NCERT Exemplar Class 9 Science Solutions Chapter 11 Work and Energy

NCERT Exemplar Class 9 Science Solutions Chapter 11 Work and Energy

Edited By Safeer PP | Updated on Sep 01, 2022 05:10 PM IST

NCERT exemplar Class 9 Science solutions chapter 11 involves the learning and understanding about work, energy, and power. Students will learn about the conservation of mechanical energy and its application in this chapter of NCERT. The NCERT exemplar Class 9 Science chapter 11 solutions are put together by our highly skilled team of subject matter experts to provide the student with study material for NCERT Class 9 Science. These NCERT exemplar Class 9 Science chapter 11 solutions provide step by step approach to the problems based on work and energy. The CBSE Class 9 Syllabus is followed in the preparation of NCERT exemplar Class 9 Science solutions chapter 11.

Question:1

When a body falls freely towards the earth, then its total energy
(a) increases
(b) decreases
(c) remains constant
(d) first increases and then decreases

Solution.

The total energy of any body is the sum of kinetic energy and potential energy.
$TE =KE + PE$
When a body falls freely towards the earth; kinetic energy of the body increases and potential energy of the body decreases.
However, by the law of conservation of energy total energy of the body remain constant.
The correct answer to this question is option C.

Question:2

A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of initial
(d) becomes 16 times that of initial

Solution.
The kinetic energy of a body increases on increasing speed.
$KE=\frac{1}{2}mv^{2}$
The potential energy of the body increases by increasing height from the surface of earth.
$PE=mgh$
In the given question, the car is increasing speed on the level road. Hence, its kinetic energy must be increasing.
The kinetic energy will be 16 times of the initial kinetic energy as velocity becomes four times.
The car's height from the ground is not changing as it is moving on a level road.
Therefore, we can say that the potential energy of the car is not changing.
Hence, the correct answer to this question is option A.

Question:3

In case of negative work the angle between the force and displacement is
(a) $0^{\circ}$ (b) $45^{\circ}$ (c) $90^{\circ}$ (d ) $180^{\circ}$

Solution.
Work done by any force is defined as:
$W=Fd\cos \theta$
This work done can be positive, negative or zero depending on the angle between force and displacement.
If the angle between them is acute, then work done will be positive because $\cos \theta$ will be positive.
If the angle between them is $90^{\circ}$, then the work done will be zero.
If the angle between them is obtuse, then the work done will be negative.
$W=Fd(\cos 180^{\circ})=-Fd$
Hence, the correct answer is option D.

Question:4

An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) Kinetic energy

Solution.

When a body falls under gravity, the body’s acceleration does not depend on mass of the body.
Therefore, acceleration for both the spheres will be same at that instant.
Their velocities will be the same at that instant too.
$v = u + at$
Momentum and kinetic energy depend on speed as well as mass.
$KE=\frac{1}{2}mv^{2}$
$P=mv$
As mass of both the spheres is different; however, speed is the same; hence their momentum and kinetic energy will be different.
Gravitational potential energy is defined as:
$PE=mgh$
Therefore, it would be different for both the spheres of different mass at same the height.
Hence, the correct answer is option A.

Question:5

A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be $(g =10 m s^{-2})$
(a) $6 \times 10^3 J$
(b) $6 J$
(c) $0.6 J$
(d) zero

Solution.
We know that work is defined as:
$W=Fd\cos \theta$
As the girl is carrying a schoolbag of 3 kg, the girl must be applying force in vertically upward direction against gravity.
The displacement of the girl is in the horizontal direction, as it is moving on a level road.
The angle between force and displacement is $90^{\circ}$.
Therefore, by the formula, work done by the girl against gravity will be zero.
$W=Fd\cos 90^{\circ}=0$
Hence, the correct answer for this question is option D.

Question:6

Which one of the following is not the unit of energy?
(a) joule
(b) newton metre
(c) kilowatt
(d) kilowatt hour

Solution.
We know that work done and energy have same unit.
The SI unit of energy is Joule.
Work done is defined as product of force and displacement; therefore, its unit will be newton meter.
$W=Fd\cos \theta$
Power is defined as work done per unit time, and its unit will be watt or kilowatt.
If we multiply power with time, we will get energy or work done.
Therefore, kilowatt-hour will be the unit of energy.
[Only kilowatt is the unit of power, and that is not unit of energy.]
Correct answer of this question is option C.

Question:7

The work done on an object does not depend upon the
(a) displacement
(b) force applied
(c) angle between force and displacement
(d) initial velocity of the object

Solution.
We know that the work done is defined as:
$W=Fd\cos \theta$
It means work done depends on force, displacement and angle between force and displacement.
Work done does not depend on any other property other than the three mentioned above.
Therefore, we can say work done does not depend on the initial velocity of the object.
Hence the correct option to this question is option D.

Question:8

Water stored in a dam possesses
(a) no energy
(b) electrical energy
(c) kinetic energy
(d) potential energy

Solution.
Water stored in a dam remains at rest.
The kinetic energy of the water stored will be zero as its speed is zero.
$KE=\frac{1}{2}mv^{2}$
Electrical energy is not possessed by stored water.
The water has mass, and it occupies the space above the surface of ground.
Therefore, the water will have gravitational potential energy.
$PE=mgh$
Hence, the correct answer to this question is option D.

Question:9

A body is falling from a height h. After it has fallen a height h/2, it will possess
(a) only potential energy
(b) only kinetic energy
(c) half potential and half kinetic energy
(d) more kinetic and less potential energy

Solution.
When a body is dropped from a height, it gets acceleration due to gravity.
Acceleration increases its speed; hence its kinetic energy keeps on increasing.
$KE=\frac{1}{2}mv^{2}$
The ball is dropped from height H, and we are analyzing the ball at height H/2.
The total energy of the ball will not change in free fall under gravity.
As the height becomes half of the initial height, the potential energy will become half.
$PE=mgh$
This decrease in potential energy, will come in the form of kinetic energy because total energy is conserved.
$TE = PE +KE$
Therefore, kinetic energy of the ball will be half of initial potential energy.
Hence, the correct answer to this question is option C.

NCERT Exemplar Class 9 Science Solutions Chapter 11-Short Answer

Question:10

A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

Solution.
We know that kinetic energy is defined as:
$KE=\frac{1}{2}mv^{2}$
We can see that the kinetic energy will be proportional to the square of speed for a given body of constant mass.
As the rocket's speed is suddenly tripled, the rocket's kinetic energy will be nine times.
$KE=\frac{1}{2}mv^{2}$
$\frac{KE_{i}}{KE_{f}}=\frac{V^{2}_{i}}{V^{2}_{f}}$

Question:11

Avinash can run with a speed of $8 ms^{-1}$ against the frictional force of 10 N, and Kapil can move with a speed of $3 ms^{-1}$ against the frictional force of 25 N. Who is more powerful and why?

Solution.
Power of anything is defined as the rate of work done by that thing.
Mathematically, power is defined as product of force and velocity.
Power =Fv
If we calculate the power of Avinash and Kapil, by the given information:
$Power_{Avinash}=10\times 8=80watt$
$Power_{Kapil}=25\times 3=75watt$
Therefore, it is quite visible that power of Avinash is more than the power of Kapil.

Question:12

A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a round-about of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto$2.0km.$ Calculate the work done by him.

Answer: $22210 J$
Solution.
The boy has to apply force in the direction of its motion against frictional force.
Therefore, the angle between force and displacement will always be $0^{\circ}$.
For calculating the work done, we have to multiply force with the distance travelled by the boy.
The magnitude of the force is always 5 newton.
Total distance travelled, will be equal to sum of 3/2 times perimeter of the circular path and distance travelled along a straight track.
$d=1500m+(3/2)(2\pi \times 100)m+2000m$
$d=4442m$
Hence work done:
$W=5\times 4442=22210J$

Question:13

Can any object have mechanical energy even if its momentum is zero? Explain.

Solution.
Mechanical energy is the sum of kinetic energy and potential energy. It is also known as total energy.
Kinetic energy is the virtue of motion which comes with speed.
$KE=\frac{1}{2}mv^{2}$
Potential energy is the virtue of position which comes with a height from the ground surface.
$PE=mgh$
As the object has zero momentum, it means it is not moving.
Momentum is defined as the product of mass and speed.
P = mv
A body at rest and some height from the ground will have potential energy or mechanical energy.

Question:14

Can any object have mechanical energy even if its momentum is zero? Explain.

Solution.
Mechanical energy consists of both potential energy and kinetic energy. Momentum is said to be zero when the velocity is zero. Hence, there is no kinetic energy but the object may have potential energy.

Question:15

The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given $g = 10 m s^{-2}$ )

$1200kg$
Solution.
The power of pump is 2 kW, it means it can give energy equal to 2000 Joule in one second.
Total time given is one minute; therefore, the motor pump's total energy will be equal to 1,20,000 joule.
$Energy=power\times time$
This energy will be equal to the potential energy of the water raised to 10 m.
Let mass of the water raised is M kg.
Therefore, its potential energy at height $H =10 m$ will be:
$PE=mgh$
$=M\times 10\times 10$
$\Rightarrow 120000=100M$
$\Rightarrow M=1200kg$
It means 1200 kg water can be raised up height 10 m, in one minute by this motor pump.

Question:16

The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A?

Answer: $0.8m$
Solution.
We will assume that the energy level (power) of a person does not change on changing planet.
Energy used by the person will increase its potential energy.
Therefore potential energy at the end will be same on planet A and earth.
We know the formula of potential energy is given as:
$PE=mgh$
As acceleration due to gravity is half on planet A. Therefore, the height raised will be twice on planet A.
Hence, height raised on planet A will be $0.8m$.

Question:17

The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Solution.
By the equation of motion, we can say:
$v^{2}=u^{2}+2as$
$\Rightarrow s=\frac{v^{2}-u^{2}}{2a}$
By Newton’s second law, we can say:
F= ma
We know that work done is defined as the product of force and displacement if the body is moving on a straight track.
$W=Fs$
$\Rightarrow W=ma\times \frac{v^{2}-u^{2}}{2a}$
$\Rightarrow W=m\times \frac{v^{2}-u^{2}}{2}$
$\Rightarrow W=\frac{mv^{2}}{2}-\frac{mu^{2}}{2}$
$\Rightarrow W=KE_{f}-KE_{i}$
Hence, work done is equal to the change in kinetic energy.

Question:18

Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example.

Solution.
Yes, it is possible. A body moves in a uniform circular motion is an example of accelerated motion. Consider the motion of the Earth around the Sun. The Earth is constantly moving in a circular path in a direction perpendicular to the gravitational pull of the Sun. So, the work done by the gravitational force is zero. Thus, the work done can be zero for an accelerated body.

Question:19

A ball is dropped from a height of 10 m. If the energy of the ball reduces by $40\%$ after striking the ground, how much high can the ball bounce back?$(g = 10 m s^{-2})$

Solution.
When the ball is dropped from 10 meters, the total energy will be in the form of potential energy.
Potential energy can be calculated as:
$PE=mgh$
After striking the ground, $40\%$ of energy is lost. Hence, it will have only $60\%$ of energy.
With the help of this $60\%$ energy, the object will attain maximum height.
Therefore, if height attained is H then we can say:

$60\%(mg(10))=mgh$

$\Rightarrow \frac{60}{100}(mg(10))=mgh$

$\Rightarrow h=6m$
The height attained is 6m.

Question:20

If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April.

Answer: $64800000J$
Solution.
The electrical energy consumed is calculated in units.
We know that one unit is equal to the energy of 1 kilowatt-hour.
Power of heater = 1200 watt =1.2 kilowatt
Per day usage = 30 minutes = 0.5 hour
Energy consumed per day = (1.2 kilowatt) (0.5 hour) = 0.6 units.
In April, total number of days = 30
Therefore, total consumption will be: Energy = (30) (0.6) = 18 units
It can be converted in joules:
$1unit=1000\times 3600J$
$\Rightarrow 18unit=(18\times 1000\times 3600)J=64800000J$

NCERT Exemplar Class 9 Science Solutions Chapter 11-Long Answer

Question:21

A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

Solution.
For any object of mass m, the kinetic energy and momentum are defined as:
$KE=\frac{1}{2}mv^{2}$
$P=mv$
Here v is the speed of the object.
We try to eliminate speed and to find out the relation between kinetic energy and momentum.
$KE=\frac{1}{2}m\left ( \frac{p}{m} \right )^{2}$
$\Rightarrow KE=\frac{1}{2}\frac{p^{2}}{m}$
If two bodies have the same momentum, then kinetic energy will be inversely proportional to the mass of the body.
Therefore, the lighter body will have more kinetic energy than the heavy body.

Question:22

An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just afte the collision.

Solution.
Mass of the car = 1000 kg
Speed of car A = 36 km/h = 10 m/s
Frictional force =100 N
We know that power is defined as:
Power = Fv
Hence, power of engine will be:
Power = (100N) (10m/s) =1000 watt.
In the collision, no external force is involved. Therefore, total momentum of both cars will be conserved.
$m_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}$
Where $m_{A}=1000kg;m_{B}=1000kg;u_{A}=10m/s;u_{B}=0m/s;v_{A}=0m/s;$
By putting the values, we will get:
$v_{B}=10m/s;$

Question:23

A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of $4 ms^{-1}$ by applying a force.
The trolley comes to rest after traversing a distance of 16 m.

(a) How much work is done on the trolley?
(b) How much work is done by the girl?

Solution
Mass of the girl = 35 kg
Mass of the trolley = 5 kg
Initial velocity of trolley = 4m/s
Total distance travelled by trolley = 16 m
We know that
$\\\;v^2-u^2 = 2as \\\; a=\frac{v^2-u^2}{2s} \\\; = \frac{-16}{2\times 16}= -0.5$

$Force=ma=40\times (-0.5) = -20N$
$\\\;work\; done\; on\; the\; trolley\;=20N\times 16m=320 J.\\\; work \;done \;by \;the\; girl =0 J.$

Question:24

Solution.
Mass of the box = 250 kg
Height attained = 1 m
(a) Work done in lifting the block will be stored as potential energy
$W=PE=mgh$
After calculating, the work done will be 2500 joule.
(b) When they hold it in the air without any movement, the displacement of box will be zero.
We know work is defined as:
$W=Fd\cos \theta$
Hence, work done will be zero.
(c) When they hold it in air without any movement, their muscles are stretched.
These stretched muscles requires energy and causes fatigue and tiredness.

Question:25

What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tons of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? $(g = 10 m s^{-2})$

Solution

Power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = F v.
Power is defined as work done per unit time and its unit will be watt or kilowatt.
If we multiply power with time, we will get energy or work done.
Therefore, kilowatt -hour will be the unit of energy.
[Only kilowatt is the unit of power and that is not unit of energy.]
In one minute all the potential energy of water can be utilised.
Energy utilised = Potential energy = mgh
$PE=(2000\times 10^{3}kg)\times 10ms^{-2}\times 20m$
$=4\times 10 ^{8}Joule$
Now time t = 1 minute = 60 seconds
Hence power will be
$power=\frac{energy}{time}=\frac{4\times 10^{8}Joule}{60seconds}=6.67\times 10^{6}watt$

Question:26

How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of $1 m s^{-1}$ vertically? $(g = 10 m s^{-2})$

Power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = F v.
Power is defined as work done per unit time and its unit will be watt or kilowatt.
If we multiply power with time, we will get energy or work done.
If a man is lifting a body with constant speed, he has to apply force equal to his weight.
Force applied: $F = mg = 10m N$
Speed attained: $V = 1 m s^{-1}$
Power used: P = 100W
By using the formula of power:
$100 = (10m) (1)$
$m = 10kg.$

Question:27

Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of $20 m s^{-1}$ ?

Solution.
Power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = F v.
Power is defined as work done per unit time and its unit will be watt (in SI unit) or kilowatt.
If we multiply power with time, we will get energy or work done.
$power=\frac{energy}{time}$
$unit of power=\frac{joule}{second}=watt$
$1 kilowatt = 1000 watt = 1000 joule/second.$
For the car:
Mass of the car: $m = 150 kg$
Speed attained: $V = 20 m s^{-1}$
Power used: $P = 500W per kg$ hence,
Total power will be:
$Power=500\times 150=75000watt$
By using the formula of power:
$75000 = (Force) (20)$
$Force = 3750 N.$

Question:28

Compare the power at which each of the following is moving upwards against the force of gravity? $(given: g = 10 m s^{-2})$
(i) a butterfly of mass $1.0 g$ that flies upward at a rate of $0.5 m s^{-1}$.
(ii) a 250 g squirrel climbing up on a tree at a rate of $0.5 m s^{-1}$

Solution.
Power of anything is defined as the rate of work done by that thing or energy supplied per unit time.
Mathematically, power is defined as the product of force and velocity.
Power = F v.
In lifting a body with constant speed, it has to apply force equal to his weight.
(i)For the Butterfly:
Mass of the butterfly: $m = 1 gram = 0.001 kg$
Speed attained: $V= 0.5 m s^{-1}$
Force required: $F = mg = 0.01 N$
By using the formula of power:
$Power=0.01\times 0.5=5\times 10^{-3}watt$
(ii)For the Squirrel:
Mass of the squirrel: m =250 gram = 0.250 kg
Speed attained: V = 0.5 m s–1
Force required: F = mg = 2.5 N
By using the formula of power:
$Power=2.5\times 0.5=1.25watt$

NCERT Exemplar Solutions Class 9 Science Chapter 11 Important Topics:

The chapter on Work and Energy covers the below-mentioned topics via NCERT exemplar Class 9 Science solutions chapter 11 :

• Work, power, and energy discussed quantitatively.
• The mathematical definition of work is given and it is proved that the work done by any force will be equal to change in kinetic energy of the system.
• NCERT exemplar Class 9 Science solutions chapter 11 discusses the concept of mechanical energy is discussed which constitutes kinetic energy and potential energy.
• In this chapter power of any machine or force is also mathematically defined which is the rate of work done by that particular force or machine.
• The importance of the angle between force and displacement in the calculation of work is clearly explained in this chapter of work and energy.

NCERT Class 9 Science Exemplar Solutions for Other Chapters:

 Chapter wise solutions Chapter 1 Matter in our Surroundings Chapter 2 Is Matter Around Us Pure? Chapter 3 Atoms and Molecules Chapter 4 Structure of the Atom Chapter 5 The Fundamental Unit of Life Chapter 6 Tissues Chapter 7 Diversity in Living Organisms Chapter 8 Motion Chapter 9 Forces and Laws of Motion Chapter 10 Gravitation Chapter 12 Sound Chapter 13 Why do We Fall ill? Chapter 14 Natural Resources Chapter 15 Improvement in Food Resources

Features of NCERT Exemplar Class 9 Science Solutions Chapter 11:

These Class 9 Science NCERT exemplar chapter 11 solutions provide the basic understanding of work, energy, and power. These NCERT exemplar Class 9 Science solutions chapter 11 can be utilized by the students of Class 9 as reference material to get a better understanding of the work and energy-based practice questions.

NCERT exemplar Class 9 Science solutions chapter 11 pdf download enables the students to study and practice NCERT exemplar Class 9 Science chapter 11 in an offline environment. This feature is provided to the students for an uninterrupted learning experience and for preparing for the JEE Main and NEET entrance exams in a better way.

Also, read - NCERT Solutions for Class 9

Check the Solutions of Questions Given in the Book

 Chapter No. Chapter Name Chapter 1 Matter in Our Surroundings Chapter 2 Is Matter Around Us Pure Chapter 3 Atoms and Molecules Chapter 4 Structure of The Atom Chapter 5 The Fundamental Unit of Life Chapter 6 Tissues Chapter 7 Diversity in Living Organisms Chapter 8 Motion Chapter 9 Force and Laws of Motion Chapter 10 Gravitation Chapter 11 Work and Energy Chapter 12 Sound Chapter 13 Why Do We Fall ill? Chapter 14 Natural Resources Chapter 15 Improvement in Food Resources

Also Check NCERT Books and NCERT Syllabus here

1. When an apple falls from the tree what will happen to its potential energy?

We know the potential energy is due to gravitational force and it is also known as gravitational potential energy. This gravitational potential energy depends on the height from the surface of earth; more height means more energy. Therefore, if an apple falls, its potential energy will decrease.

2. A boy raises 50 blocks of one KG up to height 5 m in 50 turns and another man raises one block of 50 KG to the same height; who has done more work and who has more power?

Both of them have raised the same amount of mass up to the same height; therefore, both of them have done equal amounts of work. The power of the man is apparently more than the power of the boy because the same work is done by man in less time.

3. How important is the chapter Work and Energy for competitive exams like JEE Advanced and NEET?

This is one of the most important chapters for JEE Main, JEE Advanced and NEET. This chapter will be dealt in Class 11 NCERT Physics in detail.

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