{"id":23890,"date":"2021-06-03T16:27:02","date_gmt":"2021-06-03T10:57:02","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=23890"},"modified":"2022-03-02T10:38:28","modified_gmt":"2022-03-02T05:08:28","slug":"ncert-solutions-for-class-9-science-chapter-11","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-science-chapter-11\/","title":{"rendered":"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy"},"content":{"rendered":"<p>These <a href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-science\/\">NCERT Solutions for Class 9 Science<\/a> Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.<\/p>\n<h2>Work and Energy NCERT Solutions for Class 9 Science Chapter 11<\/h2>\n<h3><span id=\"Class_8_Science_Chapter_1_Crop_Production_and_Management_Exercise_Questions_and_Answers\">Class 9 Scien<\/span><span id=\"Class_8_Science_Chapter_1_Crop_Production_and_Management_Exercise_Questions_and_Answers\">ce Chapter 11 Work and Energy InText Questions and Answers<\/span><\/h3>\n<p>Question 1.<br \/>\nA 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?<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23891\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-1.png?resize=331%2C68&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 1\" width=\"331\" height=\"68\" srcset=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-1.png?w=331&amp;ssl=1 331w, https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-1.png?resize=300%2C62&amp;ssl=1 300w\" sizes=\"(max-width: 331px) 100vw, 331px\" data-recalc-dims=\"1\" \/><br \/>\nAnswer:<br \/>\nWhen a force F acts on an object to displace it through a distance S in its direction, then the work done W on the body by the force is given by:<br \/>\nWork done = Force x Displacement<br \/>\nW = F \u00d7 S<br \/>\nWhere,<br \/>\nF = 7N<br \/>\nS = 8m<br \/>\nTherefore, work done, W = 7 \u00d7 8<br \/>\n= 56 Nm<br \/>\n= 56 J<\/p>\n<p>Question 2.<br \/>\nWhen do we say that work is done?<br \/>\nAnswer:<br \/>\nWork is done whenever the given conditions are satisfied:<\/p>\n<ul>\n<li>A force acts On the body.<\/li>\n<li>There is a displacement of the body caused by the applied force along the direction of the applied force.<\/li>\n<\/ul>\n<p>Question 3.<br \/>\nWrite an expression for the work done when a force is acting on an object in the direction of its displacement.<br \/>\nAnswer:<br \/>\nWhen a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression:<br \/>\nWork done = Force \u00d7 Displacement<br \/>\nW = F \u00d7 s<\/p>\n<p>Question 4.<br \/>\nDefine 1 J of work.<br \/>\nAnswer:<br \/>\n1 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.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 5.<br \/>\nA 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?<br \/>\nAnswer:<br \/>\nWork done by the bullocks is given by the expression:<br \/>\nWork done = Force \u00d7 Displacement<br \/>\nW = F \u00d7 d<br \/>\nWhere,<br \/>\nApplied force, F = 140 N<br \/>\nDisplacement, d = 15m<br \/>\nW = 140 \u00d7 15 = 2100 J<br \/>\nHence, 2100 J of work is done in ploughing the length of the field.<\/p>\n<p>Question 6.<br \/>\nWhat is the kinetic energy of an object?<br \/>\nAnswer:<br \/>\nKinetic energy is the energy possessed by a body by the virtue of its motion. Every moving object possesses kinetic energy. A body uses kinetic energy to do work. Kinetic energy of hammer is used in driving a nail into a log of wood, kinetic energy of air is used to run windmills, etc.<\/p>\n<p>Question 7.<br \/>\nWrite an expression for the kinetic energy of an object.<br \/>\nAnswer:<br \/>\nIf a body of mass &#8216;m&#8217; is moving with a velocity v, then its kinetic energy E<sub>k<\/sub> is given by the expression,<br \/>\nE<sub>k<\/sub> = \\(\\frac {1}{2}\\)mv<sup>2<\/sup><br \/>\nIts SI unit is Joule (J j.<\/p>\n<p>Question 7.<br \/>\nThe kinetic energy of an object of mass, m moving with a velocity of 5 m s<sup>-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?<br \/>\nAnswer:<br \/>\nExpression for kinetic energy is<br \/>\nE<sub>K<\/sub> = \\(\\frac {1}{2}\\)mv<sup>2<\/sup><br \/>\nm = Mass of the object<br \/>\nv = Velocity of the object = 5ms<sup>-1<\/sup><br \/>\nGiven that kinetic energy, E<sub>K<\/sub> = 25 J<br \/>\n(i) If the velocity of an object is doubled, then v = 5 \u00d7 2 = 10 ms<sup>-1<\/sup>.<br \/>\nTherefore, its kinetic energy becomes 4 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 \u00d7 4 = 100 J.<br \/>\n(ii) If velocity is increased three times, then its kinetic energy becomes 9 times its original value, because it is proportional to the square of the velocity. Hence, kinetic energy = 25 \u00d7 9 = 225 J.<\/p>\n<p>Question 8.<br \/>\nWhat is power?<br \/>\nAnswer:<br \/>\nPower is 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 the expression,<br \/>\nPower = \\(\\frac{\\text { Work }}{\\text { Time }}=\\frac{\\text { Energy }}{\\text { Time }}\\)<br \/>\nP = \\(\\frac{\\mathrm{W}}{\\mathrm{T}}\\)<br \/>\nIt is expressed in watt (W).<\/p>\n<p>Question 9.<br \/>\nDefine 1 watt of power:<br \/>\nAnswer:<br \/>\nA body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s, i.e.,<br \/>\n1W = \\(\\frac{1 \\mathrm{~J}}{1 \\mathrm{~s}}\\)<\/p>\n<p>Question 10.<br \/>\nA lamp consumes 1000 J of electrical energy in 10 s. What is its power?<br \/>\nAnswer:<br \/>\nPower is given by the expression,<br \/>\nWork done = \\(\\frac{\\text { Workdone }}{\\text { Time }}\\)<br \/>\nWork done = Energy consumed by the lamp = 1000 J<br \/>\nTime = 10 s<br \/>\nPower = \\(\\frac{1000}{10}\\) = 100J S<sup>-1<\/sup> = 100 W<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 11.<br \/>\nDefine average power.<br \/>\nAnswer:<br \/>\nA body can do different amount of work in different time intervals. Hence, it is better to define average power. Average power is obtained by dividing the total amount of work done in the total time taken to do this work.<br \/>\nAverage Power = \\(\\frac{\\text { Total work done }}{\\text { Total time taken }}\\)<\/p>\n<h3><span id=\"Class_8_Science_Chapter_1_Crop_Production_and_Management_Exercise_Questions_and_Answers\">Class 9 Science Chapter 11 Work and Energy Textbook Questions and Answers<\/span><\/h3>\n<p>Question 1.<br \/>\nLook at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term \u2018work\u2019.<\/p>\n<ul>\n<li>Suma is swimming in a pond.<\/li>\n<li>A donkey is carrying a load on its back.<\/li>\n<li>A windmill is lifting water from a well.<\/li>\n<li>A green plant is carrying out photosynthesis.<\/li>\n<li>An engine is pulling a train.<\/li>\n<li>Food grains are getting dried in the sun.<\/li>\n<li>A sailboat is moving due to wind energy.<\/li>\n<\/ul>\n<p>Answer:<br \/>\nWork is done whenever the given two conditions are satisfied:<br \/>\n(i) A force acts on the body.<br \/>\n(ii) There is a displacement of the body by the application of force in or opposite to the direction of force.<\/p>\n<p>(a) While swimming, Suma applies a force to push the water backwards. Therefore, Suma swims in the forward direction caused by the forward reaction of water. Here, the force causes a displacement. Hence, work is done by Seema while swimming.<\/p>\n<p>(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.<\/p>\n<p>(c) A windmill works against the gravitational force to lift water. Hence, work is done by the windmill in lifting water from the well.<\/p>\n<p>(d) In this case, there is no displacement of the leaves of the plant. Therefore, the work done is zero.<\/p>\n<p>(e) An engine applies force to pull the train. This allows the train to move in the direction of force. Therefore, there is a displacement in the train in the same direction. Hence, work is done by the engine on the train.<\/p>\n<p>(f) Foodgrains do not move in the presence of solar energy. Hence, the work done is zero during the process of food grains getting dried in the Sun.<\/p>\n<p>(g) Wind energy applies a force on the sailboat to push it in the forward direction. Therefore, there is a displacement in the boat in the direction of force. Hence, work is done by wind on the boat.<\/p>\n<p>Question 2.<br \/>\nAn 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?<br \/>\nAnswer:<br \/>\nWork 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.<br \/>\nWork done by gravity is given by the expression,<br \/>\nW = mgh<br \/>\nWhere,<br \/>\nh = Vertical displacement = 0<br \/>\nW = mg \u00d7 0 = 0 J<br \/>\nTherefore, the work done by gravity on the given object is zero joule.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nA battery lights a bulb. Describe the energy changes involved in the process.<br \/>\nAnswer:<br \/>\nWhen a bulb is connected to a battery, then the chemical energy of the battery is transferred into electrical energy. When the bulb receives this electrical energy, then it converts it into light and heat energy. Hence, the transformation of energy in the given situation can be shown as:<br \/>\nChemical Energy \u2192 electrical Energy \u2192 Light Energy + Heat energy<\/p>\n<p>Question 4.<br \/>\nCertain 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.<br \/>\nAnswer:<br \/>\nKinetic energy is given by the expression,<br \/>\n(Ek)v = \\(\\frac {1}{2}\\)mv2<br \/>\nWhere,<br \/>\nEk = Kinetic energy of the object moving with a velocity, v<br \/>\nm = Mass of the object<br \/>\n(i) Kinetic energy when the object was moving with a velocity 5 m s-1<br \/>\n(Ek)5 = \\(\\frac {1}{2}\\) \u00d7 20 \u00d7 (5)2 = 250J<br \/>\nii. Kinetic energy when the object was moving with a velocity 2 ms-1<br \/>\n(Ek)2= \\(\\frac {1}{2}\\) \u00d7 20 \u00d7 (2)2 = 40J<br \/>\nWork done by force is equal to the change in kinetic energy.<br \/>\nTherefore, work done by force = (Ek)2 &#8211; (Ek)5<br \/>\n= 40 &#8211; 250 = -210 J<br \/>\nThe negative sign indicates that the force is acting in the direction opposite to the motion of the object.<\/p>\n<p>Question 5.<br \/>\nA 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.<br \/>\nAnswer:<br \/>\nWork 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,<br \/>\nW = mgh<br \/>\nWhere,<br \/>\nVertical displacement, h = 0<br \/>\n\u2234 W = mg \u00d7 0 = 0<br \/>\nHence, the work done by gravity on the body is zero.<\/p>\n<p>Question 6.<br \/>\nThe potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?<br \/>\nAnswer:<br \/>\nNo. The process does not violate the law of conservation of energy. This is because when the body falls from a height, then 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. During the process, total mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is not violated.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 7.<br \/>\nWhat are the various energy transformations that occur when you are riding a bicycle?<br \/>\nAnswer:<br \/>\nWhile riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kinetic energy of the bicycle. Heat energy heats the rider&#8217;s body. Kinetic energy provides a velocity to the bicycle. The transformation can be shown as:<br \/>\nMuscular Energy \u2192 Kinetic Energy + Heat Energy<br \/>\nDuring the transformation, the total energy remains conserved.<\/p>\n<p>Question 8.<br \/>\nDoes 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?<br \/>\nAnswer:<br \/>\nWhen we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there is no loss of energy because muscular energy is transferred into heat energy, which causes our body to become hot.<\/p>\n<p>Question 9.<br \/>\nA certain household has consumed 250 units of energy during a month. How much energy is this in joules?<br \/>\nAnswer:<br \/>\n1 unit of energy is equal to 1 kilowatt hour (kWh).<br \/>\n1 unit = 1 kWh<br \/>\n1 kWh = 3.6 \u00d7 10<sup>6<\/sup> J<br \/>\nTherefore, 250 units of energy = 250 \u00d7 3.6 \u00d7 10<sup>6<\/sup> = 9 \u00d7 10<sup>8<\/sup> J<\/p>\n<p>Question 10.<br \/>\nAn 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.<br \/>\nAnswer:<br \/>\nGravitational potential energy is given by the expression,<br \/>\nW = mgh Where,<br \/>\nh = Vertical displacement = 5 m<br \/>\nm = Mass of the object = 40 kg<br \/>\ng = Acceleration due to gravity = 9.8 ms<sup>-2<\/sup><br \/>\n\u2234 W = 40 \u00d7 5 \u00d7 9.8 = 1960 J.<br \/>\nAt half-way down, the potential energy of the object will be \\(\\frac{1960}{2}\\) = 980 J.<br \/>\nAt this point, the object has an equal amount of potential and kinetic energy. This is due to the law of conservation of energy. Hence, half-way down, the kinetic energy of the object will be 980 J.<\/p>\n<p>Question 11.<br \/>\nWhat is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.<br \/>\nAnswer:<br \/>\nWork is done whenever the given two conditions are satisfied:<\/p>\n<ul>\n<li>A force acts on the body.<\/li>\n<li>There is a displacement of the body by the application of force in or opposite to the direction of force.<\/li>\n<\/ul>\n<p>If the direction of force is perpendicular to displacement, then the work done is zero.<br \/>\nWhen 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 on the satellite by the Earth is zero.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nCan there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.<br \/>\nAnswer:<br \/>\nYes. For a uniformly moving object.<br \/>\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.<\/p>\n<p>Question 13.<br \/>\nA 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.<br \/>\nAnswer:<br \/>\nWork is done whenever the given two conditions are satisfied:<\/p>\n<ul>\n<li>A force acts on the body.<\/li>\n<li>There is a displacement of the body by the application of force in or opposite to the direction of force.<\/li>\n<\/ul>\n<p>When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Although, force of gravity is acting on the bundle, the person is not applying any force on it. Hence, in the absence of force, work done by the person on the bundle is zero.<\/p>\n<p>Question 14.<br \/>\nAn electric heater is rated 1500 W. How much energy does if use in 10 hours?<br \/>\nAnswer:<br \/>\nEnergy consumed by an electric heater can be obtained with the help of the expression,<br \/>\nP = \\(\\frac{W}{T}\\) Where,<br \/>\nPower rating of the heater, P = 1500 W = 1.5 kW<br \/>\nTime for which the heater has operated, T = 10 h<br \/>\nWork done = Energy consumed by the heater<br \/>\nTherefore, energy consumed = Power \u00d7 Time<br \/>\n= 1.5 \u00d7 10 = 15 kWh<br \/>\nHence, the energy consumed by the heater in 10 h is 15 kWh.<\/p>\n<p>Question 15.<br \/>\nIllustrate 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?<br \/>\nAnswer:<br \/>\nThe law of conservation of energy states that energy can be neither created nor destroyed. It can only be converted from one form to another.<\/p>\n<p>Consider the case of an oscillating pendulum.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23892\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-2.png?resize=316%2C256&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 2\" width=\"316\" height=\"256\" srcset=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-2.png?w=316&amp;ssl=1 316w, https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-2.png?resize=300%2C243&amp;ssl=1 300w\" sizes=\"(max-width: 316px) 100vw, 316px\" data-recalc-dims=\"1\" \/><br \/>\nWhen 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.<\/p>\n<p>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.<\/p>\n<p>The law of conservation of energy is not violated because the energy lost by the pendulum to overcome friction is gained by its surroundings. Hence, the total energy of the pendulum and the surrounding system remain conserved.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 16.<br \/>\nAn object of mass, m is moving with a constant velocity, v. How much work should be done oh the object in order to bring the object to rest?<br \/>\nAnswer:<br \/>\nKinetic energy of an object of mass, m moving with a velocity, v is given by the expression,<br \/>\nE<sub>k<\/sub> = \\(\\frac {1}{2}\\)mv<sup>2<\/sup><br \/>\nTo bring the object to rest, \\(\\frac {1}{2}\\)mv<sup>2<\/sup> amount of work is required to be done on the object.<\/p>\n<p>Question 17.<br \/>\nCalculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km\/h?<br \/>\nAnswer:<br \/>\nKinetic energy, Ek = \\(\\frac {1}{2}\\)mv<sup>2<\/sup><br \/>\nWhere,<br \/>\nMass of car, m = 1500 kg<br \/>\nVelocity of car, v = 60 km\/h = 60 \u00d7 \\(\\frac {5}{18}\\)ms-1<br \/>\n\u2234 Ex = \\(\\frac {1}{2}\\) \u00d7 1500 \u00d7 \\(\\left(60 \\times \\frac{5}{18}\\right)^{2}\\) =20.8 \u00d7 104J<br \/>\nHence, 20.8 \u00d7 104 J of work is required to stop the car.<\/p>\n<p>Question 18.<br \/>\nIn 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.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23893\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-3.png?resize=418%2C111&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 3\" width=\"418\" height=\"111\" srcset=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-3.png?w=418&amp;ssl=1 418w, https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-3.png?resize=300%2C80&amp;ssl=1 300w\" sizes=\"(max-width: 418px) 100vw, 418px\" data-recalc-dims=\"1\" \/><br \/>\nAnswer:<br \/>\nWork is done whenever the given two conditions are satisfied:<\/p>\n<ul>\n<li>A force acts on the body.<\/li>\n<li>There is a displacement of the body by the application of force in or opposite to the direction of force.<\/li>\n<\/ul>\n<p>Case I:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23894\" src=\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-4.png?resize=218%2C119&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 4\" width=\"218\" height=\"119\" data-recalc-dims=\"1\" \/><br \/>\nIn this case, the direction of force acting on the block is perpendicular to the displacement.<br \/>\nTherefore, work done by force on the block will be zero.<\/p>\n<p>Case II:<br \/>\nIn this case, the direction of force acting on the block is in the direction of displacement. Therefore, work done by force on the block will be positive.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23895\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-5.png?resize=285%2C69&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 5\" width=\"285\" height=\"69\" data-recalc-dims=\"1\" \/><br \/>\nIn this case, the direction of force acting on the block is in the direction of displacement.<br \/>\nTherefore, work done by force on the block will be positive.<\/p>\n<p>Case III:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23896\" src=\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-6.png?resize=289%2C70&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 6\" width=\"289\" height=\"70\" data-recalc-dims=\"1\" \/><br \/>\nIn this case, the direction of force acting on the block is opposite to the direction of displacement. Therefore, work done by force on the block will be negative.<\/p>\n<p>Question 19.<br \/>\nSoni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?<br \/>\nAnswer:<br \/>\nAcceleration 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. 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.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 20.<br \/>\nFind the energy in kW h consumed in 10 hours by four devices of power 500 W each.<br \/>\nAnswer:<br \/>\nEnergy consumed by an electric device<br \/>\ncan be obtained with the help of the expression for power,<br \/>\nP = \\(\\frac{W}{T}\\)<br \/>\nWhere,<br \/>\nPower rating of the device, P = 500 W = 0.50 kW<br \/>\nTime for which the device runs, T = 10 h<br \/>\nWork done = Energy consumed by the device<br \/>\nTherefore, energy consumed = Power \u00d7 Time<br \/>\n= 0.50 \u00d7 10 = 5 kWh<br \/>\nHence, the energy consumed by four equal rating devices in 10 h will be 4 \u00d7 5 kWh = 20 kWh = 20 Units.<\/p>\n<p>Question 21.<br \/>\nA freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?<br \/>\nAnswer:<br \/>\nWhen 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 gets converted into kinetic energy. As the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. It can also deform the ground depending upon the nature of the ground and the amount of kinetic energy possessed by the object.<\/p>\n<h3><span id=\"Class_8_Science_Chapter_1_Crop_Production_and_Management_Exercise_Questions_and_Answers\">Class 9 Science Chapter 11 Work and Energy Additional Important Questions and Answers<\/span><\/h3>\n<p><span style=\"color: #0000ff;\">Multiple Choice Questions<\/span><br \/>\n<span style=\"color: #0000ff;\">Choose the correct option:<\/span><\/p>\n<p>Question 1.<br \/>\nWhen a body falls freely towards the earth, then its total energy<br \/>\n(a) increases<br \/>\n(b) decreases<br \/>\n(c) remains constant<br \/>\n(d) first increases and then decreases<br \/>\nAnswer:<br \/>\n(c) remains constant<\/p>\n<p>Question 2.<br \/>\nA car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car<br \/>\n(a) does not change<br \/>\n(b) becomes twice to that of initial<br \/>\n(c) becomes 4 times that of initial<br \/>\n(d) becomes 16 times that of initial<br \/>\nAnswer:<br \/>\n(a) does not change<\/p>\n<p>Question 3.<br \/>\nIn case of negative work the angle between the force and displacement is<br \/>\n(a) 00<br \/>\n(b) 450<br \/>\n(c) 900<br \/>\n(d) 1800<br \/>\nAnswer:<br \/>\n(d) 1800<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nAn iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same<br \/>\n(a) acceleration<br \/>\n(b) momenta<br \/>\n(c) potential energy<br \/>\n(d) kinetic energy<br \/>\nAnswer:<br \/>\n(a) acceleration<\/p>\n<p>Question 5.<br \/>\nA girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g =10 ms<sup>-2<\/sup>)<br \/>\n(a) 6 \u00d7 103 J<br \/>\n(b) 6J<br \/>\n(c) 0.6 J<br \/>\n(d) zero<br \/>\nAnswer:<br \/>\n(d) zero<\/p>\n<p>Question 6.<br \/>\nWhich one of the following is not the unit of energy?<br \/>\n(a) joule<br \/>\n(b) newton metre<br \/>\n(c) kilowatt<br \/>\n(d) kilowatt hour<br \/>\nAnswer:<br \/>\n(c) kilowatt<\/p>\n<p>Question 7.<br \/>\nThe work done on an object does not depend upon the<br \/>\n(a) displacement<br \/>\n(b) force applied<br \/>\n(c) angle between force and displacement<br \/>\n(d) initial velocity of the object<br \/>\nAnswer:<br \/>\n(d) initial velocity of the object<\/p>\n<p>Question 8.<br \/>\nWater stored in a dam possesses<br \/>\n(a) no energy<br \/>\n(b) electrical energy<br \/>\n(c) kinetic energy<br \/>\n(d) potential energy<br \/>\nAnswer:<br \/>\n(d) potential energy<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nA body is falling from a height h. After it has fallen a height h\/2, it will possess<br \/>\n(a) only potential energy<br \/>\n(b) only kinetic energy<br \/>\n(c) half potential and half kinetic energy<br \/>\n(d) more kinetic and less potential energy<br \/>\nAnswer:<br \/>\n(c) half potential and half kinetic energy<\/p>\n<p><span style=\"color: #0000ff;\">Very Short Answer Type Questions<\/span><\/p>\n<p>Question 1.<br \/>\nA boy climbs up the staircase to reach to second, floor from first floor. Has he done any work?<br \/>\nAnswer:<br \/>\nYes, he has done work against the gravitational force by climbing up.<\/p>\n<p>Question 2.<br \/>\nIn the following activities, has work been done or not? Why?<br \/>\n(i) A girl pulls a trolly and trolley moves.<br \/>\n(ii) A book is lifted through a height &#8216;h&#8217;.<br \/>\nAnswer:<br \/>\n(i) When a girl pulls a trolley, work is done because trolley experiences a displacement from its mean position.<br \/>\n(ii) When a book is lifted through a height &#8216;h&#8217; because book has gained displacement in its height from the ground level.<\/p>\n<p>Question 3.<br \/>\nIs work a scalar or vector quantity?<br \/>\nAnswer:<br \/>\nAlthough work done = Force \u00d7 Displacement, yet it is taken as scalar quantity.<\/p>\n<p>Question 4.<br \/>\nWhat is the work done in lifting an object of mass &#8216;m&#8217; through a height &#8216;h&#8217; from the ground?<br \/>\nAnswer:<br \/>\nForce acting on body, F = m \u00d7 g<br \/>\nDisplacement = h.<br \/>\nTherefore, workdone = Force \u00d7 Displacement<br \/>\n= mg \u00d7 h = mgh<\/p>\n<p>Question 5.<br \/>\nWhat is the angle between the line of force and the direction of displacement for maximum work?<br \/>\nAnswer:<br \/>\nWith W = F.s. cos \u03b8, \u03b8 = 0 for the maximum work.<\/p>\n<p>Question 6.<br \/>\nDefine energy. What is the S.I. unit?<br \/>\nAnswer:<br \/>\nEnergy is the capacity to do work. Its S.I. unit is joule.<\/p>\n<p>Question 7.<br \/>\nWhich physical quantity has the unit Js<sup>-1<\/sup> ?<br \/>\nAnswer:<br \/>\nPower.<\/p>\n<p>Question 8.<br \/>\nWhich of the two out of work, power and energy have same unit?<br \/>\nAnswer:<br \/>\nWork and energy.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nWhat are the two forms of mechanical energy?<br \/>\nAnswer:<br \/>\nThe two forms of mechanical energy are kinetic energy and potential energy.<\/p>\n<p>Question 10.<br \/>\nState the relation in kinetic energy of a body in motion with respect to its (i) mass and (ii) velocity.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Kinetic energy of a body is directly proportional to its mass.<\/li>\n<li>Kinetic energy of a body is directly proportional to the square of its velocity.<\/li>\n<\/ul>\n<p>Question 11.<br \/>\nState the relation in potential energy of a body at rest with respect to its mass and height.<br \/>\nAnswer:<br \/>\nPotential energy of a body is directly proportional to both the mass and height.<\/p>\n<p>Question 12.<br \/>\nWhat is the relationship between Watt and Horsepower?<br \/>\nAnswer:<br \/>\n1 Horsepower = 746 Watt.<\/p>\n<p>Question 13.<br \/>\nThe work done by the heart for each beat is 1J. Calculate the power of the heart if it beats 72 times in a minute.<br \/>\nAnswer:<br \/>\nTotal work done in minute = 1J \u00d7 72 = 72J<br \/>\nTime = 1 minute = 1 \u00d7 60 = 6 seconds<br \/>\nPower = \\(\\frac{\\text { Work done }}{\\text { Time }}\\) = \\(\\frac {72}{60}\\) =1.2 Watt<\/p>\n<p>Question 14.<br \/>\nWhat type of energy transformation takes place in (i) dynamo and (ii) Electric motor?<br \/>\nAnswer:<br \/>\n(i) Dymamo\u2014Mechanical energy into electrical energy,<br \/>\n(ii) Electric motor\u2014Electrical energy into mechanical energy<\/p>\n<p>Question 15.<br \/>\nWhich type of energy is possessed by stretched rubber band?<br \/>\nAnswer:<br \/>\nElastic potential energy<\/p>\n<p>Question 16.<br \/>\nA rubber ball and a leather cricket ball are moving with same velocity. Which will have greater kinetic energy? Why ?<br \/>\nAnswer:<br \/>\nThe leather cricket ball will have greater kinetic energy because of its large mass as compared to rubber ball.<\/p>\n<p>Question 17.<br \/>\nWhich will require more force to be pushed forward, a truck in rest or a slowly rolling truck?<br \/>\nAnswer:<br \/>\nA truck in state of rest would require more force because of greater static friction than rolling friction.<\/p>\n<p>Question 18.<br \/>\nWhat happens to kinetic energy of a body in motion when it comes to rest?<br \/>\nAnswer:<br \/>\nKinetic energy of body-in motion gets transformed into its potential energy when brought to rest while some is lost in overcoming friction as heat.<\/p>\n<p><span style=\"color: #0000ff;\">Short Answer Type Questions<\/span><\/p>\n<p>Question 1.<br \/>\nA student when preparing for examinations draws diagram, read hooks and discuss problems with his friends. In the process does he do any work?<br \/>\nAnswer:<br \/>\nNo work is done by the student in reading or discussing problems with his friends as long as he remains sitting at one position to avoid any displacement.<\/p>\n<p>Question 2.<br \/>\nWhat is work, a scalar or vector quantity? why?<br \/>\nAnswer:<br \/>\nWork is a scalar quantity because on the application of force, the work done is in the direction of force. Hence, there is no change in the direction for work to be called a vector quantity.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nDerive an expression for the work done on an object when the force is acting on the body in a direction making an angle &#8216;\u03b8&#8217; with the direction of displacement of the object.<br \/>\nAnswer:<br \/>\nLet a force F acts on a body of mass &#8216;m&#8217; by making an angle &#8216;\u03b8&#8217; with the direction of the displacement of the object.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23897\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-7.png?resize=409%2C133&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 7\" width=\"409\" height=\"133\" srcset=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-7.png?w=409&amp;ssl=1 409w, https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-7.png?resize=300%2C98&amp;ssl=1 300w\" sizes=\"(max-width: 409px) 100vw, 409px\" data-recalc-dims=\"1\" \/><br \/>\nThe component of the force in direction of the displacement = F cos \u03b8<br \/>\nLet displacement of body = s<br \/>\nWork done on the object = (F cos \u03b8) \u00d7 s = F \u00d7 s cos \u03b8<\/p>\n<p>Question 4.<br \/>\nWhat is the workdone against force of gravity if a person carries a 30 kg load on his head?<br \/>\nAnswer:<br \/>\nWork done against force of gravity W = F. s. cos \u03b8<br \/>\nWith \u03b8 = 90\u00b0 and cos \u03b8 = 0<br \/>\nW = 0 \u00d7 s = 0<br \/>\nHence, total work done will be zero.<\/p>\n<p>Question 5.<br \/>\nWhat do you mean by negative work and positive work? Give example.<br \/>\nAnswer:<br \/>\nWhen a force acts on a body in a direction opposite to the direction of displacement<br \/>\nWork done = &#8211; F cos \u03b8 \u00d7 s<br \/>\nThe negative sign indicates that work done is negative.<\/p>\n<p>When a force acts on a body in a direction of the displacement of body, the work done is positive.<br \/>\nWhen a bullock pulls a cart, the work done is called positive while when a player tries to stop a moving ball, the work done is called negative.<\/p>\n<p>Question 6.<br \/>\nIf a ball tied to a string is whirled in a circular orbit with centre being the hand holding the free end of string, then what is the work done on the ball?<br \/>\nAnswer:<br \/>\nThe force acting on the ball is along the radius of the circular path i.e. the direction of the force is perpendicular to the direction of motion of the ball.<br \/>\nTherefore, work done,<br \/>\nW = F. cos \u03b8 \u00d7 s<br \/>\n= O \u00d7 s = 0<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23898\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-8.png?resize=207%2C98&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 8\" width=\"207\" height=\"98\" data-recalc-dims=\"1\" \/><br \/>\nHence, work done on the ball is zero.<\/p>\n<p>Question 7.<br \/>\nTake a toy car, wind it using key. Place the car on the ground and answer the following questions:<br \/>\n(i) Did the car move? Why?<br \/>\n(ii) Does the energy acquired by the car depends upon the number of times you turn the key?<br \/>\n(iii) From where did it get energy?<br \/>\nAnswer:<br \/>\n(i) The car would move because the potential energy acquired by the car, due to widing of the spring of car got transformed into its kinetic energy.<\/p>\n<p>(ii) The potential energy gained by the car depends upon the number of turns of the widing of the key with increasing number of turns, the greater potential energy is acquired by the car. This acquired potential energy later changes into car&#8217;s kinetic energy.<\/p>\n<p>(iii) Car got the energy to move freon the spring inside it.<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nTwo bodies having equal masses are kept at a height of 20m and 30m respectively. What is the ratio of their potential energy?<br \/>\nAnswer:<br \/>\nRatio of potential energies = mgh<sub>1<\/sub> : mgh<sub>2<\/sub><br \/>\nmg \u00d7 20 : mg \u00d7 30<br \/>\n= 20 : 30<br \/>\n= 2 : 3.<\/p>\n<p>Question 9.<br \/>\nState the type of energy changes taking place in the following:<br \/>\n(i) A body is thrown up in air.<br \/>\n(ii) In a green leaf during photosynthesis.<br \/>\n(iii) In an oscillation of a pendulum.<br \/>\nAnswer:<br \/>\n(i) When a body is thrown in upward direction, the muscular energy which is chemical potential energy is transformed into the kinetic energy of the sail. As the ball keeps rising, kinetic energy continues to change into potential energy. When at the maximum height, the body has no kinetic energy but the maximum potential energy.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23899\" src=\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-9.png?resize=288%2C224&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 9\" width=\"288\" height=\"224\" data-recalc-dims=\"1\" \/><br \/>\n(ii) In a leaf during photosynthesis, solar energy changes into chemical potential energy.<\/p>\n<p>(iii) An oscilating pendulum has the maximum potential energy at its peak position where it temporarily come to state of rest while at its mean position, it has maximum kinetic energy.<\/p>\n<p>Question 10.<br \/>\nSuppose a hammer which falls purely on a nail placed on a wood has a mass of 1 kg. If it falls from a height of 1m, how much kinetic energy will it have just before hitting the nail?<br \/>\nAnswer:<br \/>\nMass of the hammer = 1 kg<br \/>\nHeight of the hammer, h = 1m<br \/>\nAcceleration due to gravity, g = 10 ms<sup>-2<\/sup><br \/>\nPotential energy at the highest point = mgh = 1 \u00d7 10 \u00d7 1 = 10 J<br \/>\nWhen the hammer falls, its whole potential energy would be converted into kinetic energy.<br \/>\nKinetic energy just before hitting the nail 10 J.<\/p>\n<p>Question 11.<br \/>\nA man whose mass is 50 kg climbs up 30 stairs in 30 seconds. If each step is 20 cm high. Calculate the power used in climbing the stairs.<br \/>\nAnswer:<br \/>\nMass of the man, m = 50 kg<br \/>\nNumber of the stairs = 30<br \/>\nHeight of one stair = 20 cm = 0.20 m<br \/>\nTotal height of stairs = 30 \u00d7 20 = 6 m<br \/>\nWorkdone in climbing the stairs=m \u00d7 g \u00d7 h<br \/>\n=50 \u00d7 10 \u00d7 6 = 3000 J<br \/>\nTune taken by man to climb up 30 stairs =30<br \/>\nPower used inclimbing stairs = \\(\\frac {3000}{30}\\) = 100 w<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nA porter lifts a luggage of 15 kg from the ground and put it on his head 2 m above the ground. Calculate the work done by him on the luggage Tig = 10 ms<sup>-2<\/sup>)<br \/>\nAnswer:<br \/>\nMass of luggage, m = 15 kg<br \/>\nHeight or displacement, h = 2m<br \/>\nWork done by porter = m.g \u00d7 h<br \/>\n= 15 \u00d7 10 \u00d7 2 = 300 J<\/p>\n<p>Question 13.<br \/>\nA force of 10 N acts on an object in the direction of displacement. If the displacement of the object is 2 m. Calculate the work done by the force.<br \/>\nAnswer:<br \/>\nForce applied, F = 10 N.<br \/>\nDisplacement, s = 2 m<br \/>\nWork done = F \u00d7 S = 10 \u00d7 2 = 20 N.<\/p>\n<p>Question 14.<br \/>\nAn object of mass 15 kg is moving with a constant velocity of 4 ms<sup>-1<\/sup>. What is the kinetic energy possessed by the body?<br \/>\nAnswer:<br \/>\nMass of the object, m = 15 kg.<br \/>\nVelocity of the object, v = 4 ms<sup>-1<\/sup><br \/>\nKinetic energy possessed by the body = \u00bd mv<sup>2<\/sup>.<br \/>\n= \u00bd \u00d7 15 \u00d7 (4)<sup>2<\/sup> = \u00bd \u00d7 15 \u00d7 16= 120 J<\/p>\n<p>Question 15.<br \/>\nWhat is the work done to increase the velocity of a car from 30 km h<sup>-1<\/sup> to 60 km h<sup>-1<\/sup> if the mass of the car is 1500 kg ?<br \/>\nAnswer:<br \/>\nMass of the car, m = 1500 kg<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23900\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-10.png?resize=267%2C137&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 10\" width=\"267\" height=\"137\" data-recalc-dims=\"1\" \/><br \/>\nWork done = change in kinetic energy<br \/>\n= \u00bd m (v<sup>2<\/sup> &#8211; u<sup>2<\/sup>)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23901\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-11.png?resize=298%2C181&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 11\" width=\"298\" height=\"181\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 16.<br \/>\nFind the energy present in any object of mass 10 kg when it is at a height of 6m above the ground. (Take g = 10 ms<sup>-2<\/sup>).<br \/>\nAnswer:<br \/>\nMass of the object, m = 10 kg<br \/>\nHeight, h = 6 m<br \/>\ng = 10 ms<sup>-2<\/sup><br \/>\nPotential energy possessed by the object = mg \u00d7 h<br \/>\n= 10 \u00d7 10 \u00d7 6 = 600 J<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 17.<br \/>\nAn object of mass 12 kg is at a certain height above the ground. If the gravitational potential energy of the object is 480 J. Find the height at which the object is with respect to the ground, (Give g =10 ms<sup>-1<\/sup>).<br \/>\nAnswer:<br \/>\nMass of the object, m = 12 kg<br \/>\nPotential energy, P.E. = 480 J<br \/>\nLet height of the object from ground = h<br \/>\nThen P.E. = m \u00d7 g \u00d7 h<br \/>\nh = \\(\\frac{\\mathrm{P.E}}{m \\times g}=\\frac{480}{12 \\times 10}=4 m\\)<\/p>\n<p>Question 18.<br \/>\nA woman pulls a bucket of water of total mass 5 kg from a well which is 10 m deep in 10s. Calculate the power used by her. (Take g = 10 ms<sup>-2<\/sup>).<br \/>\nAnswer:<br \/>\nMass of the bucket, m = 5 kg<br \/>\nHeight, h = 10 m<br \/>\ng = 10 ms<sup>-2<\/sup><br \/>\nEnergy consumed in pulling the bucket = m \u00d7 g \u00d7 h<br \/>\n= 5 \u00d7 10 \u00d7 10 = 500 J<br \/>\nTime taken = 10s<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23902\" src=\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-12.png?resize=314%2C73&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 12\" width=\"314\" height=\"73\" srcset=\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-12.png?w=314&amp;ssl=1 314w, https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-12.png?resize=300%2C70&amp;ssl=1 300w\" sizes=\"(max-width: 314px) 100vw, 314px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 19.<br \/>\nTwo girls A and B each of weight 400 N climb up a rope to a height of 8m. Girl A takes 20 seconds while girl B takes 50 seconds to accomplish their task. What is the power spent by each girl?<br \/>\nAnswer:<br \/>\n(i) Weight of the girl &#8216;A&#8217; = 400 N.<br \/>\nDisplacement (height) h = 8m<br \/>\nWork done by girl A = mg \u00d7 h<br \/>\n=400 \u00d7 8 = 3200 J<br \/>\nTime taken by girl A, t = 20 seconds<br \/>\nPower spent by girl A,<br \/>\n= \\(\\frac{\\text { Workdone }}{\\text { Time taken }}=\\frac{3200}{20}\\) = 160 W<\/p>\n<p>(ii) Weight of the girl B, = 400 N<br \/>\nDisplacement (height), h = 8 m<br \/>\n= 400 \u00d7 8 = 3200 J<br \/>\nTime taken, t = 50 seconds<br \/>\nPower spent by girl B<br \/>\n= \\(\\frac{\\text { Workdone }}{\\text { Time taken }}=\\frac{3200}{20}\\) = 64 W<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 20.<br \/>\nA boy of 50 kg runs up a staircase of 45 steps in 9 seconds. If the height of each step is 15 cm. Calculate the power of the boy.<br \/>\nAnswer:<br \/>\nMass of the boy, m = 50 kg<br \/>\nHeight of each step = 15 cm = 0.15 m<br \/>\nNumber of steps in stair case = 45<br \/>\nTotal height of stair case = 45 \u00d7 0.15 = 6.75 m<br \/>\nTime taken to climb, t = 9 seconds<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23903\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-13.png?resize=300%2C105&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 13\" width=\"300\" height=\"105\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 21.<br \/>\nA boy pulls a toy car with a force of SON through a string which makes an angle of 30\u00b0 with the horizontal so as to move the toy 2 m horizontally. Calculate the work done by the boy on the toy car.<br \/>\nAnswer:<br \/>\nForce applied by the boy = F = SON<br \/>\nAngle of direction of force with the direction of displacement, \u03b8 = 30\u00b0<br \/>\nDisplacement, S = 2m<br \/>\nWork done = F \u00d7 s cos \u03b8<br \/>\nWork done = 50 \u00d7 2 cos 30\u00b0<br \/>\n= 50 \u00d7 2 \u00d7 \\(\\frac{\\sqrt{3}}{2}\\) = 50\\(\\sqrt{3}\\)J<br \/>\n= 50 \u00d7 1.732 = 86.5 J<\/p>\n<p>Question 22.<br \/>\nCalculate the velocity of an object of mass 100 g moving with a kinetic energy of 200 J.<br \/>\nAnswer:<br \/>\nMass of the object = 100g = 0.1 kg<br \/>\nKinetic Energy (K.E.) = 20 J<br \/>\nLet velocity of the object = v<br \/>\nK.E. = \u00bd mu<sup>2<\/sup><br \/>\n20 = \u00bd \u00d7 0.1 \u00d7 v<sup>2<\/sup><br \/>\nor v<sup>2<\/sup> = 400<br \/>\nv = 20 ms<sup>-1<\/sup><\/p>\n<p>Question 23.<br \/>\nCalculate the power of a pump which can lift 200 kg of water to store it in a tank at height of 19m in 25 seconds. (Take g = 10 ms<sup>-2<\/sup>).<br \/>\nAnswer:<br \/>\nMass of water, m =200 kg<br \/>\nHeight, h = 19 m<br \/>\nTime taken to lift the water, t = 25 s<br \/>\nPower of the pump = \\(\\frac{\\text { Workdone }}{\\text { Time taken }}\\)<br \/>\n= \\(\\frac{m \\times g \\times h}{t}=\\frac{200 \\times 10 \\times 19}{25}\\)<br \/>\n= 1520 W<\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 24.<br \/>\nCalculate the power of a pump in horse power (H.P.) which can lift 600 kg of water into the water tank at a height of 40m in 10 minutes.<br \/>\nAnswer:<br \/>\nMass of water m =600 kg<br \/>\nHeight, h =40 m<br \/>\ng = 10ms<sup>-2<\/sup><br \/>\nTime taken by pump, t = 10 min = 600 s<br \/>\nPower of the pump = \\(\\frac{\\text { Work done }}{\\text { Time taken }}=\\frac{m g}{600}\\)<br \/>\n= \\(\\frac{600 \\times 10 \\times 40}{600}\\) = 400 W<br \/>\nor = \\(\\frac {400}{746}\\)H.P = 0.52H.P. (\u2235 1H.P. = 746W)<\/p>\n<p>Question 25.<br \/>\nAn object of mass 2 kg is thrown vertically upward with an initial velocity of 20m\/ s. What will-be the potential energy at the highest point (Take g= 10 ms<sup>-2<\/sup>).<br \/>\nAnswer:<br \/>\nGiven u =20 m\/s<br \/>\ng = -10 ms<sup>-2<\/sup><br \/>\nv = 0<br \/>\nLet maximum height of the object = h<br \/>\nWe know that<br \/>\nv<sup>2<\/sup> &#8211; u<sup>2<\/sup> = 2gh<br \/>\nh = \\(\\frac{v^{2}-u^{2}}{2 g}=\\frac{0-(20)^{2}}{2 \\times(-10)}=\\frac{-400}{-20}\\) = 20 m<br \/>\nPotential energy of the object = mgh.<br \/>\n= 2 \u00d7 10 \u00d7 20 = 400 J<\/p>\n<p><span style=\"color: #0000ff;\">Long Answer Type Questions<\/span><\/p>\n<p>Question 1.<br \/>\nDefine kinetic energy and potential energy ? Give some examples in each case.<br \/>\nAnswer:<br \/>\nKinetic energy: The energy possessed by an object by virtue of its motion is known as its kinetic energy.<br \/>\nKinetic energy (E<sub>k<\/sub>) = \u00bd mv<sup>2<\/sup>.<br \/>\nKinetic energy is directly proportional to it is mass and its is directly proportional to the square of its velocity.<\/p>\n<p>Example:<\/p>\n<ol>\n<li>The energy possessed by a moving wind can move the blade of a wind-mill due to its kinetic energy.<\/li>\n<li>The flowing water also have kinetic energy which can make the water-mill to work.<\/li>\n<li>The speeding car, moving bullet, rotating wheel all have kinetic energy and can do work.<\/li>\n<\/ol>\n<p>Potential Energy: The energy posessed by a an object by virute of its position or change in configuration is known as its potential energy.<br \/>\nPotential energy =m \u00d7 g \u00d7 h<br \/>\nwhere m = mass of the object, h = height, g = acceleration due to gravity.<\/p>\n<p>Example:<\/p>\n<ol>\n<li>When a body is lifted to a certain height against the force of gravitation, the work done on the body is stored in the body in the form of potential energy.<\/li>\n<li>The water stored at a height in a dam possesses potential energy<\/li>\n<li>The energy possessed b a stretched rubber or stretched spring or compressed spring is also an example of potential energy.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nWhat is gravitational potential energy and elastic potential energy? Give examples.<br \/>\nAnswer:<br \/>\n(i) Gravitational potential energy : When a body is lifted to a certain height then die work done in raising the object from the ground to that point against gravity is stored in the body and is called gravitational potential energy.<br \/>\nThe graviational potential energy of the object = m.g.h<\/p>\n<p>The work done by the gravity depends upon the difference in height of initial position and final position and not on the path along which object is moved.<\/p>\n<p>Example: Any object lifted from the ground level or any other reference level possess potential energy.<\/p>\n<p>(ii) Elastic potential energy : The energy possesed by an object due to change in its shape is known as elastic potential energy It is mainly associated with the compression or extension of an object.<\/p>\n<p>Example :<\/p>\n<ol>\n<li>A stretched spring or rubber possesses elastic potential energy.<\/li>\n<li>A compressed spring also possesses elastic kinetic energy<\/li>\n<li>Derive an expression for the kinetic energy of an object of mass &#8216;m&#8217; moving with the velocity V.<\/li>\n<\/ol>\n<p>Or<\/p>\n<p>Derive the expression of kinetic energy (Ek = \u00bd mv<sup>2<\/sup>, where m is the mass of the object and v is the constant velocity with which the body is moving.<br \/>\nAnswer:<br \/>\nThe kinetic energy of an object is the energy possessed by the object by virtue of its motion. The kinetic energ of a body moving with a vertain velocity is equal to the work done on it to make it acquire that velocity from rest.<\/p>\n<p>Consider an object of mass &#8216;tri at rest. Letitbe displaced through a distance &#8216;S&#8217; when a constant force &#8216;F acts on its in the direction of displacement. The work done is given by<br \/>\nW = F \u00d7 S &#8230;&#8230;..(i)<br \/>\nThe work done on the object will cause a change in its velocity. Let velocity of the object changed from zero (position of rest) to v. Let &#8216;a&#8217; be the acceleration produced in the body, then<\/p>\n<p>According to Newtons second law of motion F = m \u00d7 a &#8230;&#8230;.(ii)<br \/>\nBut we know that<br \/>\nv<sup>2<\/sup> &#8211; u<sup>2<\/sup> = 2as<br \/>\nor s = \\(\\frac{v^{2}-u^{2}}{2 a}\\)<br \/>\nObject starts moving from rest, therefore u = 0<br \/>\nor s = \\(\\frac{v^{2}}{2 a}\\) &#8230;&#8230;.(iii)<br \/>\nNow substituting the value of &#8216;F&#8217; and &#8216;s&#8217; from equation (ii) and (iii) in equation (i) we get<br \/>\nW = m \u00d7 a\\(\\left(\\frac{v^{2}}{2 a}\\right)\\)<br \/>\n= \u00bd mv<sup>2<\/sup><br \/>\nNow, the work done is stored in the form of kinetic energy of the object.<br \/>\nHence, kinetic energy (E<sub>k<\/sub>) = \u00bd mv<sup>2<\/sup><\/p>\n<p><img loading=\"lazy\" class=\"alignnone\" src=\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy\" width=\"170\" height=\"17\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 4.<br \/>\nWhat is the law of conservation of energy ? Prove it when an object falls freely from a certain height under the force of gravity.<br \/>\nAnswer:<br \/>\nLaw of conservation of energy: Energy can neither be created nor destroyed but it can be changed from&#8217;one form into another form. The total energy after and before the transformation always remains constant i.e. total energy of a closed system remains constant.<\/p>\n<p>Let an object of mass &#8216;m&#8217; is dropped freely from ascertain height &#8216;h&#8217; from the ground.<br \/>\n(i) The potential energy of the body at point A = mgh<br \/>\nKinetic energy at A = 0 (Because velocity is zero at A).<br \/>\nTotal energy at point A (highest point)<br \/>\n= mgh + 0 mgh<\/p>\n<p>(ii) After a certain interval of time it reaches at B. It moves through a distance V due to force of gravitation in downward direction.<br \/>\nThe potential energy at B = m.g. (h &#8211; x)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23904\" src=\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2021\/06\/NCERT-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-14.png?resize=94%2C168&#038;ssl=1\" alt=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy 14\" width=\"94\" height=\"168\" data-recalc-dims=\"1\" \/><br \/>\nLet the velocity of the body at B = v<br \/>\nWe know that v<sup>2<\/sup> &#8211; u<sup>2<\/sup> = 2 gs<br \/>\nv<sup>2<\/sup> = 2g(x)(u = 0)<br \/>\nKinetic energ at B = \u00bd mv<sup>2<\/sup><br \/>\n= \u00bd m(2gx)<br \/>\n= mgx<\/p>\n<p>Total energy at B = Potential energ + kinetic energy<br \/>\n= m.g. (h &#8211; x) + mgx<br \/>\n= m.g.h. &#8211; mgx + mgx<br \/>\n= mgh.<\/p>\n<p>(iii) As the fall continues, the potential energy goes on changing into kinetic energy. The potential energu would decrease while kinetic energy would increase, when the object is about the reach the ground then, h &#8211; 0.<br \/>\nThe potential energy at C = mg \u00d7 h = mg \u00d7 0 = 0.<br \/>\nLet v be the velocity of the object just before reaching the ground.<br \/>\nu = 0, s = 0<br \/>\nv<sup>2<\/sup> &#8211; u<sup>2<\/sup> = 2gh<br \/>\nor v<sup>2<\/sup> &#8211; 0 = 2gh<br \/>\nKinetic energy of the object = \u00bd mv<sup>-1<\/sup><br \/>\n= \u00bd m (2gh) &#8211; mgh<br \/>\nTotal energy = mgh + 0 = mgh.<\/p>\n<p>The total energy i.e. the sum of potential energy and kinetic energy at all the three point is same. Similarly, the total energy at any point will be the same. This proves the law of conservation of energy.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>These NCERT Solutions for Class 9 Science Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams. Work and Energy NCERT Solutions for Class 9 Science Chapter 11 Class 9 Science Chapter 11 Work and Energy InText Questions and Answers Question &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-science-chapter-11\/\"> <span class=\"screen-reader-text\">NCERT Solutions for Class 9 Science Chapter 11 Work and Energy<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.8 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 9 Science Chapter 11 Work and Energy - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-science-chapter-11\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Science Chapter 11 Work and Energy - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Science Chapter 11 Work and Energy Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams. 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