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**Q.1 Â Â Look at the activities listed below. Reason out whether or not work is done in the light of yourÂ understanding of the term â€˜workâ€™.Suma is swimming in a pond.Â A donkey is carrying a load on its back.Â A wind mill is lifting water from a well.Â A green plant is carrying out photosynthesis.An engine is pulling a train.Â Food grains are getting dried in the sun.Â A sailboat is moving due to wind energy.
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Â Â Â Â Â Â (b) While carrying a load, the donkey has to apply a force in the upward direction.But, is placement of the load is in the forward direction. Since, displacement is perpendicular to force,the work done is zero.

**Q.2 Â Â ****An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground.Â The initial and the final points of the path of the object lie on the same horizontal line. What is the workÂ done by the force of gravity on the object?**

* Sol.*Â Â Â Â 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.Â Work done by gravity is given by the expression,

Â Â Â Â Â Â W = mgh

Â Â Â Â Â Â Where,

Â Â Â Â Â Â Â h = Vertical displacement = 0

Â Â Â Â Â Â Â W = mg Ã— 0 = 0 J

Â Â Â Â Â Â Â Therefore, the work done by gravity on the given object is zero joule.

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

* Sol.*Â Â Â When a bulb is connected to a battery, then the chemical energy of the battery is transferred intoÂ 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:

**Q.4 Â Â Certain force acting on a 20 kg mass changes its velocity from 5 Â to 2 .Â Calculate the work done by the force.**

* Sol.Â Â Â Â *Kinetic energy is given by the expression,

**Q.5 Â Â Â 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,Â what is the work done on the object by the gravitational force? Explain your answer.**

* Sol. Â Â *Work done by gravity depends only on the vertical displacement of the body. It does not depend upon theÂ path of the body. Therefore, work done by gravity is given by the expression,

Â Â Â Â Â Â W = mgh

Â Â Â Â Â Â Where,

Â Â Â Â Â Â Vertical displacement, h = 0

Â Â Â Â Â Â âˆ´W = mg Ã— 0 = 0

Â Â Â Â Â Â Hence, the work done by gravity on the body is zero.

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

* Sol.Â Â Â * Â No. The process does not violate the law of conservation of energy. This is because when the body fallsÂ from a height, then its potential energy changes into kinetic energy progressively. A decrease in theÂ potential energy is equal to an increase in the kinetic energy of the body. During the process, totalÂ mechanical energy of the body remains conserved. Therefore, the law of conservation of energy is notÂ violated.

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

* Sol.Â Â Â *While riding a bicycle, the muscular energy of the rider gets transferred into heat energy and kineticÂ energy of the bicycle. Heat energy heats the riderâ€™s body. Kinetic energy provides a velocity to the bicycle.Â The transformation can be shown as:Â During the transformation, the total energy remains conserved.

**Q.8 Â Â Does the transfer of energy take place when you push a huge rock with all your might and fail to move it?Â Where is the energy you spend going?**

* Sol. Â Â Â Â *When we push a huge rock, there is no transfer of muscular energy to the stationary rock. Also, there isÂ no loss of energy because muscular energy is transferred into heat energy, which causes our body toÂ become hot.

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

* Sol. Â Â *1 unit of energy is equal to 1 kilowatt hour (kWh).

Â Â Â Â Â Â 1 unit = 1 kWh

Â Â Â Â Â Â 1 kWh = 3.6 Ã— Â JÂ

Â Â Â Â Â Â Therefore, 250 units of energy = 250 Ã— 3.6 Ã—Â Â = 9 Ã—Â

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

* Sol.Â Â Â Â Â *Work is done whenever the given two conditions are satisfied:

Â Â Â Â Â Â Â (i) A force acts on the body.

Â Â Â Â Â Â Â (ii) There is a displacement of the body by the application of force in or opposite to the direction of force.Â If the direction of force is perpendicular to displacement, then the work done is zero.

When a satellite moves around the Earth, then the direction of force of gravity on the satellite is

perpendicular to its displacement. Hence, the work done on the satellite by the Ear this zero.

**Q.12 Â Â Can there be displacement of an object in the absence of any force acting on it? Think. Discuss thisÂ question with your friends and teacher.**

* Sol.Â Â Â Â Â Â *Yes. For a uniformly moving object.Â Suppose an object is moving with constant velocity. The net force acting on it is zero. But, there is aÂ displacement along the motion of the object. Hence, there can be a displacement without a force.

**Q.13 Â Â A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work orÂ not? Justify your answer.**

* Sol.Â Â Â Â Â *Work is done whenever the given two conditions are satisfied:

Â Â Â Â Â Â Â (i) A force acts on the body.

Â Â Â Â Â Â Â (ii) There is a displacement of the body by the application of force in or opposite to the direction of force.Â 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 theÂ absence of force, work done by the person on the bundle is zero.

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

*Sol.Â Â Â Â Â Â *Energy consumed by an electric heater can be obtained with the help of the expression,Where,

Â Â Â Â Â Â Â Power rating of the heater, P = 1500 W = 1.5 kW

Â Â Â Â Â Â Â Time for which the heater has operated, T = 10 h

Â Â Â Â Â Â Â Work done = Energy consumed by the heater

Â Â Â Â Â Â Â Therefore, energy consumed = Power Ã— Time

Â Â Â Â Â Â Â = 1.5 Ã— 10 = 15 kWh

Â Â Â Â Â Â Â Hence, the energy consumed by the heater in 10 h is 15 kWh.

**Q.15 Â Â Illustrate the law of conservation of energy by discussing the energy changes which occur when we drawÂ a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? WhatÂ happens to its energy eventually? Is it a violation of the law of conservation of energy?**

*Sol.Â Â Â Â *Â The law of conservation of energy states that energy can be neither created nor destroyed. It can only be Â converted from one form to another. Consider the case of an oscillating pendulum.Â When a pendulum moves from its mean position P to either of its extreme positions A or B, it risesÂ through a height h above the mean level P. At this point, the kinetic energy of the bob changes completelyÂ into potential energy. The kinetic energy becomes zero, and the bob possesses only potential energy. As itÂ moves towards point P, its potential energy decreases progressively. Accordingly, thekinetic energyÂ increases. As the bob reaches point P, its potential energy becomes zero and the bob possesses onlyÂ kinetic energy. This process is repeated as long as the pendulum oscillates.Â The bob does not oscillate forever. It comes to rest because air resistance resists its motion. TheÂ pendulum loses its kinetic energy to overcome this friction and stops after some time.Â The law of conservation of energy is not violated because the energy lost by the pendulum to overcomeÂ friction is gained by its surroundings. Hence, the total energy of the pendulum and the surroundingÂ system remain conserved.

**Q.16 Â Â An object of mass, m is moving with a constant velocity, v. How much work should be done on the object Â in order to bring the object to rest?**

*Sol.Â Â Â Â Â Â *Kinetic energy of an object of mass, m moving with a velocity, v is given by the expression,

Â Â Â Â Â Â Â Hence, 20.8 Ã— 104 J of work is required to stop the car.

**Q.18 Â Â In each of the following a force, F is acting on an object of mass, m. The direction of displacement is fromÂ west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work doneÂ by the force is negative, positive or zero.**

* Sol.*Â Â Â Â Â Work is done whenever the given two conditions are satisfied:

Â Â Â Â Â Â Â (i) A force acts on the body.

Â Â Â Â Â Â Â (ii) There is a displacement of the body by the application of force in or opposite to the direction of force.Â In this case, the direction of force acting on the block is perpendicular to theÂ displacement. Therefore,Â work done by force on the block will be zero.Â In this case, the direction of force acting on the block is in the direction of displacement. Therefore, workÂ done by force on the block will be positive.Â In this case, the direction of force acting on the block is opposite to the direction of displacement.Â Therefore, work done by force on the block will be negative.Â

**Q.19 Â Â Soni says that the acceleration in an object could be zero even when several forces are acting on it. DoÂ you agree with her? Why?**

*Sol.Â Â Â Â Â Â *Acceleration in an object could be zero even when several forces are acting on it.

This happens when all the forces cancel out each other i.e., the net force acting on the object is zero. For aÂ uniformly moving object, the net force acting on the object is zero. Hence, the acceleration of the object is Â zero. Hence, Soni is right.

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

*Sol.Â Â Â Â Â Â *Energy consumed by an electric device can be obtained with the help of the expression for power,

Â Â Â Â Â Â Â Where,

Â Â Â Â Â Â Â Power rating of the device, P = 500 W = 0.50 kW

Â Â Â Â Â Â Â Time for which the device runs, T = 10 h

Â Â Â Â Â Â Â Work done = Energy consumed by the device

Â Â Â Â Â Â Â Therefore, energy consumed = Power Ã— Time

Â Â Â Â Â Â Â = 0.50 Ã— 10 = 5 kWh

Â Â Â Â Â Â Â Hence, the energy consumed by four equal rating devices in 10 h will be 4 Ã— 5 kWh

Â Â Â Â Â Â Â = 20 kWh = 20 Units.

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

*Sol.Â Â Â Â Â Â *When an object falls freely towards the ground, its potential energy decreases and kinetic energyÂ increases. As the object touches the ground, all its potential energy gets converted into kinetic energy. AsÂ the object hits the hard ground, all its kinetic energy gets converted into heat energy and sound energy. ItÂ can also deform the ground depending upon the nature of the ground and the amount of kinetic energyÂ possessed by the object.

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