{"id":5046,"date":"2023-03-25T05:26:53","date_gmt":"2023-03-25T05:26:53","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=5046"},"modified":"2023-03-28T06:16:05","modified_gmt":"2023-03-28T06:16:05","slug":"ncert-solutions-class-9-science-chapter-11-work-and-energy","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-science-chapter-11-work-and-energy\/","title":{"rendered":"NCERT Solutions Class 9 Science Chapter 11 Work and Energy"},"content":{"rendered":"

NCERT Solutions Class 9 Science\u00a0<\/span><\/h2>\n

The NCERT Solutions in English Language for Class 9 Science Chapter – 11 (Work and Energy) <\/strong>has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter – 11 (Work and Energy)\u00a0<\/span><\/h2>\n

Questions<\/strong><\/span><\/p>\n

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?
\n<\/strong><\/span>Answer – <\/strong><\/span>We know that if force\u00a0F<\/i>\u00a0acting on an object to displace it through a distance\u00a0S<\/i>\u00a0in one direction, then the work done\u00a0W<\/i>\u00a0on the body by the force is given by:
\nWork done = Force \u00d7 Displacement
\nW<\/i>\u00a0=\u00a0F<\/i>\u00a0\u00d7\u00a0S
\n<\/i>Where,
\nF<\/i>\u00a0= 7 N
\nS<\/i>\u00a0= 8 m
\nTherefore, work done,\u00a0W<\/i>\u00a0= 7 \u00d7 8
\n= 56 Nm
\n= 56 J<\/span><\/p>\n

Questions <\/strong><\/span><\/p>\n

1. When do we say that work is done?
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>Work is completed whenever the given conditions are satisfied:
\n<\/span>(i)<\/strong> A force acts on the body.
\n<\/span>(ii)<\/strong> There\u2019s a displacement of the body by applying force in or opposite to the direction of the force.<\/span><\/p>\n

2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>When a force F displaces a body through a distance S within the direction of the applied force, then the work done W on the body is given by the expression:
\n<\/span>W = F \u00d7 S<\/span><\/p>\n

3. Define 1 J of work.
\n<\/strong><\/span>Answer –\u00a0<\/strong>1 J is the amount of work done by a force of 1 N on an object that displaces it through a distance of 1 m in the direction of the applied force.<\/span><\/p>\n

4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. <\/strong><\/span>How much work is done in ploughing the length of the field?
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>Work done by the bullocks is given by the expression:
\n<\/span>W = F \u00d7 d
\n<\/span>Where,
\n<\/span>Applied force, F = 140 N
\n<\/span>Displacement, d = 15 m
\n<\/span>W = 140 \u00d7 15 = 2100 J
\n<\/span>Therefore, 2100 J of work is done in ploughing the length of the field.<\/span><\/p>\n

Questions <\/strong><\/span><\/p>\n

1. What is the kinetic energy of an object?
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>The kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
\nThe heavier a thing is, and the faster it moves, the more kinetic energy it has.<\/span><\/p>\n

2. Write an expression for the kinetic energy of an object.
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>If a body of mass m is moving with a speed v, then its K. E. Ek<\/sub>\u00a0is given by the expression,
\n<\/span>Ek<\/sub> = \"\\frac{1}{2}\" m v2
\n<\/sup><\/span>Its SI unit is Joule (J).<\/span><\/p>\n

3. The kinetic energy of an object of mass, m moving with a velocity of 5 ms-1<\/sup> 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?
\n<\/strong><\/span>Answer –
\n<\/strong><\/span>K.E. of the object = 25J
\n<\/span>Velocity of the object (v) = 5 m\/s
\n<\/span>K.E. = \"\\frac{1}{2}\" mv2
\n<\/sup><\/span>25 = \"\\frac{1}{2}\" m (5)2
\n<\/sup><\/span>50 = 25 \u00d7 m
\n<\/span>m = 50\/25
\n<\/span>m = 2 kg
\n<\/span>Now, when velocity is doubled
\n<\/span>v = 10 m\/s
\n<\/span>m = 2 kg
\n<\/span>K.E. = \"\\frac{1}{2}\" \u00d7 2 \u00d7 (10)2
\n<\/sup><\/span>K.E. = 102
\n<\/sup><\/span>K.E. = 100 J
\n<\/span>When velocity is increased three times, then
\n<\/span>v = 15 m\/s
\n<\/span>m = 2 kg
\n<\/span>K.E. = \"\\frac{1}{2}\" \u00d7 2 \u00d7(15)2
\n<\/sup><\/span>K.E. = (15)2
\n<\/sup><\/span>K.E. = 225 J<\/span><\/p>\n

Questions<\/strong><\/span><\/p>\n

1. What is power?
\n<\/strong><\/span>Answer –\u00a0<\/strong>Power is the rate of doing work or the rate of transfer of energy. If\u00a0W\u00a0<\/i>is the amount of work done in time\u00a0t<\/i>, then power is given by the expression,
\nPower= Work \/ Time = Energy \/ Time
\nP\u00a0<\/i>=\u00a0W<\/i>\/T
\n<\/i>It is expressed in watt (W<\/i>).<\/span><\/p>\n

2. Define 1 watt of power.
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>A body is claimed to possess power of one watt if it works at the speed of 1 joule in 1 s.
\n<\/span>That is, <\/span>One W = 1 J\/1 S<\/span><\/p>\n

3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>Power = Work\/Time
\n<\/span>P = W\/T
\n<\/span>Time = 10 s
\n<\/span>Work done = Energy consumed by the lamp = 1000 J
\n<\/span>Power = 1000\/10 = 100 Js-1<\/sup> =100 W
\n<\/span>Hence, the power of the lamp is 100 W<\/span><\/p>\n

4. Define average power.
\n<\/strong><\/span>Answer –\u00a0<\/strong>The average Power of an agent may be defined as the total work done by it in the total time taken.
\nAverage Power = Total Work Done \/ Total time taken<\/span><\/p>\n

Exercises<\/strong><\/span><\/span><\/p>\n

1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term \u2018work\u2019.
\n<\/strong><\/span>(a)<\/strong> Suma is swimming in a pond.
\n<\/span>(b)<\/strong> A donkey is carrying a load on its back.
\n<\/span>(c)<\/strong> A wind-mill is lifting water from a well.
\n<\/span>(d)<\/strong> A green plant is carrying out photosynthesis.
\n<\/span>(e)<\/strong> An engine is pulling a train.
\n<\/span>(f)<\/strong> Food grains are getting dried in the sun.
\n<\/span>(g)<\/strong> A sailboat is moving due to wind energy.
\n<\/span>Answer –\u00a0 <\/strong><\/span>Work is finished whenever the given 2 conditions are satisfied:
\n<\/span>(i)<\/strong> A force acts on the body.
\n<\/span>(ii)<\/strong> There\u2019s a displacement of the body by applying force in or opposite to the direction of the force.<\/span><\/p>\n

(a)<\/strong> Work is done because the displacement of swimmer takes place in the direction of applied force.<\/span><\/p>\n

(b)<\/strong> A donkey is not doing work against gravity. No work is done as the displacement of load does not take place in the direction of applied force.<\/span><\/p>\n

(c)<\/strong> Work is done, as the displacement takes place in the direction of force.<\/span><\/p>\n

(d)<\/strong> No work is done, because no displacement takes place.<\/span><\/p>\n

(e)<\/strong> When an engine pulls a train the angle between distance and force becomes 0\u00b0 and work will be done.<\/span><\/p>\n

(f)<\/strong> No work is done in this process as the grains are not moving or covering some distance when being dried in the sun, hence no work is done.<\/span><\/p>\n

(g)<\/strong> When wind energy is applied to the boat it starts moving in the direction of the force applied by the wind. So the angle between distance and force becomes 0\u00b0 and some work will be done.<\/span><\/p>\n

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?
\n<\/strong><\/span>Answer –\u00a0<\/strong>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\/heights of the object, which is zero.
\nWork done by gravity is given by the expression,
\nW<\/i>=\u00a0m\u00d7g\u00d7h<\/i>
\nWhere,
\nh <\/i>= Vertical
\ndisplacement = 0
\nW<\/i>\u00a0=\u00a0mg\u00a0<\/i>\u00d7 0 = 0 J
\nTherefore, the work done by gravity on the given object is zero joule.<\/span><\/p>\n

3. A battery lights a bulb. Describe the energy changes involved in the process.
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>When a bulb is connected to a battery, then the energy of the battery is transferred into voltage. Once the bulb receives this voltage, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:
\n<\/span>Chemical Energy \u2192 Electrical Energy \u2192 Light Energy + Heat Energy.<\/span><\/p>\n

4. Certain force acting on a 20 kg mass changes its velocity from 5 m s-1<\/sup>\u00a0to 2 m s-1<\/sup>. Calculate the work done by the force.
\n<\/strong><\/span>Answer –
\n<\/strong><\/span>Initial velocity u = 5 m\/s
\n<\/span>Mass of the body = 20kg
\n<\/span>Final velocity v = 2 m\/s
\n<\/span>The initial kinetic energy
\n<\/span>Ei<\/sub> = \"\\frac{1}{2}\" mu2<\/sup> = \"\\frac{1}{2}\" \u00d7 20\u00a0\u00d7 (5)2
\n<\/sup><\/span>= 10\u00a0\u00d7 25
\n<\/span>= 250 J<\/span><\/p>\n

Final kinetic energy
\n<\/span>Ef<\/sub> = \"\\frac{1}{2}\" mv2<\/sup> = \"\\frac{1}{2}\" \u00d7 20\u00a0\u00d7 (2)2
\n<\/sup><\/span>= 10\u00a0\u00d7 4
\n<\/span>\u00a0= 40 J
\n<\/span>Therefore,
\n<\/span>Work done = Change in kinetic energy
\n<\/span>Work done = Ef<\/sub>\u00a0\u2013 Ei
\n<\/sub><\/span>Work done = 40 J \u2013 250 J
\n<\/span>Work done = -210 J<\/span><\/p>\n

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.
\n<\/strong><\/span>Answer –\u00a0<\/strong>Work done by gravity depends only on the vertical displacement of the body. It does not depend upon the path of the body. Therefore, work done by gravity is given by the expression,
\nW<\/i>=\u00a0m\u00d7g\u00d7h<\/i>
\nWhere,
\nVertical displacement,\u00a0h\u00a0<\/i>= 0
\n\u2234W=\u00a0mg<\/i> \u00d7 0 = 0
\nHence, the work done by gravity on the body is zero.<\/span><\/p>\n

6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
\n<\/strong><\/span>Answer –\u00a0<\/strong>It does not violate the law of conservation of energy. Whatever, is the decrease in Potential energy due to loss of height, same is the increase in the Kinetic energy due to increase in velocity of the body.<\/span><\/p>\n

7. What are the various energy transformations that occur when you are riding a bicycle?
\n<\/strong><\/span>Answer – <\/strong>The chemical energy of the food changes into heat and then to muscular energy. On paddling, the muscular energy changes into mechanical energy.
\n<\/span>Muscular energy \u2192 mechanical energy + heat<\/span><\/p>\n

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?
\n<\/strong><\/span>Answer –\u00a0<\/strong>Energy transfer does not take place as no displacement takes place in the direction of applied force. The energy spent is used to overcome inertia of rest of the rock.<\/span><\/p>\n

9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
\n<\/strong><\/span>Answer –
\n<\/strong><\/span>1 unit of energy = 1kWh
\n<\/span>Energy (E) = 250 units
\n<\/span>1 unit = 1 kWh
\n<\/span>1 kWh = 3.6 x 106<\/sup> J
\n<\/span>Therefore, 250 units of energy = 250 \u00d7 3.6 \u00d7 106
\n<\/sup><\/span>= 9 \u00d7 108<\/sup>\u00a0J.<\/span><\/p>\n

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.
\n<\/strong><\/span>Answer –
\n<\/strong><\/span>Mass (m) = 40 kg
\n<\/span>Acceleration due to gravity (g)= 10m\/s\u00b2
\n<\/span>Height (h)= 5m
\n<\/span>Potential energy= m \u00d7 g\u00d7 h
\n<\/span>P.E= 40 \u00d7 10 \u00d7 5 = 2000J
\n<\/span>Potential energy = 2000J ( 2000 joules)
\n<\/span>At a height of 5 metres, the object has a potential energy of 2000 J.
\n<\/span>When this object is allowed to fall and it is halfway down, its height above the ground will be half of 5 m= 5\/2= 2.5m.
\n<\/span>P.E at Halfway down= m \u00d7 g \u00d7 h
\n<\/span>P.E= 40\u00d7 10 \u00d7 2.5= 1000J
\n<\/span>[h= 2.5 m]
\n<\/span>Potential Energy halfway down= 1000 joules.
\n<\/span>According to the law of conservation of energy:
\n<\/span>Total potential energy= potential energy halfway down+ kinetic energy halfway down
\n<\/span>2000 = 1000 + K.E halfway down
\n<\/span>K.E at halfway down= 2000- 1000= 1000 J
\n<\/span>Kinetic energy at halfway down= 1000 joules.<\/span><\/p>\n

11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
\n<\/strong><\/span>Answer – <\/strong><\/span>Work is completed whenever the given two conditions are satisfied:
\n<\/span>(i)<\/strong> A force acts on the body.
\n<\/span>(ii)<\/strong> There\u2019s a displacement of the body by applying force in or opposite to the direction of the force.
\n<\/span>If the direction of force is perpendicular to displacement, then the work done is zero. When a satellite moves around the Earth, then the force of gravity on the satellite is perpendicular to its displacement. Therefore, the work done on the satellite by the Earth is zero.<\/span><\/p>\n

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
\n<\/strong><\/span>Answer –\u00a0<\/strong>Yes, For a uniformly moving object.
\nSuppose an object is moving with constant velocity. The net force acting on it is zero. But, there is a displacement along the motion of the object. Hence, there can be a displacement without a force.<\/span><\/p>\n

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.
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>Work is completed whenever the given 2 conditions are satisfied.
\n<\/span>(i)<\/strong> A force acts on the body.
\n<\/span>(ii)<\/strong> There\u2019s a displacement of the body by applying force in or opposite to the direction of the force.
\n<\/span>When an individual holds a bundle of hay over his head, there is no displacement in the hay bundle. Although the force of gravity is acting on the bundle, the person isn\u2019t applying any force on it. Therefore, in the absence of force, work done by the person on the bundle is zero.\u00a0<\/span><\/p>\n

14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
\n<\/strong><\/span>Answer –\u00a0 <\/strong><\/span>Power of the heater = 1500 W = 1.5 kW
\n<\/span>Time taken = 10 hours
\n<\/span>Energy consumed by an electric heater can be obtained with the help of the expression,
\n<\/span>Power = Energy consumed \/ Time taken
\n<\/span>Hence,
\n<\/span>Energy consumed = Power x Time taken
\n<\/span>Energy consumed = 1.5 \u00d7 10
\n<\/span>Energy consumed = 15 kWh
\n<\/span>Therefore, the energy consumed by the heater in 10 hours is 15 kWh.<\/span><\/p>\n

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?
\n<\/strong><\/span>Answer –\u00a0 <\/strong>The law of conservation of energy states that energy can be neither created nor destroyed. It can only be converted from one form to another. Consider the case of an oscillating pendulum.
\n\"NCERT
\n<\/span>When a pendulum moves from its mean position P to either of its extreme positions A or B, it rises through a height h above the mean level P. At this point, the kinetic energy of the bob changes completely into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As it moves towards point P, its potential energy decreases progressively. Accordingly, the kinetic energy increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses only kinetic energy. This process is repeated as long as the pendulum oscillates. The bob does not oscillate forever. It comes to rest because air resistance resists its motion. The pendulum loses its kinetic energy to overcome this friction and stops after some time. The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain.<\/span><\/p>\n

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?
\n<\/strong><\/span>Answer –\u00a0<\/strong>Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,\u00a0E<\/span><\/span><\/span>k <\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/sub><\/span>= \"\\frac{1}{2}\"<\/span><\/span>m<\/span><\/span>v<\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/sup><\/span><\/span><\/span><\/span>. To bring the object to rest, \"\\frac{1}{2}\"<\/span><\/span>m<\/span><\/span>v<\/span><\/span><\/span>2\u00a0<\/span><\/span><\/span><\/span><\/span><\/sup><\/span><\/span><\/span><\/span><\/span>amount of work is required to be done on the object.<\/span><\/p>\n

17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km\/h.
\n<\/strong><\/span>Answer –\u00a0 <\/strong>Kinetic energy, Ek<\/sub> = \"\\frac{1}{2}\"<\/span><\/span><\/span><\/span><\/span>mv2
\n<\/sup>Where,
\nMass of car, m = 1500 kg
\nVelocity of car, v = 60 km\/h = 60 \u00d7 \"\\frac{5}{18}\" ms-1<\/sup>
\n<\/span><\/p>\n

\u2234 Ek <\/sub>= \"\\frac{1}{2}\" \u00d7 1500 \u00d7 (60 \u00d7 \"\\frac{5}{18}\")2<\/sup><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n
= 20.8 \u00d7 104<\/sup> J<\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n
Hence, 20.8 \u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 104<\/sup>\u00a0J of work is required to stop the car.<\/div>\n
<\/div>\n
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.
\n\"NCERT
\n<\/strong><\/span>Answer – <\/strong><\/span><\/div>\n

Case I
\n<\/span><\/strong>In this case, the direction of force functioning on the block is perpendicular to the displacement. Therefore, work done by force on the block will be zero.<\/span><\/p>\n

Case II
\n<\/span><\/strong>In this case, the direction of force functioning on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.<\/span><\/p>\n

Case III
\n<\/span><\/strong>In this case, the direction of force functioning on the block is contrary to the direction of displacement. Therefore, work done by force on the block will be negative.<\/span><\/p>\n

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?
\n<\/strong><\/span>Answer –\u00a0<\/strong><\/span>Acceleration in an object could be zero even when many forces work on it. This happens when all the forces get rid of one another, i.e., the online force working on the object is zero. For a uniformly moving object, the online force working on the it is zero. Hence, the acceleration of the thing is zero. Hence, Soni is correct.<\/span><\/p>\n

20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
\n<\/strong><\/span>Answer –
\n<\/strong><\/span>Power rating of the device (P) = 500 W = 0.50 kW
\n<\/span>Time for which the device runs (T) = 10 h
\n<\/span>Energy consumed by an electric device can be obtained by the expression
\n<\/span>Power = Energy consumed\/Time taken
\n<\/span>\u2234 Energy consumed = Power \u00d7 Time
\n<\/span>Energy consumed = 0.50 \u00d7 10
\n<\/span>Energy consumed = 5 kWh
\n<\/span>Thus, the energy consumed by four equal rating devices in 10 h will be
\n<\/span>= 4 \u00d7 5 kWh
\n<\/span>= 20 kWh<\/span><\/p>\n

21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
\n<\/strong><\/span>Answer –\u00a0 <\/strong><\/span>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. Since the object hits the ground, all its kinetic energy becomes heat energy and sound energy. It can also deform the ground depending upon the ground\u2019s nature and the amount of kinetic energy possessed by the object.<\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/p>\n

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NCERT Solutions Class 9 Science\u00a0 The NCERT Solutions in English Language for Class 9 Science Chapter – 11 (Work and Energy) has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter – 11 (Work and Energy)\u00a0 Questions 1. A force of 7 N acts on an object. The displacement […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[729],"tags":[730,752,753,733,5],"class_list":["post-5046","post","type-post","status-publish","format-standard","hentry","category-class-9-science","tag-class-9-ncert-solutions","tag-ncert-class-9-science-chapter-11-work-and-energy","tag-ncert-solutions-class-9-science-chapter-11-in-english","tag-ncert-solutions-class-9-science-in-english","tag-ncert-solutions-in-english"],"yoast_head":"\nNCERT Solutions Class 9 Science Chapter 11 Work and Energy | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 9 Science\u00a0The NCERT Solutions in English Language for Class 9 Science Chapter - 11 (Work and Energy) has been provided here to help the students in solving the questions from this exercise.\u00a0Chapter - 11 (Work and Energy)\u00a0\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-science-chapter-11-work-and-energy\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 9 Science Chapter 11 Work and Energy\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions Class 9 Science\u00a0The NCERT Solutions in English Language for Class 9 Science Chapter - 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