**Q **1. What kind of energy is possessed by the following?

(a) A stone kept on the roof top.

(b) Water stored in the reservoir of a dam.

(c) A stretched rubber band.

**Solution**

(a) Potential energy

(b) Potential energy

(c) Potential energy

**Q **2. Explain how wind mills are used to generate electricity?

**Solution**

In a wind mill, kinetic energy of wind is used to move the blades of fan which in turn move the coil of generator thereby converting kinetic energy to electrical energy. Thus, in the above example there is only transformation of energy from one form to another such that the total energy of the system remains conserved.

**Q **3. Explain the principle behind the wind up toys.

**Solution**

Wind up toys have a spring inside it which gets wound up with the help of winding keys. So when we wind-up the spring of a toy car by using a winding key, then some work is done by us. As a result, the spring gets coiled up more tightly. The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy

(i.e potential energy due to change in shape). When we release the winding key the spring tries to attain its original shape. During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

**Q **4. Write the form of energy possessed by the body in the following situations:

(a) a coconut falling from tree

(b) an object raised to a certain height

(c) blowing wind

(d) a child driving a bicycle on road

**Solution**

(a) Kinetic Energy and potential energy

(b) Potential Energy

(c) Kinetic Energy

(d) Kinetic Energy

**Q **5. An engine pumps 40 kg of water per second. If water comes out with a velocity of 3m/s, what is the power of the engine?

**Solution**

**Q **6. A porter lifts a luggage of 15 kg from the ground and puts it on the head 1.7 m above the ground. Find the work done by the porter on the luggage. Take g = 10 m/s^{2}.

**Solution**

Work done by the porter on the luggage = Potential energy of the luggage W = m x g x h = 15 x 10 x 1.7 = 255 J

**Q **7. An electric bulb is rated 15 watts. What does it mean? What is the energy consumed in joules if it is used for 10 minutes?

**Solution**

If power of an electric bulb is 15 W, it consumes 15 joules of energy per second. Energy consumed by the bulb in 10 minutes = 15W x 600s = 9000 joules

**Q **8. A ball is dropped from a height of 10m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10ms^{-2})

**Solution**

The potential energy at height of 10 m = mgh = m x 10 x 10 = (100m) J On striking the ground, the potential energy gets converted into kinetic energy. Kinetic energy = (100m) J Energy is reduced by 40%, so the remaining energy = 60% x 100m = (60m) J The height upto which the ball can bounce back is given by h’, such that mgh’ = 60m m x 10 x h’ = 60 x m h’ = 60/10 = 6 m

**Q **9. A man whose mass is 70 kg climbs from top of a hill station ‘A’ to another hill station ‘B’. The height of hill station A is 1200 m above the sea level and that of B is 1800 m above the sea level. Calculate the work done in going from A to B (g=10m/s^{2})

**Solution**

PE at A = mgh_{1} = 70 x 10 x 1200 = 84×10^{4} J PE at B = mgh_{2} = 70 x 10 x 1800 = 126×10^{4} J Work done = 126×10^{4 }– 84×10^{4} = 42×10^{4} J

**Q **10. Is it possible that an object is in the state of accelerated motion due to an external force acting on it but no work is being done by the force? Explain with an example.

**Solution**

Yes, it is possible. A body in uniform circular motion is an example of accelerated motion. Consider the motion of the Earth around the Sun. The Earth is constantly moving in a circular path in a direction perpendicular to the gravitational pull of the Sun. So, the work done by the gravitational force is zero. Thus, the work done can be zero for an accelerated body.

**Q **11. Give examples of atmospheric phenomena that have large kinetic energy?

**Solution**

Hurricanes, tornadoes and jet stream are some of the atmospheric phenomena having large kinetic energy on account of their high speeds.

**Q **12. (a) “An arrow and the stretched string on the bow is said to possess energy”. Comment.

(b) A bag of wheat weighs 50 kg. Calculate the height to which it should be raised so that its potential energy is 5000J.

**Solution**

(a) (i) The stretched string possesses potential energy. When the bow string is released, arrow flies off the bow. The potential energy stored in the bow is used in the form of kinetic energy of the arrow.

(b) P.E = mgh 5000 = 50 x 10 x h h = 5000/500 = 10 m

**Q **13. When is work said to be done against the force of gravity? State and define SI unit of work.

**Solution**

When the body is raised to a height, the work is said to be done against the force of gravity. S.I. Unit of work is joule. One joule of work is said to be done when 1N force moves a body through a distance of 1m.

**Q **14. What is the work done by frictional force on an object when dragged along a rough surface?

**Solution**

Frictional force is always opposing the relative motion of the body. When a body is dragged along a rough surface, the frictional force will be acting in the direction opposite to the displacement. The angle between the frictional force and the displacement of the body will be 180^{o}. Thus, the work done by the frictional force will be negative.

**Q **15. If the velocity of an object is doubled, what is the change in its kinetic energy?

**Solution**

Kinetic energy is proportional to the square of velocity. Hence, kinetic energy will become 4 times when velocity is doubled.

**Q **16. When an object moves on a circular path, what is the work done?

**Solution**

Work done is zero, because displacement (along tangent) is always perpendicular to the direction of force (along centre).

**Q **17. State the energy transformation in the following :

(i) heat engine

(ii) electric motor

**Solution**

(i) In a heat engine heat energy gets converted into mechanical energy

(ii) In an electric motor electrical energy gets converted into mechanical energy

**Q **18. What kind of energy does a flying bird or a flying airplane has during their course of action?

**Solution**

A flying bird or a flying airplane has kinetic energy as well as potential energy during their course of action. The flying bird and airplane have kinetic energy because they are moving. They have potential energy at any point of time in their flight due to their height above ground at that particular point of time. Thus, a body can have both potential and kinetic energy at the same time.

**Q **19. When we burst a cracker, is it correct to say that the chemical energy of crackers get used up or are destroyed as we are left with just ashes after bursting cracker?

**Solution**

Its wrong to say that energy gets used up or destroyed. According to law of conservation of energy, energy can neither be created nor destroyed it just converts from one form to other. On bursting a cracker the chemical energy of crackers transform into heat and light energy.

**Q **20. Given below are a few situations. Study them and state in which of the cases, work is said to be done. Give reason for your answer.

(i) A person pushing a huge rock but the rock does not move. (ii) A bullock pulling a cart upto 1 km on road.

(iii) A girl pulling a trolley for about 2 m distance.

(iv) A person standing with a heavy bag on his head.

**Solution**

(i) No work is done, as there is zero displacement

(ii) Yes, the work is said to be done as the cart moves through a distance.

(iii) Yes, the work is said to be done as the trolley moves through a distance.

(iv) No work is done, as there is no displacement of the person or the bag.

**Q **21. Two masses m and 2m are dropped from heights h and 2h. On reaching ground, which will have a greater kinetic energy and why?

**Solution**

Body with mass 2m will have a greater kinetic energy, because it has greater height and mass, and hence greater potential energy. This potential energy is converted to kinetic energy as the mass falls down. Hence, mass 2m has greater kinetic energy.

**Q **22. A mass m is dropped from a height h. At half way to the ground, what is the energy?

**Solution**

At maximum height h, potential energy is maximum and kinetic energy is zero. PE = mgh Now, according to conservation of energy, the energy at all the points is same along the line of motion. Thus, at midway also, the total energy is mgh.

**Q **23. Which formula do we use to calculate the work done when a body is moved against gravity ?

**Solution**

When the work is done against gravity, then the amount of work done is equal to the weight of the body times the vertical distance through which it is lifted. Work done = Weight of body x vertical distance against gravity W = mg x h= mgh Where, W – Work m – Mass g – acceleration due to gravity h – vertical distance

**Q **24. An object of mass 20 kg is dropped from a height ‘h’ metres as shown in the tables below. (g = 10 m/s^{2}) Table A Height from which an object is dropped ‘h’ (m) Velocity of an object falling from the height ‘v^{2}’ (m/s^{2}) 10 0 2 80 Table B Height from which an object is dropped ‘h’ (m) Velocity of an object falling from the height ‘v^{2}’ (m/s^{2}) 10 0 2 40 Which of the above tables has wrong data in terms of the law of conservation of energy of an object falling from height ‘h’ metres?

**Solution**

Table A: Height from which an object is dropped ‘h’ (m) Velocity of an object falling from the height ‘v^{2}’ (m/s^{2}) Potential energy mgh (J) Kinetic energy ½mv^{2} (J) TE = PE + KE 10 0 20 × 10 ×10 = 2000 ½ × 20 × 0 = 0 2000 2 80 20 × 10 ×2 = 400 ½ × 20 × 80 = 800 1200 Table B: Height from which an object is dropped ‘h’ (m) Velocity of an object falling from the height ‘v^{2}’ (m/s^{2}) Potential energy mgh (J) Kinetic energy ½mv^{2} (J) TE = PE + KE 10 0 20 × 10 × 10 = 2000 ½ × 20 × 0 = 0 2000 2 40 20 × 10 × 2 = 400 ½ × 20 × 40 = 400 800 According to the law of conservation of energy, the sum of the potential energy and kinetic energy of the object should remain the same at every point during its fall.

**Q **25. Define the SI unit of Work.

**Solution**

The SI unit of work is Joule (J) Work = Force x Distance 1 Joule = 1 newton x 1 metre Therefore, we define 1 Joule as the amount of work done when a force of one newton moves a body through a distance of 1 metre in the direction of force applied.

**Q **26. A microwave uses 450 KJ of electric energy in 5 minutes. Calculate its power rating.

**Solution**

Electric energy consumed, E = 450 KJ = 450 x 1000 J = 450000 J Time taken, t = 5 minutes = 5 x 60 sec = 300 sec Power, P = E/t = 450000/300 = 1500 watts

**Q **27. Two bodies of same mass start from rest and move with velocities of v and 2v respectively. Find the ratio of their kinetic energies.

**Solution**

For body1: K.E._{1} = ½ mv^{2} For body 2: K.E._{2} = ½ m(2v)^{2} K.E._{2} = 4 (½ mv^{2}) Thus, we have K.E._{1}: K.E._{2} = 1:4

**Q **28. State the factors on which work done depends.

**Solution**

Work done by a force depends on: 1. Magnitude of the force applied. 2. The distance through which the body moves in the direction of force.

**Q **29. Derive an expression for the work done when a force is acting on an object in the direction of its displacement. A pair of bullocks exerts a force of 150 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

**Solution**

When 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:Work done = Force × DisplacementW = F × s Work done by the bullocks is given by the expression:Work done = Force × DisplacementW = F × sWhere,Applied force, F = 150 NDisplacement, s = 15 mW = 150 × 15 = 2250 J Hence, 2250 J of work is done in ploughing the length of the field.

**Q **30. Home electricity bills come in units called Kilowatt-hour. How many joules is equal to 1 kilowatt hour. Why kilowatt hours are used instead of joules in the bill?

**Solution**

1 Kilowatt hour = 36, 00,000 joules = 3.6 x 10^{6} J One kilowatt hour is the amount of electrical energy consumed when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour. Kilowatt hour is used for measuring electricity consumed, as joule represents a very small quantity of energy and therefore it is inconvenient to use joules where large quantities of energy is involved.