Question 3.
\nWrite an expression for the work done when a force is acting on an object in the direction of its displacement.
\nAnswer:
\nLet a constant force F act on an object. Let the object be displaced through a distance S in the direction of the force. Let W be work done. We define work to be equal to the product of the force and displacement.
\nWork done = force \u00d7 displacement W = F \u00d7 S.<\/p>\n
Question 4.
\nDefine 1 J of work.
\nAnswer:
\n1 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.<\/p>\n
Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 1. Answer:<\/p>\n Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Fill in the blanks: Question 2. Question 3. Question 4. Question 5. Answer the following questions:<\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. <\/p>\n","protected":false},"excerpt":{"rendered":" Karnataka Board\u00a0Class 9 Science Chapter 11 Work and Energy KSEEB Solutions Class 9 Science Chapter 11 Intext Questions Question 1. A force of 7N acts on an object. The displacement is, say 8m, in the direction of the force. Let us take it that they force acts on the object through the displacement. What is …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2821"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2821"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2821\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2821"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2821"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nA 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?
\nAnswer:
\nForce exerted, F = 140N.
\nDisplacement, S = 15m
\nWork done in ploughing the field
\nW = F \u00d7 S
\n= 140 \u00d7 15
\n= 2100J
\n= 2.1 \u00d7 103<\/sup> J.<\/p>\n
\nWhat is the kinetic energy of an object?
\nAnswer:
\nObjects in motion possess energy, we call this energy kinetic energy.<\/p>\n
\nWrite an expression for the kinetic energy of an object.
\nAnswer:
\nKE = \\(\\frac{1}{2}\\) mv2<\/sup>.<\/p>\n
\nThe kinetic energy of an object of mass, m moving with a velocity of 5 ms-1<\/sub> 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?
\nAnswer:
\nKinetic energy, k = \\(\\frac{1}{2}\\) mv2<\/sup>.
\nWhere m = mass of the object
\nV = velocity of the object.
\nHere mass (m) is the same in both cases (\u2235 object is same)
\n\\(\\frac{\\mathbf{k}_{1}}{\\mathbf{k}_{2}}=\\left(\\frac{\\mathbf{v}_{1}}{\\mathbf{v}_{2}}\\right)^{2}\\)
\nInitial kinetic energy k1<\/sub> = 25J
\nInitial velocity V2<\/sub> = 5 ms-1<\/sup>
\nNew kinetic Energy K2<\/sub> = ?
\nNew velocity v2<\/sub> = 3v1<\/sub> = 3 \u00d7 5 = 10 ms-1<\/sup>
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-1.png\")
\n\u2234 When velocity is increased three times its K.E. is 225J.<\/p>\n
\nWhat is power?
\nAnswer:
\nPower is defined as the rate of doing work or the rate of transfer of energy w
\nP = \\(\\frac{w}{t}\\)<\/p>\n
\nDefine 1 watt of power.
\nAnswer:
\nPower is 1 W when the rate of consumption of energy is 1 JS-1<\/sup>.<\/p>\n
\nA lamp consumes 1000 J of electrical energy in 10s. What is its power?
\nAnswer:
\nPower = 1000 J Work w = 1000J
\nTime = 10 s
\nPower of lamp, P = w\/t
\n= 1000\/10 = 100W
\n\u2234 Power of a lamp = 100 KW.<\/p>\n
\nDefine average power.
\nAnswer:
\nWe obtain average power by dividing the total energy consumed by the total time taken.<\/p>\nKSEEB Solutions for Class 9 Science Chapter 11 Textbook Exercises<\/h3>\n
\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\n
\n
\nAn object was 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?
\nAnswer:
\nWork done by the force of gravity, W = mgh.
\nWhere h = difference in height of initial and final positions of the object.
\nAccording to the question, the initial and final positions of the object lie in the same horizontal line. So h = 0.
\n\u2234 Work done W = mg \u00d7 0 = 0<\/p>\n
\nA battery lights a bulb. Describe the energy changes involved in the process.
\nAnswer:
\nIn the case given in the question, the battery has chemical energy which is converted into energy. Electric energy provided to the bulb further converted into light energy.<\/p>\n
\nA certain force acting on a 20 kg mass changes its velocity from 5 ms-1<\/sub> to 2 ms-1<\/sub>. Calculate the work done by the force.
\nAnswer:
\nMass, m = 20 kg.
\nInitial velocity, u = 5 ms-1<\/sup>
\nFinal velocity v = 2ms-1<\/sup>
\nWork done by the force = change in kinetic energy
\n= Final kinetic energy – Initial kinetic energy
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-2.png\")
<\/p>\n
\nA mass of 10kg 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.
\nAnswer:
\nWork done by gravitational force W = mgh
\nWhere h = Difference in the heights of initial and final positions of the object.
\nHere both the initial and final positions are on the same horizontal line.
\nSo there is no difference in height i.e., h = 0.
\n\u2234 work done W = mg \u00d7 0 = 0.<\/p>\n
\nThe potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
\nAnswer:
\nThe total mechanical energy remains constant \/ as P.E. of the freely falling object decreases. Its kinetic energy and the kinetic energy increases on account of an increases its velocity) the law of conservation of energy is not violated.<\/p>\n
\nWhat are the various energy transformations that occur when you are riding a bicycle?
\nAnswer:<\/p>\n\n
\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?
\nAnswer:
\nWhen we push a huge rock and fall to move it. The energy spent in doing so is absorbed by the.rock. This energy is converted into potential energy of the configuration of the rock which results in its deformation. However t his deformation is not visible on account of the huge size of the rock.<\/p>\n
\nA certain household has consumed 250 units of energy during a month. How much energy is this in Joules?
\nAnswer:
\nEnergy consumed w = 250 units
\n= 250 kwh
\n= 250 \u00d7 1000 W \u00d7 3600 S
\n= 250 \u00d7 1000 J\/S \u00d7 3600 S
\n= 9 \u00d7 108<\/sup> J.<\/p>\n
\nAn object of mass 40 kg is raised to a height of 5m above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is halfway down.
\nAnswer:
\nMass m = 40 kg.
\nheight h = 5 m.
\nPotential energy PE = mgh = 40 \u00d7 9.8 \u00d7 5 = 1960J.
\nKE at half way down = PE at halfway down = \\(\\mathrm{mg} \\frac{\\mathrm{h}}{2}\\)
\n= 40 \u00d7 9.8 \u00d7 \\(\\frac{5}{2}\\) = 980 J.<\/p>\n
\nWhat is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
\nAnswer:
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-3.png\")
\nThe satellite round the earth moves in a circular orbit. Here the force of gravity acts towards the centre of the earth and displacement of the satellite is along the tangent of the circular path that means therefore and displacement are perpendicular to each other.
\nSo, work done, W = F.S cos\u03b8
\n= F \u00d7 S cos 90\u00b0
\n= F \u00d7 S \u00d7 0
\n= 0
\nThat is, no work is done by the force of gravity.<\/p>\n
\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.
\nAnswer:
\nIf an object moves with a constant velocity (i.e, there is no acceleration) then no force acts on it. As the object is moving ie it is displaced from one position to another position.<\/p>\n
\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.
\nAnswer:
\nThe person has no movement ie his displacement is zero. So the person had done no work (\u2235 work is done only when the object is displaced).<\/p>\n
\nAn electric heater is rated 1500 W. How much energy does it use in 10 hours?
\nAnswer:
\nTime = 10 h = 10 \u00d7 60 min
\n= 10 \u00d7 60 \u00d7 60 S
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-4.png\")
\nEnergy = power \u00d7 time
\n= 1500 \u00d7 10 \u00d7 60 \u00d7 60
\n= 5.4 \u00d7 107<\/sup> J.<\/p>\n
\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 comes to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
\nAnswer:
\nWhen the pendulum oscillates in the air, the air friction opposes its motion. So some part of the kinetic energy of the pendulum is used to overcome this friction. With the passage of time, the kinetic energy of the pendulum goes on decreasing and finally becomes zero. The kinetic energy of the pendulum is transferred to the atmosphere. So energy is being transferred ie it is converted into one form to another. So here is no violation of the law of conservation of energy.<\/p>\n
\nAn 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?
\nAnswer:
\nThe work done on the object to bring the object to rest
\n= change in kinetic energy
\n= Final kinetic energy – Initial Kinetic energy
\nHere final kinetic energy is zero because the object is brought to rest.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-5.png\")
<\/p>\n
\nCalculate the work required to be done to stop a car of 1500 kg moving at the velocity of 60km\/h?
\nAnswer:
\nMass, m = 1500 kg.
\nInitial velocity, u = 60 kmh-1<\/sup>
\n= 60 \u00d7 \\(\\frac{5}{18}\\) = 16.67 ms-1<\/sup>
\nFinal velocity, v = 0
\n(\u2235 the car comes to rest)
\nWork done to stop the car = change in kinetic energy
\n= Final kinetic energy – Initial Kinetic energy
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-6.png\")
<\/p>\n
\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 diagram carefully and state whether the work done by the force is negative, positive or zero.
\n![\"KSSEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/05\/KSSEB-Solutions-for-Class-9-Science-Chapter-11-Work-and-Energy-7.png\")
\nAnswer:
\nCas i) The force and displacement are perpendicular to each other. So S = 90\u00b0
\nWork done = FScos \u03b8\u00b0
\n= FScos 90\u00b0
\n= FS \u00d7 0 = 0
\n(\u2235 cos90\u00b0 = 0)
\nCas ii) The force and displacement are in the same direction so \u03b8 = 0\u00b0
\nWork done = FScos \u03b8\u00b0
\n= FScos 90\u00b0
\n= FS \u00d7 1 = FS
\n(\u2235 cos \u03b8\u00b0 = 1)
\nThat is the work done is positive.
\nCas iii) The force and displacement in opposite direction \u03b8 = 180\u00b0
\nWorkdone = FScos \u03b8\u00b0
\n= FScos 180\u00b0
\n= FS \u00d7 -1 = FS
\n(\u2235 cos 180\u00b0 = -1)
\nThat is workdone is negative.<\/p>\n
\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?
\nAnswer:
\nYes, I agree with Soni, the acceleration of an object can be zero even when several forces are acting on it if the resultant of all the forces acting is zero.<\/p>\n
\nFind the energy in kWh consumed in 10 hours by four devices of power 500 w each.
\nAnswer:
\nTotal power p = 500w \u00d7 4 = 2000w.
\nTime, t = 10h
\nEnergy = p \u00d7 t.
\n= 2000w \u00d7 10h
\n= 2kwh \u00d7 10h = 20kwh.
\n\u2234 Energy = 20kwh.<\/p>\n
\nA freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
\nAnswer:
\nP.E. of the configuration of the body and the ground (the body may be deformed and the ground may at the place of collision).
\nThis process energy in which the kinetic energy of a freely falling body is lost in an unproductive chain of energy charges is called dissipation of energy.<\/p>\nKSEEB Solutions for Class 9 Science Chapter 11 Additional Questions and Answer<\/h3>\n
\nQuestion 1.
\nWork done = ___________
\nAnswer:
\nforce, displacement<\/p>\n
\nAn object having the capability to do work is said to possess __________
\nAnswer:
\nenergy.<\/p>\n
\n__________ is the energy possessed by an object due to its motion.
\nAnswer:
\nKinetic energy<\/p>\n
\nP.E
\nAnswer:
\nmgh.<\/p>\n
\nLarger unit of energy is called
\nAnswer:
\nkilojoule.<\/p>\n
\nCan there be displacement of an object in the absence of any force acting on it?
\nAnswer :
\nIn the absence of any force on the object i.e., F = 0, ma, (as F = ma) since m \u2260 0 and a = 0. In such a case the object is either at rest or in a state of uniform motion in a straight line in the latter case there is a displacement . of the object without any force acting on it.<\/p>\n
\nHow does a bullet pierce a target?
\nAnswer :
\nA bullet moves with large velocity and as such possesses a lot of kinetic energy the work in piercing the target is derived from the kinetic energy of the bullet.<\/p>\n
\nWhy do some engines require fuels like petrol and diesel?
\nAnswer :
\nInternal combustion heat engines use the chemical energy of fossil fuel (petrol and diesel) for their operation. These engines first convert the chemical energy of the fuels into heat energy. Which is later on converted into mechanical energy.<\/p>\n
\nCalculate the work done by a body. By the force of 5 N makes to move through a distance of 12 m.
\nAnswer :
\nW = F x s = 5 x 12 = 60 joules<\/p>\n
\nWhen force 6 N applied on a wall, the wall remains in the same position, calculate the work done.
\nAnswer :
\nW = F x s
\nW – 6 x s = 6 x 0 = 0 Joules No work is done on the body<\/p>\n
\nElectricity is the most convenient form of energy. Why?
\nAnswer :
\nIt can be converted into other forms of energy easily.
\nIt can be produced by different means. It is ecofriendly<\/p>\n
\nCalculate the power experienced by a source that can do work of 50 joules in 5 seconds.
\nAnswer :
\nPower = \\($\\frac{\\text { Workdone }}{\\text { Time taken }}$\\)
\n\\(=\\frac{50}{5} \\frac{\\text { Joules }}{\\text { Second }}\\) = 10 J\/s = 10 watts<\/p>\nKSEEB Solutions for Class 9 Science<\/a><\/h5>\n