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Case Study Questions Class 9 Science Work and Energy

Case study questions class 9 science chapter 11 work and energy.

CBSE Class 9 Case Study Questions Science Work and Energy. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Work and Energy.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Work and Energy

(1 ) Work done by force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction. Work done is negative when the force acts opposite to the direction of displacement. Work done is positive when the force is in the direction of displacement.The unit of work is newton-metre (N m)or joule (J).

(i) Work done is

(a) Scalar quantity

(b) Vector quantity

(d) None of these

(ii) When force acts against the direction of displacement then work done will be

(a) positive

(b) negative

(iii) SI unit of work is

(a) Joule(J)

(b) Newton meter(N-m)

(iv)You are lifting stone from floor. Work is done by theforce exerted by you on the stone. Theobject moves upwards. The force youexerted is in the direction ofdisplacement. However, there is theforce of gravity acting on the object. Which one of these forces is doingpositive work?

Which one is doing negative work?

(v) Define 1J of work.

Answer key-1

(iv) Here work done by you is positive work as work is being done in the direction of displacement unlike in case of gravitational force which acts in downward direction against the direction of displacement which is in upward direction.

(v) When 1 Newton of force acts on body and body displaces from its position by 1 meter then the work done is said to be 1 joule (J).

(2) A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet, blowing wind, a rotating wheel, a speeding stone can do work. How does a bullet pierce the target? How does the wind move the blades of a windmill? Objects in motion possess energy. We call this energy kinetic energy.

Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is

KE = ½ *mv 2

The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J).

(i) Energy possessed by body which is in motion is called

(a) Potential energy

(b) Kinetic energy

(ii) Which of the following has same unit?

(a) Potential energy and Force

(b) Kinetic energy and work

(iii) Kinetic energy depends

(a) Inversely on velocity of body

(b) Directly on square of velocity of body

(iv) Define kinetic energy of body. Give its SI unit

(v) Is kinetic energy scalar or vector? Justify your answer

Answer key-2

(iv) Energy possessed by object due to its motion is called as kinetic energy. Its SI unit is N-m or Joule(J).

(v) kinetic energy is scalar quantity as it is a work done and work done is scalar quantity hence kinetic energy is also scalar quantity and doesn’t have any direction.

(3) Lift an object through a certain height. The object can now do work. It begins to fall when released. This implies that it has acquired some energy. If raised to a greater height it can do more work and hence possesses more energy. From where did it get the energy? In the above situations, the energy gets stored due to the work done on the object. The energy transferred to an object is storedas potential energy if it is not used to cause a change in the velocity or speed of the object.An object increases its energy when raisedthrough a height. This is because work isdone on it against gravity while it is being raised. The energy present in such an objectis the gravitational potential energy.The gravitational potential energy of anobject at a point above the ground is definedas the work done in raising it from the ground by height h

to that point against gravity.Let the work done on the object against gravity beW. That is,

work done, W = force × displacement

Therefore potential energy (PE)= mg*h.

(i) Energy possessed by body due to its position is called

(ii) SI unit of potential energy is

(iii)You do work while winding the key of a toy car. The energy transferred to the spring inside is stored as

(iv)Find the energy possessed by an object of mass 5kg when it is at a height of 10 m above the ground. Given, g = 9.8 m/s 2 .

(v)Find the work done by Gravity on an object of mass 5 kg which moves from height 10m to ground when it is released from height of 10 m. Given, g = 9.8 m/s 2 .

Answer key-3

(iv) we have potential energy as

=5 ×9.8 ×10

(v) work done, W = force × displacement

= 5 ×9.8 ×10

(4) The form of energy can be changed from one form to another. What happens to the totalenergy of a system during or after the process?Whenever energy gets transformed, the totalenergy remains unchanged. This is the law ofconservation of energy. According to this law, energy can only be converted from one form to another it can neither be created nor destroyed. The total energy before and after the transformation remains the same.The lawof conservation of energy is valid inall situations and for all kinds of transformations. Thus during motion the sum of the potential energy and kinetic energy of the object would be the same at all points. That is, potential energy + kinetic energy = constant.Andcalled as mechanical energy.

(i) Which of the energy conversion occur in electric iron?

(a) Electric energy converted into heat energy

(b) Electric energy converted into light energy

(ii) When ball drops from height which of the energy conversion takes place

(a) Gravitational potential energy converted into kinetic energy

(b) Kinetic energy converted into Gravitational potential energy

(iii) When ball is thrown vertically upward which of the following quantity remains constant?

(iv) State law of conservation of energy.

(v) In hydroelectric power plant which energy conversion happens?

Answer key-4

(iv) This is the law ofconservation of energy. According to this law,energy can only be converted from one form to another it can neither be created nor destroyed. The total energy before and after the transformation remains the same.

(v) In hydroelectric power plant potential energy of water reservoir is converted into electric energy.

(5 ) A more powerful vehiclewould complete a journey in a shorter timethan a less powerful one. We talk of the powerof machines like motorbikes and motorcars.The speed with which these vehicles changeenergy or do work is a basis for theirclassification. Power measures the speed ofwork done, that is, how fast or slow work isdone. Power is defined as the rate of doingwork or the rate of transfer of energy. If anagent does a work W in time t, then power isgiven by

P= work/time

P= W/T. The unit of power is watt.

(i) The rate of doing work is defined as

(ii) Total energy consumed divided by total time taken is called as

(a) Average power

(b) Instantaneous power

(iii) Let A and B having same weight start climbing the rope and reach height of 10m. Let A takes 10sec while B takes 12sec then work done

(a) By both will be same

(b) By A is more than work done by B

(iv) Define 1 Watt of power

(v) An electric bulb of 20W is used for 5h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Answer key-5

(iv) A power is said to be 1 watt when 1 joule of work is done within 1 second of time.

(v) Power of electric bulb = 20 W

Time used, t = 5 h

Energy = power × time taken

= 0.02kW × 5 h

= 0.10 kW h

= 0.10‘units’.

The energy consumed by the bulb 0.10 units

Jaru mitaya

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Class 9 Science Case Study Questions Chapter 11 Work and Energy

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Case study Questions in Class 9 Science Chapter 11 are very important to solve for your exam. Class 9 Science Chapter 11 Class 9 Science Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 11 Work and Energy

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Work and Energy Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 11 Work and Energy

Case Study/Passage-Based Questions

Case Study 1: The figure shows a watch glass embedded in clay. A tiny spherical ball is placed at edge B at a height h above the center A

case study based questions class 9 physics work and energy

The kinetic energy of the ball, when it reaches point A is (a) zero (b) maximum (c) minimum (d) can’t say

Answer: (b) maximum

The ball comes to rest because of (a) frictional force (b) gravitational force (c) both (a) and (b) (d) none of these

Answer: (c) both (a) and (b)

The energy possessed by the ball at point C is (a) potential energy (b) kinetic energy (c) both potential and kinetic energy (d) heat energy.

Answer: (a) potential energy

Case Study 2: The principle of conservation of energy states that the energy in a system can neither be created nor be destroyed. It can only be transformed from one form to another, but the total energy of the system remains constant. Conservation of electrical energy to various forms or vice versa along with devices is illustrated in the figure given below.

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

Answer: (d) potential energy.

A battery lights a bulb. Describe the energy changes involved in the process. (a) Chemical energy →Light energy → Electrical energy (b) Electrical energy → Chemical energy → Electrical energy (c) Chemical energy → Electrical energy → Light energy (d) None of these.

Answer: (c) Chemical energy → Electrical energy → Light energy

Name a machine that transforms muscular energy into useful mechanical work. (a) A microphone (b) Bicycle (c) Electric torch (d) An electric bell

Answer: (b) Bicycle

A body is falling from a height of 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.

Answer: (c) half potential and half kinetic energy

Case Study 3: An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.4 m s –1 . For this purpose, a motor with the required horsepower is used

The power of motor in hp is (a) 2.33 (b) 2.43 (c) 2.53 (d) 2.63

Answer: (d) 2.63

Case Study 4: Work and energy are fundamental concepts in physics that help us understand the physical world and the processes happening around us. Work is done when a force is applied to an object, and the object moves in the direction of the applied force. It is calculated as the product of force and displacement. The unit of work is joule (J). Energy, on the other hand, is the ability to do work. It exists in different forms, such as kinetic energy, potential energy, and various other forms like thermal energy, electrical energy, and chemical energy. The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. Understanding the concepts of work and energy helps us analyze the efficiency of machines, calculate the amount of work done, and comprehend various physical phenomena.

When is work considered to be done on an object? a) When a force is applied to the object b) When the object moves in the direction of the applied force c) When the object remains stationary d) When the object changes its shape Answer: b) When the object moves in the direction of the applied force

How is work calculated? a) Force multiplied by velocity b) Force multiplied by acceleration c) Force multiplied by displacement d) Force divided by time Answer: c) Force multiplied by displacement

What is the unit of work? a) Newton (N) b) Meter (m) c) Joule (J) d) Watt (W) Answer: c) Joule (J)

What is energy? a) The ability to do work b) The force applied to an object c) The distance traveled by an object d) The mass of an object Answer: a) The ability to do work

According to the law of conservation of energy, what happens to energy? a) It can be created b) It can be destroyed c) It can be transformed from one form to another d) It remains constant Answer: c) It can be transformed from one form to another

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Class 9th Science - Work and Energy Case Study Questions and Answers 2022 - 2023

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QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Work and Energy, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Work and energy case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

(ii) What is the value of total energy of the bob at position A ?

(iii) What is the value of kinetic energy of the bob at mean position 'O' ?

(iv) What is the value of kinetic energy and potential energy of the bob at the position 'P' whose height above 'O' is 2 cm ?

(v) What is kinetic energy? (a) Energy acquired due to motion (b) Energy acquired due to rest (c) Sum of Potential and mechanical energy (d) It is the energy stored inside a body.

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Work and energy case study questions with answer key answer keys.

(i) (a) 0.05 J The work done in raising the bob through a height of 5 cm (against the gravitational attraction) gets stored in the bob in the form of its potential energy. PE = mgh = 0.1 x 10 x 5 x 10-2 = 0.05 J (ii) (b) 0.05 J At position A, PE = 0.05 J, KE = 0 So, Total energy = 0.05 J (iii) (c) 0.05 J At mean position, potential energy is zero, hence KE at O = 0.05 J (iv) (d) P.E. = 0.02 J and K.E. = 0.03 J PE at P = mgh = 0.1 x 10 x 2 x 10-2 = 0.02 J K.E = Total energy – PE = 0.05 – 0.02 = 0.03 J (v) (a) Energy acquired due to motion

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NCERT Solutions for Class 9 Science Chapter 11 - Work And Energy

NCERT Solutions
Chapter 11 Work And Energy

NCERT Solutions for Class 9 Science Chapter 11 - Work and Energy - Free PDF Download

The class 9 chapter 11 science is now made easier for the students with the comprehensive NCERT Class 9 science solutions chapter 11 PDF. Every topic and subtopic is meticulously elaborated for students to understand clearly. You can now find the PDFs of these explanations on the Vedantu website. These solutions are the one-stop guide to preparing for your exams with ease. Now, refer to them to get in-depth knowledge on essential topics of Work and Energy to ace your exams with ease. Vedantu is a platform that provides free NCERT Solutions and other study materials for students. Maths Students who are looking for better solutions, they can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Access NCERT Answers for Class 9 Science Chapter 11 – Work and energy

Intext exercise 1.

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?

Ans: In the above question it is given that:

Force is $F=7N$.

Displacement is $S=8m$

Work done is given by the formula:

$\text{Work done = Force }\!\!\times\!\!\text{ Displacement}$

$W=F\times S=7\times 8=56Nm$

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

Intext Exercise 2

1. When do we say that work is done?

Ans: Work is said to be done if the following two conditions are satisfied:

The force must act on the object

The object should be displaced.

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

Ans: Consider a constant force $F$ acting on an object. If the object is displaced by a distance $s$ in the direction of the force (Fig. 11.1) and \[W\] is the work done. Hence work done is equal to the product of the force and displacement.

$\Rightarrow W=F\times S$

3. Define 1 J of work.

Ans: $1J$ of work is the amount of work done on an object when a force of $1N$ displaces it by $1m$ along the line of action of the force.

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?

Applied force is $F=140N$.

Displacement is $s=15m$.

We know that

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

Therefore, work done in ploughing the length of the field is $2100J$.

Intext Exercise 3

1. What is the kinetic energy of an object?

Ans: The energy possessed by an object due to its motion is called kinetic energy.

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

Ans: Consider a body of mass $'m'$ to be moving with a velocity $'v'$. Hence its kinetic energy ${{E}_{k}}$ is given by:

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

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

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?

Velocity of the object is $5m/s$.

Kinetic energy of the object is ${{E}_{k}}=25J$.

Mass of the object is $m$.

We know that,

Case 1: When the velocity of an object is doubled,

Kinetic Energy is given by:

${{E}_{k}}\propto {{v}^{2}}$

Hence, when $v=10m/s$

${{E}_{k}}=25\times 4=100J$

The kinetic energy becomes 4 times its initial value.

Case 2: If velocity is increased three times, then its kinetic energy will be nine times its original value, as ${{E}_{k}}\propto {{v}^{2}}$.

Hence, kinetic energy will be:

${{E}_{k}}=25\times 9=225J$

Thus, kinetic energy becomes nine times its initial value.

Intext Exercise 4

1. What is power?

Ans: Power is defined as the rate of doing work or the rate of transfer of energy. If W is the amount of work done in time t, then power is given by:$\text{Pow}er=\frac{Work\text{ }done}{Time}=\frac{Energy}{Time}$

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

The S.I. unit power is watt $\left( W \right)$.

2. Define 1 watt of power.

Ans: The power of an agent, which does work at the rate of 1 joule per second is defined as $1$ watt. It is also said that power is $1W$ when the rate of consumption of energy is $1J/s$.

$1W=\frac{1J}{1s}$

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

Energy consumed is $E=1000J$

Time $T=10s$

$\text{Pow}er=\frac{Work\text{ }done}{Time}=\frac{Energy}{Time}$

$\Rightarrow Power=\frac{1000}{10}=100W$

Hence, the power generated is$100W$.

4. Define average power.

Ans: Average power is defined as the power obtained by dividing the total amount of work done in the total time taken to do this work.$\text{Average power}=\frac{\text{Total work done}}{\text{total time taken}}$

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.

Ans: According to the definition, work is done whenever the following two conditions are satisfied:

A force acts on the body.

There is a displacement of the body by the application of force in or opposite to the direction of force.

In this case of swimming, Suma applies a force to push the water backwards. This allows Suma to swim in the forward direction by the forward response of water. Hence, the force causes a displacement. Thus, while swimming work is done by Suma.

In this case the donkey applies a force in the upward direction while carrying a load. But the displacement of the load is in the forward direction. As the displacement is perpendicular to force, the work done is zero.

In this case work is done by the windmill in lifting water from the well as a windmill works against the gravitational force to lift water.

In this case there is no displacement of the leaves of the plant. Hence, the work done is zero.

In this case an engine applies force to pull the train. This makes the train move in the direction of the applied force. Thus, there is a displacement in the train in the same direction. Hence the work is done by the engine on the train.

In this case during the process of food grains getting dried in the sun, no work is done as food grains do not displace in the presence of solar energy.

In this case wind energy applies a force on the sailboat to push it in forward direction. Therefore, the displacement of the boat is along the direction of the force. Hence, the work is done by wind on the boat.

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?

Ans: In the above question the 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.

Work done by the force of gravity on an object depends only on vertical displacement. Vertical displacement is given by the difference in the initial and final positions (height) of the object, which is zero.

Work done by gravity is expressed as, \[W\text{ }=\text{ }mgh\]

Where,

m is the mass of the object,

g is the acceleration due to gravity

h is the vertical displacement, which is zero.

\[\Rightarrow W\text{ }=\text{ }mg\left( 0 \right)=0J\]

Thus, the work done by gravity on the given object is zero joules.

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

Ans: Once a bulb is connected to a battery its chemical energy is converted into electrical energy. When the bulb receives this electrical energy, it is converted into light and heat energy. Therefore, the transformation of energy in the given situation can be shown as:\[\text{Chemical Energy}\to \text{Electrical Energy}\to \text{Light Energy}\to \text{Heat energy}\]

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

Ans: Change in kinetic energy is defined as the work done.

It is given that a 20 kg mass changes its velocity from $5m/s$ to $2m/s$.

Kinetic energy is given by the expression,

$\left( {{E}_{k}} \right)v=\frac{1}{2}m{{v}^{2}}$

${{E}_{k}}=$ Kinetic energy of the object moving with a velocity, v

$m=$ Mass of the object

Kinetic energy when the object was moving with a velocity $5m/s$,${{\left( {{E}_{k}} \right)}_{5}}=\frac{1}{2}\times 20\times {{\left( 5 \right)}^{2}}=250J$

Kinetic energy when the object was moving with a velocity $2m/s$,${{\left( {{E}_{k}} \right)}_{2}}=\frac{1}{2}\times 20\times {{\left( 2 \right)}^{2}}=40J$

Therefore, work done by force $=40-250=-210J$

Here, the negative sign shows that the force is acting in the opposite direction of the motion of the object.

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.

Ans: Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Hence, work done by gravity is given by:

\[W\text{ }=\text{ }mgh\]

Vertical displacement, $h=0$

Thus, the work done by gravity on the object is zero joules.

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

Ans: When the body drops from a height, its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equivalent to an increase in the kinetic energy of the body.

During this process, the total mechanical energy of the body is conserved. Therefore, the law of conservation of energy is not violated.

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

Ans: When we ride a bicycle, our muscular energy is transferred into the bicycle's heat energy and kinetic energy. Heat energy heats the body and kinetic energy provides a velocity to the bicycle. The transformation can be shown as:\[Muscular\text{ Energy}\to Kinetic\text{ Energy}\to \text{Heat energy}\]

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?

Ans: When we push a huge rock there is no transfer of energy because there is no transfer of muscular energy to the stationary rock. Instead muscular energy is transferred into heat energy, which causes our body to become hot.

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

Ans: We know that \[1\] unit of energy is equal to \[1\] kilowatt hour (kWh).\[1\text{ }unit\text{ }=\text{ }1\text{ }kWh\]

$1kWh=3.6\times {{10}^{6}}J$

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.

\[h\text{ }=\text{ }Vertical\text{ }displacement\text{ }=\text{ }5\text{ }m\]

\[m\text{ }=\text{ }Mass\text{ }of\text{ }the\text{ }object\text{ }=\text{ }40\text{ }kg\]\[g\text{ }=\text{ }Acceleration\text{ }due\text{ }to\text{ }gravity\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\]

Gravitational potential energy is given by the expression,

\[\Rightarrow W\text{ }=\text{ }40\times 5\times 9.8\text{ }=\text{ }1960\text{ }J\]

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

At this point, the object has potential energy equal to kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be \[980J\].

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

Ans: When the direction of force is perpendicular to displacement, the work done is zero. If a satellite moves around the Earth 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.

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.

Ans: Yes, there be displacement of an object in the absence of any force acting on it by moving with uniform velocity. Suppose an object is moving with constant velocity, then net force acting on it is zero. Hence, there can be a displacement without 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.

Ans: If a person holds a bundle of hay over his head, gravitational force is acting on the hay downwards. But, there is no displacement in the bundle of hay in the direction of force. Hence, no work is done.

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

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

\[P\text{ }=\text{ }1500\text{ }W\text{ }=\text{ }1.5\text{ }kW\]

\[T\text{ }=\text{ }10\text{ }hrs\]

\[Work\text{ }done\text{ }=\text{ }Energy\text{ }consumed\text{ }by\text{ }the\text{ }heater\]

Hence, \[energy\text{ }consumed\text{ }=\text{ }Power\text{ }\times \text{ }Time=1.5\times 10=15kWh\]= 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?

Ans: According to the “law of conservation of energy”: Energy can be neither created nor destroyed. It can only be converted 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 the extreme point, bob comes to rest and the kinetic energy of the bob is transformed into potential energy.

As it moves towards point P, its potential energy decreases progressively. Simultaneously, the kinetic energy increases.

As the bob reaches point P, its potential energy is converted to kinetic energy. At this point 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 due to some friction in the air.

The pendulum loses energy overcoming this friction. After all the energy is lost, it comes to rest. The law of conservation of energy holds here as 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 is conserved.

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?

Ans: When a body is in motion it possesses kinetic energy. Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,

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?

Mass of car, \[m\text{ }=\text{ }1500\text{ }kg\]

Velocity of car, \[v\text{ }=\text{ }60\text{ }km/h\text{ }=60\times \frac{5}{18}m/s\]

Kinetic energy is given by:

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

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:

Force and displacement are perpendicular

Here the direction of force acting on the block is perpendicular to the displacement.

Hence, work done by force on the block will be zero.

Here the direction of force acting on the block is in the direction of displacement. So, work done by force on the block will be positive.

Force and displacement are anti-parallel

Here the direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block 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?

Ans: Acceleration of an object will be zero when net force acting on an object will be zero. Net force can be zero even when there are multiple forces acting on the body. For a uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is zero. Hence, Soni is right.

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

\[P\text{ }=\text{ }500\text{ }W\text{ }=\text{ }0.50\text{ }kW\]

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

\[Work\text{ }done\text{ }=\text{ }Energy\text{ }consumed\text{ }by\text{ }the\text{ }device\]\[Energy\text{ }consumed\text{ }by\text{ }each\text{ }device\text{ }=\text{ }Power\text{ }\times \text{ }Time\]\[\text{Work done}=\text{ }0.50\text{ }\times \text{ }10\text{ }=\text{ }5\text{ }kWh\]

Hence, the energy consumed by four equal rating devices in \[10\text{ }hrs\] will be \[4\times 5kWh=20kWh\]

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

Ans: Once the object hits the hard ground, the kinetic energy gets converted into heat energy and sound energy. Further, entire energy is lost to the environment. This energy can also deform the ground with respect to the nature of the ground and the amount of kinetic energy possessed by the object.

NCERT Solutions for Class 9 Science - Free PDF Download

You can opt for Chapter 11 - Work and Energy NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of 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

Along with this, students can also view additional study materials provided by Vedantu, for Class 9

NCERT Solutions For Class 9

Revision Notes for Class 9

CBSE Class 9 Syllabus

To get hands-on an elaborate presentation of Class 9 work and energy solutions, go through the free PDFs offered on the official webpage of Vedantu. Grasp all essential topics for preparing well for your exam.

List of Topics Covered Under NCERT Solutions for Class 9 Science Chapter 11 - Work and Energy

The scientific conception of Work

Work done by a constant force

Forms of energy

Kinetic energy

Potential energy

Potential energy of an object at a height

Law of conservation of energy

Rate of Doing Work

Commercial unit of Work

A Glance About The Class 9 Science Chapter 11 - Work and Energy

The work done is the product of force acting on an object and displacement of the object

The capacity of an object to do work is called Energy

The energy is of two types, kinetic and potential energy

Chemical energy, Mechanical energy, Light energy..etc are the forms of energy.

An object of mass that moves with constant velocity is called Kinetic energy

The energy which is transferred and stored to the object, while work is done is called Potential energy.

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

The NCERT solutions for class 9 science chapter 11 involve brief explanations of work and energy. They offer insights into the very concept of work and how ‘work’ is defined in our everyday life. Discover some prime examples through the below-mentioned sub-topics to know more about work.

11.1.1 Not Much ‘Work’ in Spite of Working Hard

NCERT Class 9 science chapter 11 solutions give students insights into distinguishing examples of work. This is explained by talking about the core subject, Kamali. She is keen on preparing for her exams. She is dedicated to following her time table and studying for exams whenever feasible. She organizes, analyses, and maps her thoughts, thereby collecting relevant study materials for studying.

Thus, all these activities that Kamali performs require a sufficient amount of energy. In standard parlance, Kamali is working hard. In a nutshell, the definition of work varies from individual to individual. Activities like talking to friends, going out with your family, playing your favourite game, or watching a movie fall under the ‘not much work despite working hard’ category.

11.1.2 Scientific Conception of Work

Activities where the subject and object both have an impact after work is being done, in general, is nothing hit the scientific conception of work. Work and Energy Class 9 NCERT Solutions deal with different situations that fall under the scientific conception of work.

From a scientific perspective, work can be defined as pushing a pebble lying down on a surface, lifting a book through a respective height, and so on. Thus, for work two be done, two conditions need to be satisfied. These include- the action of force on an object and the displacement of the item.

11.1.3 Work Done By a Constant Force

Ever wondered how work is defined by science? One can easily understand this by a simple demonstration as follows. Take a situation wherein the force is acting in the very direction of displacement.

Now, consider a constant force F acting on a specific object. Let this object be displayed by a particular distance S in the exact direction of the force. Here:

Work Done = Force x Displacement

Dive into the ncert solutions for class 9 chapter 11 to know more.

11.2 Energy

We must all understand by now that energy is a necessity for life itself. The need for energy seems to increase at an exceeding rate. However, the question lies, where does energy come from?

Well, the sun constitutes the largest natural source of energy for living creatures. Several energy sources are primarily extracted from the sun. Additionally, we can also derive energy from nuclei of atoms, tides, and the interior of the earth.

11.2.1 Forms of Energy

Our world offers a myriad of forms of energy. These include- potential energy + kinetic energy (mechanical energy), light energy, chemical energy, electrical energy, and heat energy.

11.2.2 Kinetic Energy

The kinetic energy of a certain object is nothing but the energy that it consists of because of its motion. It is known as the work required to accelerate the respective body of a mass from rest to its actual velocity.

11.2.3 Potential Energy

The energy possessed by the respective body mainly due to the change of shape or position is known as potential energy.

11.2.4 Potential Energy of An Object At A Height

A particular object enhanced its energy after being raised by a height. This happens because work is being done on it in opposition to gravity while being raised. This energy available in a relevant object is known as gravitational potential energy.

11.2.5 Are Various Energy Forms Inconvertible?

Is energy conversion from one form to another possible? Go through NCERT Solutions for Class 9 Science Chapter 11 to understand how energy conversion to different forms occurs.

11.2.6 Law of Conservation of Energy

The law of conservation of energy states that energy is only transformed from one form to another. It cannot be created, not destroyed. The sum total of energy before and after the transformation is always constant.

11.3 Rate of Doing Work

Power is primarily defined as the overall rate of doing work. The SI unit of power is defined as a watt.

11.3.1 Commercial Unit of Energy

Joule is an extremely small unit and is thus not reliable to define definite and vast quantities of energy. This is why a bigger unit called kilowatt hour (kW h) is used. The energy accessed in about an hour at the rate of 1kW is known as 1 kW h.

Key Features of NCERT Solutions for Class 9 Science Chapter 11

Key Features of NCERT Solutions, These solutions are designed to help students achieve proficiency in their studies. They are crafted by experienced educators who excel in teaching class 9 Science. Some of the features include:

Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

Clear and structured presentation for easy comprehension.

Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

Visual aids like diagrams and illustrations to simplify complex concepts.

Additional tips and insights to enhance students' performance.

Chapter summaries for quick revision.

Online accessibility and downloadable resources for flexible study and revision.

The NCERT Solutions for Class 9 chapter 11 , Science - Word and Energy, provided by Vedantu, is a valuable tool for Class 9 students. It helps introduce Science concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for Class 9 Students to understand the material. By using Vedantu's resources, Students can develop a deeper understanding of NCERT concepts. These solutions are a helpful aid for class 9 students, empowering them to excel in their studies and develop a genuine appreciation for Work and Energy.

FAQs on NCERT Solutions for Class 9 Science Chapter 11 - Work And Energy

1. How Many Exercises Does this Chapter Include?

Work and energy is a vast chapter that includes about 5 exercises in total. These are mentioned below.

Exercise 11.1 - 1 Question on work and power.

Exercise 11.2 - 4 questions on energy and units of energy.

Exercise 11.3 - 3 questions on different forms of energy.

Exercise 11.4 - 4 questions on kinetic energy and its expression.

Exercise 11.5 - 21 questions on potential energy and its expression and conservation of energy.

2. Why Should you Rely on Vedantu NCERT Solutions?

NCERT solutions in class 9 Chapter 11 by Vedantu aim to help students get definite clarity on what they should focus on while preparing for their exams. Along with brief summaries on every topic and subtopic, you will also have access to different question banks that you can refer to. Further, all the questions and answers and explanations are precise, easy to understand, and reliable for quick preparation of exams. With the right study material, you can take the benefit of the doubt and gain essential knowledge on the concepts of work and energy.

3. What is the kinetic energy of an object in Chapter 11 of NCERT Solutions for Class 9 Science?

The Kinetic Energy of a particular object is the energy that it has because of its motion. It is known as the work required to accelerate the respective body of a given mass from rest to its actual velocity. Thus, any object in motion always has kinetic energy. For instance, a person walking or running, a thrown baseball, a crumb falling from a tabletop are all examples of Kinetic Energy. To know more about it, visit Vedantu website or Vedantu Mobile app. The solutions provided by Vedantu are free of cost.

4. What is work according to Chapter 11 of Class 9 Science NCERT?

Work done is as the product of two components- force (F) acting in the direction of the displacement and the magnitude of displacement (s) of the object. Thus, the two conditions required for work to be done are the action of force on an object and the displacement of the object. The formula for work done is: Work Done = Force x Displacement, W = Fs. It is a scalar quantity (because it has only magnitude and no direction) and it is measured in Joules (J).

5. What is energy according to Chapter 11 of Class 9 Science NCERT?

A body's Energy is defined, in Physics, as the capacity of that body to perform work. Energy has only magnitude and no direction i.e., it is a scalar quantity and is measured in Joules (J). The Sun constitutes the largest natural source of energy for all living creatures. Some other energy sources are nuclei of atoms, tides, and the interior of the Earth. There are several types of energy, such as Potential Energy, Kinetic Energy, Mechanical Energy, Light Energy, Chemical Energy, Electrical Energy, and Heat Energy.

6. What is the formula of Energy according to Chapter 11 of NCERT Solutions for Class 9 Science?

A body's Energy, in Physics, is defined as the capacity of that body to perform work. Energy’s SI unit is Joules. It is scalar in nature as it has only magnitude and no direction. The formula for Potential Energy, Ep, of an object is equivalent to the work done over an object of mass 'm' to raise it by a height 'h' when 'g' is the acceleration due to gravity. Therefore, Ep = mgh.

7. What is Work and Energy in Physics of Class 9?

In physics, work is defined as the sum of two factors: the force acting in the direction of displacement (F) and the amount of the object's displacement (s). The capacity of a body to produce work is defined as its energy in physics. Both Work and Energy are scalar quantities that are measured in Joules (J).

NCERT Solutions for Class 9

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Unit 4: Work & Energy

About this unit.

In this chapter, we will define the terms 'work' and 'energy'. We will learn how to calculate them, and use them to look at our world in a very different way.

Intro to work (Opens a modal)
Positive & negative work (Opens a modal)
Work done on lifting/falling things - Solved numerical (Opens a modal)
Energy intro (kinetic & potential) (Opens a modal)
Kinetic energy derivation (Opens a modal)
Gravitational potential energy derivation (Opens a modal)
Work done by gravity (path independent) (Opens a modal)
Using the kinetic energy equation 4 questions Practice

Work energy theorem

Work-energy theorem (Opens a modal)
Work done from kinetic energy - solved example (Opens a modal)
Calculating change in kinetic energy from a force 4 questions Practice

Law of conservation of energy

Law of energy conservation (Opens a modal)
Energy conservation - solved example (Opens a modal)
Power (Opens a modal)
Relating power and energy 4 questions Practice

Commercial unit of energy

Commercial unit of electrical energy (Opens a modal)
Solved example - Cost of operation of electrical device (Opens a modal)

myCBSEguide

Class 9 Science Case...

Class 9 Science Case Study Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

due to its high compressibility
large volumes of a gas can be compressed into a small cylinder
transported easily
all of these
shape, volume
volume, shape
shape, size
size, shape
the presence of dissolved carbon dioxide in water
the presence of dissolved oxygen in the water
the presence of dissolved Nitrogen in the water
liquid particles move freely
liquid have greater space between each other
both (a) and (b)
none of these
Only gases behave like fluids
Gases and solids behave like fluids
Gases and liquids behave like fluids
Only liquids are fluids

Answer Key:

(d) all of these
(a) shape, volume
(b) the presence of dissolved oxygen in the water
(c) both (a) and (b)
(c) Gases and liquids behave like fluids

Class 9 science case study question 2

12/32 times
18 g of O 2
18 g of CO 2
18 g of CH 4
1 g of CO 2
1 g of CH 4 CH 4
2 moles of H2O
20 moles of water
6.022 × 1023 molecules of water
1.2044 × 1025 molecules of water
(I) and (IV)
(II) and (III)
(II) and (IV)
Sulphate molecule
Ozone molecule
Phosphorus molecule
Methane molecule
(c) 8/3 times
(d) 18g of CH 4
(c) 1g of H 2
(d) (II) and (IV)
(c) phosphorus molecule

Class 9 science case study question 3

collenchyma
chlorenchyma
It performs photosynthesis
It helps the aquatic plant to float
It provides mechanical support
Sclerenchyma
Collenchyma
Epithelial tissue
Parenchyma tissues have intercellular spaces.
Collenchymatous tissues are irregularly thickened at corners.
Apical and intercalary meristems are permanent tissues.
Meristematic tissues, in its early stage, lack vacuoles, muscles
(I) and (II)
(III) and (I)
Transpiration
Provides mechanical support
Provides strength to the plant parts
None of these
(a) Collenchyma
(b) help aquatic plant to float
(b) Sclerenchyma
(d) Only (III)
(c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units: Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms: Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

Science-Textbook for class IX-NCERT Publication
Assessment of Practical Skills in Science-Class IX – CBSE Publication
Laboratory Manual-Science-Class IX, NCERT Publication
Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

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myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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Case Study Questions of Chapter 11 Work and Energy PDF Download

Case study Questions on Class 9 Science Chapter 11 are very important to solve for your exam. Class 9 Science Chapter 11 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 11 Work and Energy

Work and Energy Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 11 Work and Energy

Case Study/Passage Based Questions

Figure shows a watch glass embedded in clay. A tiny spherical ball is placed at the edge B at a height h above the centre A

The kinetic energy of the ball, when it reaches at point A is (a) zero (b) maximum (c) minimum (d) can’t say

Answer: (b) maximum

The ball comes to rest because of (a) frictional force (b) gravitational force (c) both (a) and (b) (d) none of these

Answer: (c) both (a) and (b)

The energy possessed by ball at point C is (a) potential energy (b) kinetic energy (c) both potential and kinetic energy (d) heat energy.

Answer: (a) potential energy

The principle of conservation of energy states that the energy in a system can neither be created nor be destroyed. It can only be transformed from one form to another, but total energy of the system remains constant. Conservation of electrical energy to various forms or vice versa along with devices is illustrated in the figure given below.

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

Answer: (d) potential energy.

Answer: (c) Chemical energy → Electrical energy → Light energy

Name a machine that transforms muscular energy into useful mechanical work. (a) Amicrophone (b) Bicycle (c) Electric torch (d) An electric bell

Answer: (b) Bicycle

Answer: (c) half potential and half kinetic energy

An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.4 m s –1 . For this purpose a motor of required horse power is used

The power of motor in hp is (a) 2.33 (b) 2.43 (c) 2.53 (d) 2.63

Answer: (d) 2.63

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 9 Science Work and Energy Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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Work and Energy

Class 9 - ncert science solutions, intext questions 1.

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Force (f) = 7 N

Displacement (S) = 8 m

Work done = Force × Displacement

Substituting we get,

W = 7 × 8 = 56 Nm or 56 J

Hence, work done = 56 J

Intext Questions 2

When do we say that work is done?

Work is said to be done when force applied on an object shows the displacement in that object. It is equal to the product of force and displacement.

Work done = force x displacement

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

Define 1 J of work.

1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

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?

Force (F) = 140 N

Displacement (S) = 15 m

W = 140 x 15 = 2100 J

Hence, 2100 J of work is done in ploughing the length of the field.

Intext Questions 3

What is the kinetic energy of an object?

The kinetic energy is the energy possessed by an object due to its motion. It increases when the speed increases.

Write an expression for the kinetic energy of an object.

Kinetic Energy (K E ) = 1 2 \dfrac{1}{2} 2 1 mv 2 . Its SI unit is Joule (J).

The kinetic energy of an object of mass m, moving with a velocity of 5 ms -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?

Kinetic Energy (K E ) = 25 J

Velocity (v) = 5 ms -1

Kinetic Energy (K E ) = 1 2 \dfrac{1}{2} 2 1 mv 2 .

25 = 1 2 \dfrac{1}{2} 2 1 x m x 5 2

25 x 2 = 25 x m

m = 50 25 \dfrac{50}{25} 25 50

When velocity is doubled :

v' = 10 ms -1

K E = 1 2 \dfrac{1}{2} 2 1 x 2 x 10 2

K.E. = 100 J

When velocity is increased three times, then

v'' = 15 ms -1

K E = 1 2 \dfrac{1}{2} 2 1 x 2 x 15 2

K.E. = 225 J

Hence, Kinetic Energy becomes 100 J when velocity is doubled and it becomes 225 J when velocity is increased three times.

Intext Questions 5

What is power?

Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by:

Power = Work done Time \dfrac{\text{Work done}}{\text{Time}} Time Work done

It is expressed in watt (W).

Define 1 watt of power.

1 watt is the power of an agent, which does work at the rate of 1 joule per second.

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

Time = 10 s

Work done = Energy consumed by the lamp = 1000 J

Power = 1000 10 \dfrac{1000}{10} 10 1000 = 100 Js -1 or 100 W

Hence, the power of the lamp is 100 W

Define average power.

Average power is defined as the ratio of total energy consumed to the total time taken by the body.

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.

(b) A donkey is carrying a load on its back.

(d) A green plant is carrying out photosynthesis.

(e) An engine is pulling a train.

(f) Food grains are getting dried in the sun.

(g) A sailboat is moving due to wind energy.

Work is said to be done when force applied on an object shows the displacement in that object.

(a) While swimming, Suma applies a force to push the water backwards. She swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, the work is done .

(b) While carrying a load, the donkey has to apply 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) A windmill works against gravity to elevate water. The windmill lift water by applying a force in an upward direction, and thus the water is moving in the same upward direction itself. Hence, work is done .

(d) No force is required when a green plant is carrying out photosynthesis and there is no displacement of plant. Hence, no work is done .

(e) When an engine is pulling a train, it is applying a force in the forward direction and the train is moving. As, displacement and force are in the same direction. Hence, work is done .

(f) As there is no force applied when grains are dried and there is no displacement as well. Hence, no work is done .

(g) When a sailboat is moving due to wind energy, it is applying force in the forward direction. So, it is moving in the forward direction. As, displacement and force are in the same direction. Hence, work is done .

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?

Work done by gravity depends on the vertical displacement of the body.

So, work done by gravity is :

Vertical displacement, h = 0 [∵ initial and the final points of the path of the object lie on the same horizontal line.]

∴ W = m × g × 0 = 0

Hence, work done by the force of gravity on the object = 0.

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

A battery converts chemical energy into electrical energy. When the bulb receives this electrical energy, it converts it into light and heat energy. Hence, energy changes involved in the process are :

Chemical Energy ⟶ Electrical Energy ⟶ Light Energy + Heat Energy.

Certain force acting on a 20 kg mass changes its velocity from 5 ms -1 to 2 ms -1 . Calculate the work done by the force.

Initial velocity u = 5 ms -1

Mass of the body = 20 kg

Final velocity v = 2 ms -1

Initial kinetic energy

E i = 1 2 \dfrac{1}{2} 2 1 mu 2

E i = 1 2 \dfrac{1}{2} 2 1 x 20 x 5 2 = 10 × 25 = 250 J

Final kinetic energy

E f = 1 2 \dfrac{1}{2} 2 1 mv 2 = 1 2 \dfrac{1}{2} 2 1 x 20 x 2 2 = 10 × 4 = 40 J

As, Work done = Change in kinetic energy = E f – E i

Work done = 40 J - 250 J

Work done = -210 J

Hence, work done by the force = -210 J.

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.

Mass (m) = 10 kg

Work done by gravity depends on the vertical displacement of the body. It is independent of the horizontal path.

So, work done by gravity is = m g h

Vertical displacement, h = 0 [∵ the line joining A and B is horizontal. ]

Hence, work done on object by gravity is zero.

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

No. When the body falls from a height, its potential energy changes into kinetic energy progressively. A decrease in the potential energy is equal to an increase in the kinetic energy of the body. Hence, throughout the process, the total mechanical energy of the body remains conserved.

Hence, the law of conservation of energy is not violated.

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

The rider's muscular energy is converted to heat energy and the bicycle's kinetic energy while riding a bicycle. Hence, energy transformations are :

Muscular energy ⟶ Kinetic energy + Heat energy

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?

When we push a huge rock and fail to move it, our muscular energy is not transferred to the rock as kinetic energy as there is no displacement of the rock. However, as per the law of conservation of energy, our muscular energy is transformed into heat energy that heats up our body and makes us sweat.

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

Energy (E) = 250 units

1 kWh = 3.6 x 10 6 J

1 unit of energy = 1 kWh

So, 250 units of energy = 250 × 3.6 × 10 6 = 9 × 10 8 J.

Hence, 250 units of energy = 9 × 10 8 J.

Question 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.

Mass (m) = 40 kg

Acceleration due to gravity = 10 ms -2

Height (h) = 5 m

Potential energy = ?

Potential energy = m g h

P.E. = 40 × 10 × 5 = 2000 J

Height when halfway down : = 5 2 \dfrac{5}{2} 2 5 = 2.5 m

Potential energy when halfway = ?

P.E. = 40 × 10 × 2.5 = 1000 J

According to the law of conservation of energy:

Total potential energy = potential energy halfway down + kinetic energy halfway down

2000 = 1000 + K.E. halfway down

K.E. at halfway down = 2000 - 1000 = 1000 J

Hence, Potential energy = 2000 J and kinetic energy at halfway down = 1000 joules .

Question 11

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

When a satellite revolves around the earth in a circular orbit the work done is zero as force of gravity acting on satellite is perpendicular to its displacement.

Question 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

Yes, a uniformly moving object can experience displacement even when no force is acting on it. According to Newton's first law of motion, an object in motion will continue in its straight-line motion unless acted upon by an external force. Therefore, displacement of an object can occur in the absence of any force acting on it.

Question 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.

Work is said to be done when force applied on an object shows the displacement in that object. In this case, as there is no displacement of the hay bundle, hence no work is done.

Question 14

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

Power of the heater = 1500 W

Converting W to kW

1000 W = 1 kW

So, 1500 W = 1500 1000 \dfrac{1500}{1000} 1000 1500 = 1.5 kW

Time taken = 10 h

Power = Energy consumed Time taken \dfrac{\text{Energy consumed}}{\text{Time taken}} Time taken Energy consumed

Energy consumed = Power x Time taken

Energy consumed = 1.5 x 10 = 15 kWh

Converting kWh to J

So, 15 kWh = 3.6 x 10 6 x 15 = 5.4 x 10 7 J

Hence, the energy consumed = 5.4 x 10 7 J

Question 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?

The law of conservation of energy states that energy can only be converted from one form to another; it can neither be created or destroyed. The total energy before and after the transformation remains the same.

Refer the figure of an oscillating pendulum bob shown below:

The kinetic energy decreases and the potential energy becomes maximum at B.

After a moment the the to and fro movement starts again.

So, from B to A, again the potential energy changes into kinetic energy and this process repeats again and again.

So, when the bob is in its state of to and fro movement it has potential energy at the extreme position B or C and kinetic energy at resting position A.

It has both the kinetic energy and potential energy at an intermediate position. However, the sum of kinetic and potential energy remain same at every point of movement.

The bob will eventually come to rest due to the frictional resistance offered by air on the surface of bob and pendulum loses its kinetic energy to overcome this friction and finally comes to rest.

The law of conservation of energy is not violated because the kinetic energy lost by the pendulum to overcome the friction is gained by surroundings. Hence, total energy of the system will remain conserved.

Question 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?

The kinetic energy of an object of mass m, moving with a velocity, v, is given by the expression,

Kinetic energy = 1 2 \dfrac{1}{2} 2 1 mv 2

In order to bring it to rest, its velocity has to be reduced to zero.

An external force has to absorb energy from the object, i.e., do negative work on it, equal to its kinetic energy = − 1 2 -\dfrac{1}{2} − 2 1 mv 2

Question 17

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 kmh -1 .

Mass (m) = 1500 kg

Initial velocity (v) = 60 kmh -1

Final velocity = 0

Converting kmh -1 to ms -1 : multiply by 5 18 \dfrac{5}{18} 18 5

60 x 5 18 \dfrac{5}{18} 18 5 = 50 3 \dfrac{50}{3} 3 50 ms -1

Work required to stop the moving car = Kinetic energy of car (K.E.)

K.E. = 1 2 \dfrac{1}{2} 2 1 mv 2

= 1 2 \dfrac{1}{2} 2 1 x 1500 x ( 50 3 ) (\dfrac{50}{3}) ( 3 50 ) 2

= 1 2 \dfrac{1}{2} 2 1 x 1500 x 2500 9 \dfrac{2500}{9} 9 2500

= 208333.3 J

Hence, work done = 208333.3 J.

Question 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.

The direction of force acting on the block is perpendicular to the displacement. Hence, work done by force on the block will be zero .

The direction of force acting on the block is in the direction of displacement. Hence, work done by force on the block will be positive .

The direction of force acting on the block is opposite to the direction of displacement. Hence, work done by force on the block will be negative.

Question 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?

Acceleration in an object could be zero even when several forces are acting on it. This happens when all the forces cancel out each other i.e., the net force acting on the object is zero.

Consider the below object:

F = 0 [∵ equal and opposite forces cancel out each other]

∴ V final = V initial

Hence, if the object was initially moving, it will continue to move with uniform velocity without acceleration.

Therefore, I agree with Soni's statement.

Question 20

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

Power rating (P) = 500 W

Converting W into kW

500 W = 500 1000 \dfrac{500}{1000} 1000 500 = 0.5 kW

Time (T) = 10 h

Energy consumed by each device = Power x Time taken

Energy consumed by each device = 0.5 x 10 = 5 kWh

Energy consumed by four devices = 4 x 5 = 20 kWh

So, 20 kWh = 3.6 x 10 6 x 20 = 7.2 x 10 7 J

Hence, the energy consumed = 7.2 x 10 7 J

Question 21

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

When an object falls freely towards the ground, its potential energy decreases, and kinetic energy increases. As the object touches the ground, all its potential energy becomes kinetic energy. When the object hits the ground, all its kinetic energy gets converted into heat energy and sound energy.

NCERT Solutions for Class 9 Science Chapter 11 Work and Energy

Chapter 11 Work and Energy NCERT Solutions for Class 9 Science

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Extra Questions for Class 9 Science Chapter 11 Work and Energy

Extra questions for Class 9 Science Chapter 11 Work and Energy with answers is given below. Our subject expert prepared these solutions as per the latest NCERT textbook. These questions will be helpful to revise the all topics and concepts. CBSE Class 9 extra questions are the most simple and conceptual questions that are prepared by subject experts for the students to study well for the final exams. By solving these extra questions, students can be very efficient in their exam preparations.

Work and Energy Class 9 Science Extra Questions and Answers

Very short answer questions.

1: What is the work done against the gravity when a body is moved horizontally along a frictionless surface? Answer: Zero

2: When displacement is in a direction opposite to the direction of force applied, what is the type of work done? Answer: Negative work. 3: A 40 kg girl is running along a circular path of radius 1 m with a uniform speed. How much work is done by the girl in completing one circle? Answer: Zero

4: Seema tried to push a heavy rock of 100 kg for 200 s but could not move it. Find the work done by Seema at the end of 200 s. Answer: Work done = 0 Since displacement, s = 0

5: Identify the kind of energy possessed by a running athlete. Answer: Kinetic energy.

Short Answer Type Questions

1: An electrical heater is rated 1200 W. How much energy does it use in 10 hours?

Answer: Electrical energy = Power × time taken = 1.2 × 10 = 12 kWh

2: If an electric appliance is rated 1000 W and is used for 2 hours. Calculate the work done in 2 hours.

Answer: Work done = Energy consumed Energy = Power × Time taken = 1000 W × 2 hour = 2000 W-hr or 2 kW-hour or 2 kWh

3: A man of mass 62 kg sums up a stair case of 65 steps in 12 s. If height of each step is 20 cm, find his power.

Answer: m = 62 kg, g = 10 m/s 2 h = 65 × 20/100 = 13 m

P.E. = mgh P.E. = 62 × 10 × 13 = 8060 J

Power (𝑃) = (𝑃.𝐸.)/𝑡 = 8060/12 = 671.67 𝑊

4: How is work done by a force measured? A porter lifts a luggage of 20 kg from the ground and puts it on his head 1.7 m above the ground. Find the work done by the porter on the luggage. (g = 10m/s 2 )

Answer: Work done is product of force and displacement W=F × s m=20 kg g=10m/s 2 h =1.7𝑚 The work done by the porter = mgh = 20 × 10 × 1.7 = 340 J

5: 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 motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 1

6: When a force retards the motion of a body, what is the nature of work done by the force? State reason. List two examples of such a situation.

Answer: The work done by the force is negative because the displacement is opposite to the direction of force applied. Example: (i) Work done by the force of friction; (ii) Work done by applying brakes.

7: When is the work done by a force said to be negative ? Give one situation in which one of the forces acting on the object is doing positive work and the other is doing negative work.

Answer: Negative work: When the force is acting opposite to the direction of the displacement, the work done by the force is said to be negative. When we lift an object, two forces act on the (i) Muscular force: Doing positive work in the direction of the displacement. (ii) Gravitational force: Doing negative work opposite to the direction of the displacement.

8: (a) Under what conditions work is said to be done? (b) A porter lifts a luggage of 1.5 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.

Answer: (a) (i) Force should be applied. (ii) Body should move in the line of action of force. (iii) Angle between force and displacement should not be 90°.

(b) Mass of luggage, m = 15 kg and displacement, s = 1.5 m. Work done, W = F×s = mg × s = 15 × 10 × 1.5 = 225 J

9: Four persons jointly lift a 250 kg box to a height of 1 m and hold it. (i) Calculate the work done by the persons in lifting the box. (ii) How much work is done for just holding the box ? (iii) Why do they get tired while holding it ? (g = 10ms 2 )

Answer: (i) F = 250 × 10 = 2500𝑁 s =1 m W = F ×s = 2500 ×1| =2500𝐽

(ii) Zero, as there is no displacement. (iii) To hold the box, men are applying a force which is opposite and equal to the gravitational force acting on the box. While applying the force muscular effort is involved, and so they feel tired.

10: A boy is pulling a cart by supplying a constant force of 8 N on a straight path of 20 m. On a round about of 10 m diameter he forgets the path and takes 1½ turns and then continues on the straight path for another 20 m. Find the net work done by the boy on the cart.

Answer: F = 8N Work done, W = F × s

W 1 = 8 × 20 = 160 J

D =10 m So radius, D/2 = 10/2 = 5m

Circumference of a circle = 2πr = 2 × 22/7 × 5 = 31.43

Distance in 1⁄2 circle = πr = 22/7 × 5 = 15.71

Total distance for 1 ½ circle = 31.43 + 15.71 = 47.14 m

W 2 = F × s = 8 × 47.14 = 376 J

W 3 = 20 × 8 = 160 J

Total work done = 160 + 376 + 160 = 696 J

Long Answer Type Questions

1: Calculate the electricity bill amount for a month of 30 days, if the following devices are used as specified: (i) 2 bulbs of 40 W for 6 hours. (ii) 2 tubelights of 50 W for 8 hours. (iii) A TV of 120 W for 6 hours.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 2

Given the cost of electricity is ₹2.50 per unit.

Answer: Given the cost of electricity is ₹2.50 per unit. (i) 2 bulbs of 40 watts for 6 hrs. Energy consumed by Bulbs E 1 = 2 × 40 × 6 = 480 W = 0.48 kWh

(ii) Energy consumed by 2 tubelights E 2 = 50 × 8 × 2 = 0.800 kWh

(iii) Energy consumed by TV E 3 = 120 × 6 = 0.720 kWh Total Energy = 0.48 + 0.80 + 0.72 = 2.00 units rate = 2.50 per unit

Cost per day = 2 × 2.50 = 5.00 Cost 30 days = 5.00 × 30 = 150

2: (i) What is meant by mechanical energy? State its two forms. State the law of conservation of energy. Give an example in which we observe a continuous change of one form of energy into another and vice-versa. (ii) Calculate the amount of work required to stop a car of 1000 kg moving with a speed of 72 km/h. Answer: (i) Sum of kinetic energy and potential energy of an object is the total mechanical energy. Its two forms are kinetic energy and potential energy. Energy can neither be created nor be destroyed but can be transformed from one form to another. One example is simple pendulum.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 3

3. A boy pushes a book by applying a force of 40N. Find the work done by this force as the book is displaced through 25 cm along the path.

Answer: Here, force acting on the book, F = 40N distance through which book is displaced, s = 25 cm = 0·25 m Work done by the force, i.e., W = F × s = (40 N) (0·25 m) = 10J

4. A ball of mass 1 kg thrown upwards, reaches a maximum height of 4 m. Calculate the work done by the force of gravity during the vertical displacement. (g = 10 m/s 2 ).

Answer: Here, force of gravity on the ball, F = mg = (1 kg) (10 m/s 2 ) = 10N vertical displacement of the ball, s = 4m Since the force and the displacement of the ball are in opposite directions, work done by the force of gravity, i.e., W= F × s = 10 × 4 = 40J Obviously, work done against the force of gravity = 40J

5. An engine pulls a train 1 km over a level track. Calculate the work done by the train given that the frictional resistance is 5 × 10 5 N.

Answer: Here, frictional resistance, F = 5 × 10 5 N distance through which the train moves, s = 1 km = 1000 m Work done by the frictional force, i.e., W = Fs = (5 × 10 5 ) × 1000 = 5 × 10 8 J (F and s are in opposite directions) Obviously, work done by the train is 5 × 10 8 J

6. A man weighing 70 kg carries a weight of 10 kg on the top of a tower 100 m high. Calculate the work done by the man. (g = 10 m/s 2 ).

Answer: Here, force exerted by the man, F = (70 + 10) = 80 kg wt = 80 × 10 = 800 N vertical displacement, s = 100 m Work done by the man, i.e., W = F × s = (800N) (100m) = 80000 J

7. How fast should a man of mass 60 kg run so that his kinetic energy is 750 J ?

Answer: Here, mass of the man, m = 60 kg kinetic energy of the man, E k = 750J If v is the velocity of the man, then

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 4

8. Find the mass of the body which has 5J of kinetic energy while moving at a speed of 2 m/s. Answer: Here, kinetic energy of the body, E k = 5J speed of the body, v = 2 m/s If m is the mass of the body, then

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 5

9. A player kicks a ball of mass 250 g at the centre of a field. The ball leaves his foot with a speed of 10 m/s, Find the work done by the player on the ball.

Answer: The ball, which is initially at rest, gains kinetic energy due to work done on it by the player.

Thus, the work done by the player on the ball, W = kinetic energy (E k ) of the ball as it leaves his foot, i.e., W = E k = mv 2

Here, m = 250 g = 0·25 kg, v = 10 m/s

W = (0·25) ×(10) 2 = 12·5 J

10. A body of mass 5 kg, initially at rest, is subjected to a force of 20N. What is the kinetic energy acquired by the body at the end of 10s?

Answer: Here, mass of the body, m = 5 kg initial velocity of the body, u = 0 force acting on the body, F = 20 N time for which the force acts, t = 10 s If a is the acceleration produced in the body,

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 6

Let v be the velocity of the body after 10 s. Clearly, v = u We know that V = u + at When u = 0 v = 0 + at ⇒ v = 4 × 10 ⇒ v = 40 m/s

Kinetic energy acquired by the body, E k = mv 2 = 5 × (40) 2 = 4000J

11. A bullet of mass 20 g moving with a velocity of 500 m/s, strikes a tree and goes out from the other side with a velocity of 400 m/s. Calculate the work done by the bullet in joule in passing through the tree.

Answer: Here, mass of the bullet, m = 20 g = 0·02 kg initial velocity of the bullet, u = 500 m/s final velocity of the bullet, v = 400 m/s If W is the work done by the bullet in passing through the tree, then according to work-energy theorem W = mu 2 – mv 2 = m(u 2 – v 2 ) = (0·02) [(500) 2 – (400) 2 ] = 900J

12. A body of mass 4 kg is taken from a height of 5 m to a height 10 m. Find the increase in potential energy.

Answer: Here, mass of the body, m = 4 kg increase in height of the body, h = (10 – 5) = 5m Increase in potential energy, E p = mgh = 4 × 10 × 5 = 200J

Initial potential energy of the body, E pi = mgh = 4 × 10 × 5 = 200J

Final potential energy of the body, E pf = mgh f = 4 × 10 × 10 = 400J

Increase in potential energy, E p = E pf – E pi = 400J – 200J = 200J

13. A 5 kg ball is thrown upwards with a speed of 10 m/s. (a) Find the potential energy when it reaches the highest point. (b) Calculate the maximum height attained by it.

Answer: (a) Here, mass of the ball, m = 5 kg, speed of the ball, v = 10 m/s

Kinetic energy of the ball, E k = mv 2 = 5 × (10) 2 = 250J

When the ball reaches the highest point, Its kinetic energy becomes zero as the entire kinetic energy is converted into its potential energy (E p ) i.e., E p = 250J ….(i)

(b) If h is the maximum height attained by the ball, E p = mgh …. (i) From eqn. (i) and (ii),

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 7

14. A 5 kg ball is dropped from a height of 10m. (a) Find the initial potential energy of the ball. (b) Find the kinetic energy just before it reaches the ground and (c) Calculate the velocity before it reaches the ground.

Answer: Here, mass of the ball, m = 5 kg Height of the ball, h = 10m

(a) Initial potential energy of the ball, E p = mgh = 5 × 10 × 10 = 500J

(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into its kinetic energy (E k ), i.e., E k = 500J

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 8

15. A body is thrown up with a kinetic energy of 10 J. If it attains a maximum height of 5 m, find the mass of the body.

Answer: Here, kinetic energy of the body, E k = 10J maximum height attained by the body, h = 5m When the body attains maximum height, its entire kinetic energy is converted into its potential energy (E p )

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 9

16. A rocket of mass 3 × 10 6 kg takes off from a launching pad and acquires a vertical velocity of 1 km/s and an altitude of 25 km. Calculate its (a) potential energy (b) kinetic energy.

Answer: Here, mass of the rocket, m = 3 × 10 6 kg velocity acquired by the rocket, v = 1 km/s = 1000 m/s height attained by the rocket, h = 25 km = 25000 m

(a) Potential energy of the rocket, E p = mgh = (3 × 10 6 ) × (10 2 ) × 25000 = 7.5 × 10 11 J

(b) Kinetic energy of the rocket, E k = mv 2 = (3 × 10 6 ) × (1000) 2 = 1.5 × 10 12 J

17. A boy of mass 40 kg runs up a flight of 50 steps, each of 10 cm high, in 5 s. Find the power developed by the boy.

Answer: Here, mass of the boy, m = 40 kg total height gained, h = 50 × 10 cm = 500 cm = 5m time taken to climb, t = 5s Work done by the boy, W = mgh = 40 × 10 × 5 = 2000J

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 10

18. What should be the power of an engine required to lift 90 metric tonnes of coal per hour from a mine whose depth is 200 m

Answer: Here, mass of the coal to be lifted, m = 90 metric tonnes = 90 × 1000 kg = 9 × 10 4 kg

height through which the coal is to be lifted, h = 200m time during which the coal is to be lifted, t = 1h = 60 × 60 = 3600 s

work done to lift the coal, i.e., W = mgh = 9 × 10 4 × 10 × 200m = 18 × 10 7 J

Power of the engine required i.e.,

19. How much time does it take to perform 500J of work at a rate of 10W ?

Answer: Here, work to be performed, W = 500J rate at which work is to be performed, i.e., power, P = 10W

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 12

20. Calculate the units of energy consumed by 100 W electric bulb in 5 hours.

Answer: Here, power of the electric bulb, P = 100 W = 0·1 kW time for which bulb is used, t = 5h As P = W/t, ⇒ W = Pt

Energy consumed by the bulb, W = Pt = 0·1 × 5 = 0·5 kWh = 0·5 units

21. A lift is designed to carry a load of 4000 kg through 10 floors of a building, averaging 6 m per floor, in 10 s. Calculate the power of the lift.

Extra Questions for Class 9 Science Chapter 11 Work Power and Energy 13

Textbook Solutions
Lakhmir Singh and Manjit Kaur Solutions
Lakhmir Singh Class 9
Class 9 Physics
Chapter 4 Work and Energy

Lakhmir Singh Solutions Class 9 Physics Chapter 4 Work and Energy

Lakhmir Singh Solutions for Class 9 Physics Chapter 4 Work and Energy are provided in this article. Work is defined as the force which is required to cause a displacement in the object. It is the product of force acting and the displacement caused in the direction of the force.

The unit of work is Newton-meter or Joule. One joule is defined as the amount of work done by 1 N force when 1 m displacement is caused. Work done can be either positive work or negative work, depending on the direction of displacement.

Work done is said to be positive work when both force and displacement are in the same direction. When the force and displacement are in the opposite direction, the work done is said to be negative. But when force and displacement are at an angle of 90 degrees, the work done is said to be zero.

Energy is defined as the ability of an object to do work. Joule is the unit of energy. There are various forms of energy, and they are as follows:

Kinetic energy, potential energy, chemical energy, electrical energy, heat energy, and light energy.

Also, according to the law of conservation of energy, energy can neither be created nor destroyed but can be transformed from one form to another.

Lakhmir Singh Solutions for Physics Class 9 Chapter 4 Work and Energy

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Access Lakhmir Singh Solutions for Class 9 Physics Chapter 4 Work and Energy

Page No: 145

Very Short Answer Type Questions

Q1. How much work is done when a body of mass m is raised to a height h above the ground?

Height above the ground = h

Work done = potential energy acquired by the body = mgh

Where g is the acceleration due to gravity.

Q2. State the SI unit of work.

SI unit of work is Joule.

Q3. Is work a scalar or a vector quantity?

Work is a scalar quantity as it requires only magnitude.

Q4. Define 1 joule of work.

The work done on a body is said to be 1J when a force of 1N is applied to move a body through a distance of 1m.

Q5. What is the condition for a force to do work on a body?

When a motion is produced in the body, work is said to be done on the body as force is applied to it.

Q6. Is energy a vector quantity?

Energy is a scalar quantity as it has only magnitude.

Q7. What are the units of a) work b) energy

a) Unit of work is Joule

b) Unit of energy is Joule.

Q8. What is the work done against gravity when a body is moved horizontally along a frictionless surface?

The work done against gravity when a body is moved horizontally along a frictionless surface is zero, as the force of gravity acts perpendicular to the direction of motion.

Q9. By how much will the kinetic energy of a body increase if its speed is doubled?

The kinetic energy is directly proportional to the square of the speed of the body. Therefore, when the kinetic energy of the body is increased, the kinetic energy will become four times the actual value.

Q10. Write an expression for the kinetic energy of a body of mass m moving with a velocity v.

Velocity = v

KE = ½ mv 2

Q11. If the speed of a body is halved, what will be the change in its kinetic energy?

The kinetic energy will become one-fourth of the actual value when the speed of the body is halved.

Q12. On what factors does the kinetic energy of a body depend?

The following are the factors on which the kinetic energy of a body depends:

a) mass of the body, m.

b) square of velocity of the body, v 2 .

Q13. Which would have a greater effect on the kinetic energy of an object: doubling the mass or doubling the velocity?

Doubling the velocity would have a greater effect on the kinetic energy of an object.

Q14. How fast should a man of 50 kg run so that his kinetic energy will be 625J?

Mass = 50kg

Kinetic energy = 625 J

Q15. State whether the following objects possess kinetic energy, potential energy, or both:

a) a man climbing a hill

b) a flying aeroplane

c) a bird running on the ground

d) a ceiling fan in the off position

e) a stretched spring lying on the ground

a) Both kinetic and potential energy

b) Both kinetic and potential energy

c) Only kinetic energy

d) Only potential energy

e) Only potential energy

Q16. Two bodies, A and B, of equal masses are kept at heights of h and 2h, respectively. What will be the ratio of their potential energies?

Let the masses of A and B = m

Height of A = h

Height of B = 2h

Potential energy for body A, PE A = mgh

Potential energy for body B, PE B = mg2h

Ratio = PE A : PE B = (mgh)/(mg2h) = ½

Q17. What is the kinetic energy of a body of mass 1 kg moving with a speed of 2 m/s?

Mass = 1 kg

Velocity = 2 m/s

KE = ½ mv 2 = 2J

Q18. Is potential energy a vector or a scalar quantity?

Potential energy is a scalar quantity as it has magnitude only.

Q19. A load of 100 kg is pulled up by 5 m. Calculate the work done.

Mass = 100 kg

Height = 5 m

g = 9.8 m/s 2

PE = mgh = 4900J

Work done = potential energy

Q20. State whether the following statement is true or false:

The potential energy of a body of mass 1 kg kept at the height of 1m is 1J.

= (1) (9.8) (1)

Q21. What happens to the potential energy of a body when its height is doubled?

Potential energy is directly proportional to height. Therefore, when the height is doubled, potential energy also gets doubled.

Q22. What kind of energy is possessed by the following:

a) a stone kept on roof-top

b) a running car

c) water stored in the reservoir of a dam

d) a compressed spring

e) a stretched rubber band.

a) Potential energy

b) Kinetic energy

c) Potential energy

d) Potential energy

e) Potential energy

Q23. Fill in the blanks with suitable words:

a) Work is measured as a product of ………. and ………..

b) The work done on a body moving in a circular path is ……………

c) 1 joule is the work done when a force of one ………….. moves an object through a distance of one ……….. in the direction of ………..

d) The ability of a body to do work is called ……….. The ability of a body to do work because of its motion is called ………..

e) The sum of the potential and kinetic energies of a body is called ………….. energy.

a) Force, distance

c) Newton, meter, force

d) Energy, kinetic energy

e) Mechanical

Short Answer Type Questions

Q24. What are the quantities on which the amount of work done depends? How are they related to work?

Following are the quantities on which the amount of work done depends:

a) magnitude of the force

b) distance through which the body moves

Force and distance are related to work done as work done is directly proportional to the force applied and the distance moved by the body.

W is the work done

F is the force applied

s is the distance travelled

Q25. Is it possible that a force is acting on a body, but still, the work done is zero? Explain, giving one example.

Yes, it is possible that a force is acting on a body, and still, the work done is zero.

When a man is pushing a wall, work done is zero, as there is no displacement.

Q26. A boy throws a rubber ball vertically upwards. What type of work, positive or negative, is done:

a) by the force applied by the boy?

b) by the gravitational force of the earth?

a) The force applied by the boy is a positive work as it is in the direction of motion of the body.

b) The work done by the gravitational force of the earth is negative as it is against the direction of motion of the body.

Q27. Write the formula for work done on a body when the body moves at an angle to the direction of force. Give the meaning of each symbol used.

W = F cos θ s

s is the distance travelled by the body

θ is the angle between the direction of force and the direction of motion of the body

Q28. How does the kinetic energy of a moving body depend on its

a) Kinetic energy is directly proportional to the speed.

b) Kinetic energy is directly proportional to the mass.

Q29. Give one example each in which a force does

a) positive work

b) negative work

c) zero work

a) Positive work: When a ball is thrown upwards, work done by the force is positive work.

b) Negative work: Work done by the gravitational force of the earth on the ball thrown, which is thrown upwards.

c) Zero work: Sliding of the box on the ground is an example of zero work as the gravitational force acting on the box.

Q30. A ball of mass 200g falls from a height of 5 meters. What is its kinetic energy when it just reaches the ground?

Mass = 200 g = 0.2kg

Height = 5m

Initial velocity, u = 0

Acceleration due to gravity, g = 9.8 m/s 2

Final velocity, v

Using third equation of motion,

v 2 – u 2 = 2gs

Kinetic energy = ½ mv 2

Kinetic energy = (½)(0.2)(98) = 9.8J

Q31. Find the momentum of a body of mass 100g having a kinetic energy of 20J.

Mass = 100g = 0.1 kg

Momentum = mv = 2 kg.m/s

Q32. Two objects having equal masses are moving with uniform velocities of 2 m/s and 6 m/s, respectively. Calculate the ratio of their kinetic energies.

Let masses of the two objects be m

KE1 = 1/2 m.(2) 2

KE2 = 1/2 m.(6) 2

Ratio = (KE1)/(KE2) = 1/9

Q33. A body of 2 kg falls from rest. What will be its kinetic energy during the fall at the end of 2s?

Mass of body = 2 kg

Time taken, t = 2 s

Acceleration due to gravity, g = 10 m/s2

Final velocity, v = ?

Using the first equation of motion,

v = u + gt = 20 m/s

KE = 1/2 mv 2 = 400 J

Q34. On a level road, a scooterist applies brakes to slow down from a speed of 10 m/s to 5 m/s. If the mass of the scooterist and the scooter be 150kg, calculate the work done by the brakes.

Mass of scooter + scooterist = 150 kg

Initial velocity, u = 10 m/s

Final velocity, v = 5 m/s

Retardation = a

Distance travelled = s

v 2 – u 2 = 2as

Work done, W = Fs

W = -5625 J

The negative sign is used to show that the force applied by the brakes is acting opposite direction of motion.

Q35. A man drops a 10 kg rock from the top of a 5m ladder. What is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground?

Mass of rock = 10 kg

Height of ladder, h = 5m

The initial velocity of rock, u = 0

g = 10 m/s 2

Using the third equation of motion

v 2 – u 2 = 2gh

KE = 1/2 mv 2 = 500J

Q36. Calculate the work done by the brakes of a car of mass 1000kg when its speed is reduced from 20 m/s to 10 m/s.

Mass of car = 1000 kg

Initial velocity, u = 20 m/s

Final velocity, v = 10 m/s

Distance covered = s

W = (1000)(150) = -150 kJ

The negative sign shows that the brakes are acting in the opposite direction of motion.

Q37. A body of mass 100kg is lifted up by 10m. Find:

a) the amount of work done

b) potential energy of the body at that height

Height, h = 10m

Acceleration due to gravity, g = 10m/s 2

a) Work done, W = mgh = 10kJ

b) Potential energy of the body = work done = 10kJ

Q38. A boy weighing 50kg climbs up a vertical height of 100m. Calculate the amount of work done by him. How much potential energy does he gain?

Height, h = 100m

Work done by the boy, W = mgh = 49kJ

Potential energy gained = work done by the boy = 49kJ

Q39. When is the work done by a force on a body:

a) positive

b) negative

Work done by a force applied on a body is:

a) When the direction of motion of the body and the force acting in the same direction, work done is positive.

b) When the direction of motion of the body and the force acting on the body are in the opposite direction, work done is negative.

c) When the force is acting at a right angle to the direction of motion of the body, the work done is zero.

Q40. To what height should a box of mass 150kg be lifted, so that its potential energy may become 7350 joules?

Mas of the box, m = 150kg

PE = 7350 J

Q41. A body of mass 2kg is thrown vertically upwards with an initial velocity of 20 m/s. What will be its potential energy at the end of 2s?

Mass of the body, m = 2kg

Height reached = h

Time, t = 2 s

Using the second equation of motion,

h = ut = 1/2 gt 2

PE after 2 s = mgh = 400J

Q42. How much work is done when a force of 1 N moves a body through a distance of 1m in its own direction?

Force, F = 1N

Distance, s = 1m

Work done, W = Fs = 1J

Q43. A car is being driven by a force of 2.5 1010 N. Travelling at a constant speed of 5 m/s, it takes 2 minutes to reach a certain place. Calculate the work done.

Force, F = 2.5 × 10 10 N

Velocity, v = 5 m/s

Time, t = 2 minutes = 120 s

Distance, s = vt = 600m

Work done, W = Fs = 15 × 10 12 J

Q44. Explain by an example that a body may possess energy even when it is not in motion.

Storing of water in the reservoir is an example of a body which possesses energy even when it is not in motion. The energy stored by this water is known as potential energy. This is dependent on the position of the body.

Q45. a) On what factors does the gravitational potential energy of a body depend?

b) Give one example of each of a body possesses both, i) kinetic energy ii) potential energy

a) Following are the factors on which the gravitational potential energy of a body depends on:

i) mass of the body

ii) height at which the body is lifted

iii) acceleration due to gravity

b) i) Energy stored in a car while its moving is an example of kinetic energy

ii) Energy stored in a stretched rubber is an example of potential energy

Q46. Give two examples where a body possesses both kinetic energy as well as potential energy.

Following are the two examples where a body possesses both kinetic energy as well as potential energy:

a) a flying aeroplane

b) A man climbing stairs

Q47. How much is the mass of a man if he has to do 2500 joules of work in climbing a tree 5m tall?

Mass of man = m

Height of tree, h = 5m

Work done, W = 2500J

Acceleration due to gravity, g = 10 m/s 2

Q48. If the work done by a force in moving an object through a distance of 20 cm is 24.2J, what is the magnitude of the force?

Work done, W = 24.2 J

Distance, s = 20cm = 0.2m

F = W/s = 121N

Q49. A boy weighing 40kg makes a high jump of 1.5m

a) What is his kinetic energy at the highest point?

b) What is his potential energy at the highest point?

Mass of boy, m = 40 kg

Height, h = 1.5m

a) At highest point velocity, v = 0

Therefore, KE = 0

b) PE = mgh = 600J

Q50. What type of energy is possessed:

a) by the stretched rubber strings of a catapult?

b) by the piece of stone which is thrown away on releasing the stretched rubber strings of the catapult?

b) Both potential and kinetic energy

Q51. A weightlifter is lifting weights of mass 200kg up to a height of 2 meters. If g = 9.8 m/s2, calculate:

a) potential energy acquired by the weight

b) work done by the weightlifter

Mass, m = 200 kg

Height, h = 2m

Acceleration due to gravity, g = 9.8 m/s2

a) Potential energy = mgh = 3920 J

b) Work done against gravity = potential energy gained by the weight

Therefore, work done, W = mgh = 3920 J

Long Answer Type Questions

Q52. a) Define the term ‘work’. Write the formula for the work done on a body when a force acts on the body in the direction of its displacement. Give the meaning of each symbol which occurs in the formula.

b) A person of mass 50kg climbs a tower of height 72 meters. Calculate the work done.

a) Work is done when an applied force produces motion in a body.

The formula for work done is:

b) Mass of the person, m = 50kg

Height of tower, h = 72m

Work done, W = mgh = 35280 J

Q53. a) When do we say that work is done? Write the formula for the work done by a body in moving up against gravity. Give the meaning of each symbol which occurs in it.

b) How much work is done when a force of 2N moves a body through a distance of 10cm in the direction of force?

a) Work is said to be done when applied force results in the motion of the body or a change in the shape and size of the body.

Work done is given as:

m is the mass of the body

g is the acceleration due to gravity

h is the height

b) Force, F = 2N

Distance, s = 10cm = 0.1m

Work done, W = Fs = 0.2 J

Q54. a) What happens to the work done when the displacement of a body is at right angles to the direction of force acting on it? Explain your answer.

b) A force of 50N acts on a body and moves it a distance of 4m on a horizontal surface. Calculate the work done if the direction of force is at an angle of 60 degree to the horizontal surface.

a) When the displacement of a body is at right angle to the direction of force acting on it, then the work done will be zero.

b) Force, F = 50N

Distance, s = 4m

The angle between the direction of force and the direction of motion, θ = 60 degree

Work done, W = F cos θ s = 100 J

Q55. a) Define the term ‘energy’ of a body. What is the SI unit of energy?

b) What are the various forms of energy?

c) Two bodies having equal masses are moving with uniform speeds of v and 2v, respectively. Find the ratio of their kinetic energies.

a) Energy is defined as the ability to do work, and joule is the SI unit of energy.

b) Following are the various forms of energy:

Kinetic energy, potential energy, nuclear energy, electrical energy, chemical energy, sound energy, light energy, and heat energy.

c) Let the mass of both bodies be m

Velocity of body 1, v1 = v

Velocity of body 2, v2 = 2v

KE1 = 1/2 mv 2 = 1/2 m(v1) 2

KE2 = 1/2 mv 2 = 1/2 m(v2) 2

Ratio = KE1 : KE2 = 1/4

Q56. a) What do you understand by the kinetic energy of a body?

b) A body is thrown vertically upwards. Its velocity goes on decreasing. What happens to its kinetic energy as its velocity becomes zero?

c) A horse and a dog are running at the same speed. If the weight of the horse is ten times that of the dog, what is the ratio of their kinetic energies?

a) Kinetic energy of a body is the energy due to motion.

b) The kinetic energy of the body will be equal to zero as kinetic energy is directly proportional to the square of the velocity. Therefore, when a body is thrown vertically upwards, the kinetic energy becomes zero.

c) Speed of horse = speed of dog = v

Weight of dog = m

Weight of horse = 10m

KE1 is the for dog = 1/2 mv 2

KE2 is for horse = 1/2 mv 2 = 1/2 (10m)v 2

Ratio = KE1 : KE2 = 10/1

Q57. a) Explain by an example what is meant by potential energy. Write down the expression for the gravitational potential energy of a body of mass m placed at a height h above the surface of the earth.

b) What is the difference between potential energy and kinetic energy?

c) A ball of mass 0.5 kg slows down from a speed of 5 m/s to that of 3 m/s. Calculate the change in kinetic energy of the ball. State your answer giving proper units.

a) The energy stored in a body due to either its position or change in shape is known as potential energy. Water stored in a reservoir is an example of potential energy.

PE is the potential energy

h is the height above the surface of the earth

b) Following are the differences between kinetic energy and potential energy:

i) Kinetic energy is defined as the energy in the body due to its motion, while potential energy is defined as the energy due to its position.

ii) Kinetic energy of a body when it is at rest is zero, whereas the potential energy of a body when it is at rest is not zero.

c) Mass of ball, m = 0.5 kg

Speed, v1 = 5 m/s

Speed, v2 = 3 m/s

KE1 = 1/2 m(v1) 2

KE2 = 1/2 m(v2) 2

Change in KE = KE1 – KE2

Q58. a) What is the difference between gravitational potential energy and elastic potential energy? Give one example of a body having gravitational potential energy and another having elastic potential energy.

b) If 784 J of work was done in lifting a 20 kg mass, calculate the height through which it was lifted.

a) Gravitational potential energy is the potential energy due to the position of the body above the ground. While elastic potential energy is the potential energy due to changes in the shape and size of the body. Storing of water in the overhead tank is an example of gravitational potential energy, while stretching of a rubber band is an example of elastic potential energy.

b) Work done, W = 784J

Mass, m = 20 kg

Multiple Choice Questions

Q59. A car is accelerated on a levelled road and acquires a velocity 4 times of its initial velocity. During this process, the potential energy of the car:

a) does not change

b) becomes twice that of the initial potential energy

c) becomes 4 times that of the initial potential energy

d) becomes 16 times that of the initial potential energy

The correct answer is a) does not change

Q60. A car is accelerated on a levelled road and attains a speed of 4 times its initial speed. In this process, the kinetic energy of the car:

b) becomes 4 times that of the initial kinetic energy

c) becomes 8 times that of the initial kinetic energy

d) becomes 16 times that of the initial kinetic energy

The correct answer is d) becomes 16 times that of initial kinetic energy

Q61. In case of negative work, the angle between the force and displacement is:

b) 45 °

c) 90 °

d) 180 °

The correct answer is d) 180°

Q62. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 3.5kg, Both spheres are dropped simultaneously from a tower. When they are 10m above the ground, they have the same:

a) acceleration

b) momentum

c) potential energy

d) kinetic energy

The correct answer is a) acceleration

Q63. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. If the value of g is 10 m/s2, the work done by the girl against the gravitational force will be:

The correct answer is c) 0 J

Q64. The work done on an object does not depend on the:

a) displacement

b) angle between force and displacement

c) force applied

d) initial velocity of the object

The correct answer is d) initial velocity of the object

Q65. Water stored in a dam possesses:

a) no energy

b) electrical energy

c) kinetic energy

d) potential energy

The correct answer is d) potential energy

Q66. The momentum of a bullet of mass 20g fired from a gun is 10 kg.m/s. The kinetic energy of this bullet in kJ will be:

The correct answer is c) 2.5

Q67. Each of the following statements describes a force acting. Which force is causing work to be done?

a) the weight of a book at rest on a table

b) the pull of a moving railway engine on its coaches

c) the tension in an elastic band wrapped around a parcel

d) the push of a person’s feet when standing on the floor

The correct answer is b) the pull of a moving railway engine on its coaches

Q68. A girl weighing 400 N climbs a vertical ladder. If the value of g is 10 m/s2, the work done by her after climbing 2 m will be:

The correct answer is b) 800 J

Q69. Which of the following does not possess the ability to do work not because of motion?

a) a sparrow flying in the sky

b) a sparrow moving slowly on the ground

c) a sparrow in the nest on a tree

d) a squirrel going up a tree

The correct answer is c) a sparrow in the nest on a tree

Q70. A stone is thrown upwards, as shown in the diagram. When it reaches P, which of the following has the greatest value for the stone?

Lakhmir Singh Solutions Class 9 Chapter 4-1

a) its acceleration

b) its kinetic energy

c) its potential energy

d) its weight

The correct answer is c) its potential energy

Also, visit Lakhmir Singh Solutions for Class 9 Physics to get complete solutions for all chapters.

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Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion

Last modified on: 2 years ago
Reading Time: 6 Minutes

Case Study/Passage Based Questions:

Question 1:

Read the following and answer any four questions from (i) to (v) given below :

In the figure below the card is flicked with a push. It was observed that the card moves ahead while coin falls in glass.

(i) Give reason for the above observation. (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (b) The coin possesses inertia of motion; it resists the change and hence falls in the glass. (c) The coin possesses inertia of rest, it accepts the change and hence falls in the glass. (d) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.

(ii) Name the law involved in this case. (a) Newton’s second law of motion. (b) Newton’s first law of motion. (c) Newton’s third law of motion. (d) Law of conservation of energy

(iii) If the above coin is replaced by a heavy five-rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler.

(iv) Name the law which provides the definition of force. (a) Law of conservation of mass (b) Newton’s third law. (c) Newton’s first law (d) Newton’s second law.

(v) State Newton’s first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

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Class Notes

Free Class Notes & Study Material

Chapter 11 Work and Energy

Last Updated on July 3, 2023 By Mrs Shilpi Nagpal

Class 9 Science Chapter 11 Work and Energy NCERT Solutions

Question Answers, Page 148

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 the displacement. What is the work done in this case?

Answer: Work done = Force × Displacement

W = 7 × 8 = 56 Nm = 56 J

1. When do we say that work is done?

Answer: Work is said to be done whenever a force acts on a body and there is a displacement of the body caused by the applied force along the direction of the applied force.

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

Answer: W= F × S

where F is the force which displaces the body through a distance of S in the direction of force applied.

3. Define 1 J of work.

Answer: When a force of 1 newton moves a body through a distance of 1 m in its own direction, then the work done is known as 1 Joules or 1 J.

4. 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?

Answer: Force = 140 N

Distance = 15 m

Work done= Force × Distance

Work done = 140 × 15 = 2100 J

1. What is the kinetic energy of an object?

Answer: The energy of a body due to its motion is called as its Kinetic energy.

For example:

1)A moving cricket ball can do work in pushing back the stamps. 2)Moving water can do work in turning the turbine for generating electricity. 3)Moving wind can do work in turning the blades of wind mill. 4)A moving hammer drives a nail into wood because of its kinetic energy. 5)A moving bullet can penetrate even a steel plate. 6)A moving bus,car,falling stone possesses kinetic energy. 7)A falling coconut, running athlete possesses kinetic energy.

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

Answer: K.E. = ½ mv 2

Where K.E. is the kinetic energy

m is the mass of the body

v is the velocity with which body is moving

3. The kinetic energy of an object of mass, m moving with a velocity of 5 ms –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?

Answer: Kinetic energy of the object = 25 J

Velocity of the object = 5 m/s

K.E. = ½mv 2

25= ½ × m ×(5) 2

If velocity is doubled , v = 5 × 2 = 10 m/s

K.E. = ½ × 2 × (10) 2

K.E. = 100 J

If velocity is tripled , v = 5 × 3= 15 m/s

K.E. = ½ × 2 × (15) 2

K.E. = 225 J

1. What is power?

Answer: Power is defined as rate of doing work .

Power = Work / Time

The S.I. unit of power is Watt.

2. Define 1 watt of power.

Answer: 1 watt is the power of an appliance which does work at rate of 1 joules per second.

1 watt = 1 joule/1 second

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

Answer: Power = Work done / Time

Work done or energy consumed = 1000 J

Time = 10 s

Power = 1000/10 = 100 J/s or 100 W

4. Define average power.

Answer: Average power of an appliance is defined as the total work done by it in the total time taken.

Average power = Total work done/ total time taken

Exercises Page 158

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: 1) Suma applies a force to push the water backwards.Therefore suma swims in the forward direction. Hence the force causes displacement. Hence , work is done by suma while swimming in a pond.

2)While carrying a load, the donkey has to apply a force in upward direction.But displacement of the load is in forward direction.Since , displacement is perpendicular to force, the work done is zero.

3) Wind mill is lifting water from a well and doing work against gravity.

4) In this case, there is no displacement and force, so no work is done.

5) An engine applies the force to pull the train.This allows the train to move in the direction of force.Therefore, there is displacement in the train in same direction.

6) During the drying of food grains, there is no force and displacement hence no work is done.

7) Wind energy applies a force on the sailboat to push it in forward direction.Therefore, there is a displacement in the boat in the direction of force.Hence work is done by wind.

Answer: Work done by the force of gravity on the object depends only on vertical displacement.

When the object move upwards, the work done by gravity is negative and when the object move downwards, the work done by gravity is positive.

Therefore the work done by the gravity on the object is zero joules.

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

Answer: When bulb is connected to a battery, then chemical energy of the battery is transferred into electrical energy.When the bulb receives this electrical signal , it converts it into heat and light energy.

4. Certain force acting on a 20 kg mass changes its velocity from 5 ms –1 to 2 ms –1 . Calculate the work done by the force.

Answer: Mass of the body = 20 kg

Initial velocity = 5 m/s

Final velocity = 2 m/s

Work done = change in kinetic energy

Work done = ½mu 2 -½mv 2

Work done = ½ m(u 2 -v 2 )

Work done = ½ × 20 (5 2 -2 2 )

Work done = 210 J

Answer: Work done on the object by the gravitational force depends only on the vertical displacement of the body and does not depend on the path followed. Therefore the work done on the object by the gravitational force is zero because direction of force is vertically downwards and displacement is horizontal.

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

Answer: Law of conservation of energy states that whenever energy gets transformed, the total energy remains unchanged.When the body falls from a height , then its potential energy changes into kinetic energy. A decrease in potential energy is equal to increase in kinetic energy of the body.Thus during the process total mechanical energy of the body remains conserved. Therefore law of conservation of energy is not violated.

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

Answer: The muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle.Heat energy heat’s the riders body whereas Kinetic energy provides a velocity to the bicycle.

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: When we push a huge rock there is no transfer of muscular energy to the stationary rock.

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

Answer: 1 unit = 3600000 J

250 units = 250 × 3600000 =9 × 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: m= 40 kg

g= 9.8 m/s 2

P.E. = 40 ×9.8 ×5

P.E. = 1960 J

When the object is half-way down, its potential energy becomes half the original energy and remaining half converted into kinetic energy.

At way down , the potential energy of the object = 1960 /2 = 980 J

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

Answer: When a satellite moves around the earth , then the direction of force of gravity on the satellite is perpendicular to its displacement.Hence the work done by the force of gravity on a satellite moving round the earth is zero.

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, there may be displacement in the absence of force.

In the absence of force , F= 0

then ma=0 , a=o but m≠0

If a= o means object is at rest or in a state of uniform motion in straight line.Thus there can be displacement of an object in the absence of any force acting on it.

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: Work is said to be done whenever a force acts on a body and there is displacement by the application of force in or opposite to the direction of force. When a person holds a bundle of hay over his head, then there will be no displacement in the bundle of hay.Hence in the absence of force, work done by the person is zero.

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

Answer: Power = work done/ Time taken

Work done or energy

Energy = Power × Time taken

Energy = 1500 × 10

Energy = 15000 KWh

Answer: When the pendulum moves from it mean position A to either of its extreme position B or C, then at this position its kinetic energy is zero and its potential energy is maximum.In this way total mechanical energy remains conserved.As it moves towards point A, its potential energy decreases and kinetic energy increases.As it reaches point A, its kinetic energy is maximum and potential energy is minimum.Again total mechanical energy remains conserved.

The bob comes to rest because of air resistance which resist its motion.The pendulum loses its kinetic energy to overcome this friction and stops after some time.The law of conservation is not violated.

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: The object in motion = Kinetic energy = ½mv 2

The kinetic energy of the object when it comes to rest= 0

Work done on object = Change in kinetic energy

= ½mv 2 – 0

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

Answer: Kinetic energy = ½mv 2

mass of the car = 1500 kg

velocity of the car= 60 km/hr = 50/3 m/s

K.E. = ½ × 1500 × (50/3) 2

K.E. = 20.8 × 10 4 J

Work done on object= change in k.E.

W.D. = 20.8 × 10 4 J

Answer: a) In case 1 , the force and displacement are perpendicular to each other , so work done is zero.

b) In case 2, the force and displacement are in same direction, so the work done is positive.

c) In case 3 , the force and displacement are in opposite direction, so the work done is 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: When all the force acting on an object cancel out each other i.e. the net force acting on the object is zero then the acceleration in an object could be zero even when several forces are acting on it.

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

Answer: Power = work done/Time taken

Power of four devices = 4 × 500 = 2000 W

Time = 10 hr

Energy consumed = power × time

= 2000 × 10

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

Answer: When an object fall freely towards the ground, its potential energy decreases and kinetic energy increases.As the object reaches the ground , all its potential energy gets converted into kinetic energy.As the object touches the ground , all its kinetic energy gets converted into heat energy and sound energy.

About Mrs Shilpi Nagpal

Author of this website, Mrs. Shilpi Nagpal is MSc (Hons, Chemistry) and BSc (Hons, Chemistry) from Delhi University, B.Ed. (I. P. University) and has many years of experience in teaching. She has started this educational website with the mindset of spreading free education to everyone. In addition to this website, author also has a Youtube channel, here is the link Class Notes Youtube Channel

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11.1.1 Not Much ‘Work’ in Spite of Working Hard

11.1.2 Scientific Conception of Work

11.1.3 Work Done By a Constant Force

11.2 Energy

11.2.1 Forms of Energy

11.2.2 Kinetic Energy

11.2.3 Potential Energy

11.2.4 Potential Energy of An Object At A Height

11.2.5 Are Various Energy Forms Inconvertible?

11.3 Rate of Doing Work

11.3.1 Commercial Unit of Energy

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