundamental concepts of physics that explain how objects move and interact.

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 displacement. What is the work done in this case?

Answer: In the above question it is given that:

Force is $F = 7 N$ .

Displacement is $S = 8 m$

Work done is given by the formula:

$Work done = Force \times Displacement$

$W = F \times S = 7 \times 8 = 56 N m$

Hence, the work done in this case is $56 J$ .

Exercise 2

1. When do we say that work is done?

Answer: If both of the following two requirements are met, work is considered completed:

-The thing must be subject to force.

-Something has to be moved.

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

Answer: If both of the following two requirements are met, work is considered completed:

The thing must be subject to force.

Something has to be moved.

Think about an object that is subject to a constant force F. In the event that the item shifts by a distance s relative to the force (Fig. 11.1) and W Is the task completed? Work completed is therefore equal to the product of force and displacement.

$Work done = Force \times Displacement$

$\Rightarrow W = F \times S$

3. Define 1 J of work.

Answer: The amount of work done on an object when a force of 1N moves it by 1m is equal to 1J of work.

along the force’s path of action.

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

Answer: In the above question it is given that:

Applied force is $F = 140 N$ .

Displacement is $s = 15 m$ .

We know that

$Work done = Force \times Displacement$

$\Rightarrow W = F \times S = 140 \times 15 = 2100 J$

Consequently, 2100J of work was required to plough the entire length of the field.

Exercise 3

1. What is the kinetic energy of an object?

Answer: Kinetic energy is the energy a thing possesses as a result of its motion.

NCERT Answers for Class 9 Science Chapter 10 Work and Energy

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

Answer: Assume that a mass ′m′ body is travelling at a velocity ′v′. Thus, the kinetic energy Ek

is provided by:

$E_{k} = \frac{1}{2} m v^{2}$

The SI unit of kinetic energy is the Joule $(J)$ .

3. The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25J.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: It is stated in the question above that:

The object’s velocity is 5 m/s.
The object’s kinetic energy is Ek=25J.
..

The object’s mass is m.
..

As we are aware,

Ek is equivalent to 12 mv2.

Case 1: When an object’s velocity doubles,

v=10 m/s

The source of kinetic energy is:

Ek is equivalent to 12 mv2.

Ek∞v2

Therefore, when v=10m/s

Ek = 25×4 = 100J

Four times as much kinetic energy is produced as initially.

Case 2: The kinetic energy of a certain object will increase nine times its initial value (Ek∞v2) if its velocity is raised three times.

Therefore, the kinetic energy is:

Ek = 25×9 = 225J

Kinetic energy increases to nine times its initial value as a result.

Intext Exercise 4

1. What is power?

Answer: The rate at which labour is completed or energy is transferred is referred to as power. Power is determined by: if W is the quantity of work completed in time t, thenPower is equal to work completed.Time= Energy

P=WT
The watt (W) is the SI unit of power.

2. Define 1 watt of power.

Answer:

One watt is the definition of an agent’s power when it operates at a rate of one joule per second. Additionally, power is defined as 1W when energy is consumed at a rate of 1J/s.

1W equals 1J1s

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

Answer: It is stated in the question above that:

The energy used is E=1000J.

Moment T=10 s

Power is equal to work completed.Time= Energy

⇒ Power = 100010 = 100W

Therefore, 100W of power is produced.

4. Define average power.

Answer: The power obtained by dividing the overall quantity of work completed in the total length of time required to do this job is known as average power.Average power is equal to the sum of all labour and time.

NCERT Exercise

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 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.

Answer: As to the definition, work is completed anytime both of the subsequent conditions are met:

1.The body is subject to a force.

2. When force is applied, the body is moved in the force’s direction or in the opposing direction.

Consequently,

-Suma exerts force to push the water backwards when swimming. Because of the water’s forward response, Suma is now able to swim forward. The force thus results in a displacement. Therefore, Suma works when swimming.

-In this instance, the donkey is carrying a load and exerting force upward. Nonetheless, the load is being displaced forward. There is no labour done since the displacement is perpendicular to the force.

-As a windmill works against gravity to lift water, it is responsible in this instance for drawing water from the well.

-The plant’s leaves are not dislocated in this instance. Thus, there has been no work done.

-In this instance, the train is being pulled by an engine. As a result, the train moves in the force’s direction. As a result, the train is moving in the same direction. Therefore, the train’s engine does the labour.

-In this instance, there is no labour involved in the process of drying food grains in the sun because food grains do not move when exposed to solar radiation.

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: When an object is thrown in the above question, it travels in a curved path before falling back to the earth at a specific angle. The object’s starting and ending points of its route are on the same horizontal line.

The amount of work that gravity does on an object is solely dependent on its vertical displacement. The difference between the object’s original and end positions (height), which is zero, determines the vertical displacement.

The formula for work done by gravity is W = mgh.
Whereas
m is the object’s mass.
G stands for gravitational acceleration.
The vertical displacement is zero, denoted by h.

⇒W = mg(0)=0J

Gravity therefore does 0 joules of work on the given object.

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

Answer: The chemical energy of a lightbulb is transformed into electrical energy once it is linked to a battery. This electrical energy is transformed into heat and light energy by the lightbulb. As a result, the energy transformation in the specified scenario can be represented as follows: Chemical Energy→Electrical Energy→Light Energy→Heat energy.

4. Certain force acting on a 20 kg mass changes its velocity from $5 m / s$ to $2 m / s$ . Calculate the work done by the force.

Answer: The work done is defined as the change in kinetic energy.

A 20 kilogramme mass is given, and it alters its velocity from 5 m/s.

up to 2 m/s

Kinetic energy can be expressed as follows:

(Ek)v equals 12 mv^2.
Whereas
Ek = The object’s kinetic energy when it moves at a certain velocity, v
m =The object’s mass
Kinetic energy at a speed of 5 meters per second for the item
Ek = 12×20×(5)2 = 250J
Kinetic energy at a speed of two meters per second for the item
(Ek)2 = 12×20×(2)2 = 40J
Thus, force worked equals 40 – 250 – 210 J.
In this instance, the object’s motion is being opposed by the force, as indicated by the negative sign.

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: The only factor affecting the work done by gravity is the body’s vertical displacement. It is independent of the body’s path. Thus, the labour that gravity does is equal to:

W is equivalent to mgh
where W = mg(0)=0 and h=0 are the vertical displacements.J
Thus, the item experiences zero joules of work from gravity.

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

Answer: The body gradually transforms its potential energy into kinetic energy when it descends from a height. The body’s kinetic energy increases in proportion to a decrease in potential energy.

The body’s whole mechanical energy is preserved during this procedure. As a result, nothing breaks the law of energy conservation.

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

Answer: Our muscle energy is converted into the heat and kinetic energy of the bicycle when we ride. The body is heated by heat energy, and the bicycle is propelled forward by kinetic energy. The change can be displayed as:

M u s c u l a r Energy \to K i n e t i c Energy \to Heat energy

The total energy is conserved in the entire transformation of energies.

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: There is no energy transfer when we push a large rock since our muscles are not exerting any force on the immobile rock. Rather, the energy from our muscles is converted into thermal energy, making our bodies hot.

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9. 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 equal to $1$

kilowatt hour (kWh).

1 u n i t = 1 k W h

Therefore,

$1 k W h = 3.6 \times 10^{6} J$

Hence, $250$ units of energy $= 250 \times 3.6 \times 10^{6} = 9 \times 10^{8} J$ .

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:

In the above question it is given that:

h = V e r t i c a l d i s p l a c e m e n t = 5 m

m = M a s s o f t h e o b j e c t = 40 k g

g = A c c e l e r a t i o n d u e t o g r a v i t y = 9.8 m / s 2

Gravitational potential energy is given by the expression,

W = m g h

\Rightarrow W = 40 \times 5 \times 9.8 = 1960 J

At half-way down, the potential energy of the object will reduce to half i.e., $\frac{1960}{2} = 980 J$ .

The object’s potential energy is now equal to its kinetic energy. The rule of conservation of energy is to blame for this. Therefore, the object’s kinetic energy will be 980J halfway down.

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

Answer: The work done is 0 when the force’s direction is perpendicular to the displacement. The direction of a satellite’s gravitational pull is perpendicular to its displacement when it orbits the planet. As a result, Earth has done zero work on the satellite.

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, an object can move with uniform velocity even in the absence of any force acting upon it. If an object moves at a constant speed, there will be no net force acting on it. Thus, a displacement is possible in the absence of a force.

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: Gravity pulls the hay downward when someone holds a bunch of hay over their head. However, the hay bundle is not moving in the force’s direction. So, nothing is accomplished.

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

Answer: Energy consumed by an electric heater is given by the expression,

$P = \frac{W}{T}$

Where,

P = 1500 W = 1.5 k W

T = 10 h r s

We know that,

W o r k d o n e = E n e r g y c o n s u m e d b y t h e h e a t e r

Hence,

e n e r g y c o n s u m e d = P o w e r \times T i m e = 1.5 \times 10 = 15 k W h

= 1.5 × 10 = 15 kWh.

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 cannot be generated or destroyed. It is limited to being transformed from one form to another. Think about the situation of a pendulum that oscillates.

A pendulum rises through a height h above the mean level P when it moves from its mean position P to either of its extreme locations A or B. When the bob reaches its extreme, its kinetic energy is converted to potential energy and it comes to rest.

Its potential energy steadily drops as it approaches point P. The kinetic energy rises concurrently.

The bob’s potential energy is transformed into kinetic energy when it approaches point P. Bob is now limited to his kinetic energy. As long as the pendulum continues to oscillate, this procedure is repeated. The oscillation of the bob is not infinite. A certain amount of air friction causes it to come to rest.

When the pendulum overcomes this friction, energy is lost. It reaches rest when it has expended all of its energy. Since the pendulum’s surroundings are gaining energy to counteract the friction, the law of conservation of energy is applicable in this situation. As a result, the entire energy of the surrounding system and the pendulum is preserved.

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: A body has kinetic energy while it is moving. The statement gives the kinetic energy of an item of mass m moving at a velocity of v.

$E_{k} = \frac{1}{2} m v^{2}$

Therefore, $\frac{1}{2} m v^{2}$ amount of work is required to be done on the object to bring the object to rest.

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

Answer: In the above question it is given that:

Mass of car,

m = 1500 k g

Velocity of car,

v = 60 k m / h = 60 \times 5 18 m / s

Kinetic energy is given by:

$E_{k} = \frac{1}{2} m v^{2}$

$\Rightarrow E_{k} = \frac{1}{2} (1500) {(60 \times \frac{5}{18})}^{2} = 20.8 \times 10^{4} J$

Hence, $20.8 \times 10^{4} J$ needs to be done to stop the car.

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.

Different positions of force and displacement w.r.t. each other

Ans: Following cases are explained below:

Case I

Force and displacement are perpendicular

In this instance, the force acting on the block is pointing in the opposite direction as the displacement.

Therefore, there will be no work done on the block by force.

Case II

towards this instance, the block is being forced towards the direction of displacement. Therefore, any work done on the block by force will be beneficial.

ase III

Force and displacement are anti-parallel

In this instance, the force exerted on the block is acting in the opposite direction as the displacement. Force on the block will therefore produce negative work.

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: When there is no net force applied on an item, its acceleration is zero. Even in situations where the body is subject to several forces, net force can still be zero. The net force acting on an item in uniform motion is zero. As a result, the object’s acceleration is zero. Thus, Soni is correct.

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

Answer: In the above question it is given that:

P = 500 W = 0.50 k W

T = 10 h r s

Energy consumed by an electric device can be obtained with the help of the expression given below:

$P = \frac{W}{T}$

Where,

W o r k d o n e = E n e r g y c o n s u m e d b y t h e d e v i c e

E n e r g y c o n s u m e d b y e a c h d e v i c e = P o w e r \times T i m e

Work done = 0.50 \times 10 = 5 k W h

Hence, the energy consumed by four equal rating devices in

10 h r s

will be

4 \times 5 k W h = 20 k W h

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

Answer: The kinetic energy of the object is transformed into heat and sound energy as soon as it strikes the hard ground. Moreover, the environment loses all of the energy. In addition, depending on the characteristics of the ground and the object’s kinetic energy, this energy has the ability to deform the ground.

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NCERT Answers for Class 9 Science Chapter 10 Work and Energy

Chapter 10 – Work and Energy

Exercise 2

Exercise 3

Intext Exercise 4

NCERT Exercise

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NCERT Answers for Class 9 Science Chapter 10 Work and Energy

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