1. A force of 7N acts us an object. The displacement is, say 8m, in the direction of the force. Let us take it that the force acts on the object through the displacement what is the work done in this case?
                             Displacement,    d   =  8m
                                             Force, F =  7N
                                                  Work =  Force ×displacement
                                              =7N×8m =  56NM

2. Define one J of work
                               The amount of work done when a force of 1N moves a body through a distance of 1m in the direction of force is equal to 1 Joule.

3. A pair of bullocks exerts a force of 140N on a plough. The field being ploughed is 15m long. How much work is done in ploughing the length of the field?
                  Here,
                    Force, F  = 40N
          Displacement, S = 15m
                  Work done = F× S
                      =140×15 = 2100 J

4. The Kinetic energy of an object of mass m moving with a velocity of 5ms-1 is 25. What will be its kinetic energy when it is Velocity is doubled? What will be its Kinetic energy when its Velocity is increased three times?
                 Mass of the object      =  m
            Velocity of the object, V   =  5ms^-1
                                         K.E   =  25J
                                         K.E   =  ¹/₂ ×m×(5)²
                                         Or M = 2Kg
                          i) When V = 2V=10ms^-1
                                           K.E =  ½  MV²
                               =1/2×2×225 =  225J
5. Look at the activities listed below Reason out whether or hot work is done in the light of your understanding of the term’ work’:

a) Suma is swimming in a pond
b) A donkey is carrying a load on his back
c) A wind mill is lifting water from a well
d) A green plant is carrying out photosynthesis
e) Food grains are getting dried in the sun
f) A sail boat is moving due to wind energy

          a) Work is done , because the displacement of swimmer takes place in the direction of applied force.
          b) If the donkey is not moving, no work is done as the displacement of load does not take place in the direction of     applied force.
         c) Work is done, as the displacement takes place in the direction of force.
         d) No work is done, as the displacement does not takes place
         e) No work is done , as no displacement takes place
         f) Work is done , as displacement takes place in the direction of applied force.

6. A battery lights a bulb. Describe the energy changes in the process
         i)  Within the electric cell of the battery , the chemical energy changes in to the electric energy.
        ii) The electric energy on flowing through the filament of the bulb, first changes into heat energy and then into the light energy.

7. A mass of 10Kg is at a point A on a label. It is moved to a point B. It the line joining A and B is horizontal, what is the work done on the object by the force of gravity?
         Work done by the gravitational force   =   mgh
                                                   Since h   = 0, (because both points A and B are at the same height)
                                               Work done  = 0

8. The potential energy of a freely, falling object decreases progressively. Does this violate the law of conservation of energy? Why?
                       No, Because the potential energy is converted in to kinetic energy during free fall. Therefore, total energy remains conserved and law of conservation of energy is not violated.

9. What are the various energy transformations that occur when you are riding a bicycle?
                      The muscular energy of the cyclist is converted into kinetic energy of the bicycle. A small amount of heart is also produced between the types and the road due force of friction.

10. Does the transfer of energy take place when you push a huge rock with all you right and foil to move it? Where is the energy you spend going?
                                    There is no transfer of energy between you and the rock. But you do work to expand and contract your muscles and to circulate blood faster than the normal ratio. Thus energy is spent on yourself.

11.  A certain household has consumed 250units of energy during a month. How much energy is this in joules?
                           Energy = 250 units =  250kwh
                                                        =  250× 1000W× 3600S
                                                        =  9×108 J

12. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer?
                                  The work done by the force of gravity on the satellite is Zero. It is because , the force of gravity acts at right angles to the direction of motion of the satellite, and hence, no displacement it caused in the direction of applied force. The force of gravity only changes the direction of motion of the satellite.

13. Can there be displacement of an object in the absence of any force acting on it ? Think discuss this question with your friends and teacher?
                           The answer is both, yes and No.
                           Yes in a sense that when an object moves in deep space from one point to another point in a straight line the displacement takes place, without the application of force.
No, in a sense that force can be Zero for displacement on the surface of earth. Some force is essential.

14. A person holds a bundle of hay over his head for 30 minutes and gets tired has he done some work or not? Justify your answer?
                             The person does not work . It is because, no displacement takes in the direction of applied force as the force acts in the vertically upward direction.

15.  An Electric heater is rated 1500w. How much energy does it use in 10 hours?
              Energy used by heater
                                           =  Power × Time
                                           = 1500W× 10h
                                          

16. Illustrate the law of conservation of energy by discussing the energy changes, which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually comes to rest ? what happens to its energy eventually? is a violation of the law of conservation of energy?
                         When the pendulum bob is pulled (say towards left), the energy supplied is store in it is the form of potential energy account of its potential energy changes in kinetic energy, such that in mean position, it have maximum kinetic and zero potential energy. As the pendulum moves further towards extreme right, its kinetic energy changes into potential energy, such that at the extreme position, it has maximum potential energy and zero kinetic energy. On moving from this extreme position to mean position, its potential energy again changes to kinetic energy. This illustrates the law of conservation of energy.
                           The bob eventually comes to rest, because during each oscillation a part of the energy possessed by it transferred to air and in overcoming friction at the point of suspension. Thus gradually the energy of the pendulum is dissipated in air.
                          The law of conservation of energy is not violated because the energy merely changes its form and is not destroyed.

17 . An object of mass ‘m’ is moving with a constant velocity ‘V’. How much work should be done on the object in order to bring the object to rest?
                             The work done to bring the object to rest = kinetic energy of the object = ½ mv2.

18. Calculate the work required to be done to stop car of 1500 kg moving at a velocity of 60 km/h.
                                   Mass of car = 1500kg
                               Velocity of car = 60 kmh-1
                                                    
                            ∴ work done       = K.E. of the car
                                                     = ½ mc²
                                                     = ½ x 1500 kg x (16.67 ms^-1)²
                                                     = ½ x 1500 kg x 277.89 m3s-2
                                                     = 208416.68J.

19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
                          Yes, we do agree, when the number of act on a body, such that they constitute balanced forces then net force acting on the body is zero in such a situation no acceleration acts on the object.

20. Find the energy in kwh consumed in 10 hours by four devices of power 500 w each.
            Total power of 4 devices
                           = 4 x 500 w = 2000 w
                           

21. Can any object have mechanical energy even it its momentum is zero? Explain.
                      Yes, mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero. Is zero. Hence, there is no kinetic energy but the object may possess potential energy.

22. Can any object have momentum even if its mechanical energy is zero? Explain.

                     No, since mechanical energy is zero, there is no potential energy and no kinetic energy kinetic energy being zero, velocity is zero hence, there will be no momentum.

23. The power or a motor pump is 2 kw. How much water per minute the pump can raise to a height of 10m? (Given g = 10 k s ^-2)
                               

24. The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high he can jump on the planet A?
                                  Since, weight of the person on planet A is half that on the earth, acceleration due to gravity there, will be ½ that on the earth. Hence he can jump double the height with the same muscular force.
The potential energy of the person will remain the same on the earth and planet A.

                      If g₁  = g then  g₂  = ¹/₂  g₁ h₁  = 0.4

                          or  h₂    =  0.4 ×  2  =  0. 8 m

25. The velocity of a body moving in straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.
                        V² = U² = 2 a.s
                  
                                 F = ma
We can write work done (w) by this force F as
                               
         = ½ mv² – ½ mu² = (K.E) – (K.E)

26. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain if with an example.
                                    Yes, it is possible, if an object is moving in a circular path. Because force is always acting perpendicular to the direction o displacement.

27. A ball is dropped from a height of 10m. If the energy of the ball reduces by 40% after striking the grand, how much high can the ball bounce back ? (g = 10 ms-2).
                                                Mgh = m × 10 × 10   = 100 mJ.
Energy is reduced by 40 % then the remaining energy is 60 MJ.
                                         

28. If an electric iron of 1200 w is used for 30 minutes every day, find electrical energy consumed in the month of April.
                                        
                                      E  = Power × time × days
                                          = 1.2 × . 5 × 30 = 18 kwh

29. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?
                                    P₁  = m₁ v₂, p₂   = m₂ v₂,
                             But P₁   = P₂ or m₁V₁ = M₂ V₂
                                          = If m₁ < m₂, then V₁ V-₂
   (K.E)₁ = ½ m₁V_-1² (K.E)² = ½ M₂V_^-2²
   (K.E)₁ = ½ (m1V1) v1       = ½ P1V_^-1
   (K.E)₁ = ½ (M₂V₂)V₂        = ½ P₂V₂
                            

                                 But V_^-1  > V_^-2
                 Therefore, (K>E)₁  > (K.E.)₂

30. An automobile engine propels a 1000 kg car along a leveled road at a speed of 36 km. Find the power if the opposing frictional force of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the sometime. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.
                M(A) = M (B) = 1000 kg, V = 36 km/h = 10 m/s
                                                 Frictional force = 100 N
Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force.
                                 
                                                 = 100N x 10 m/s = 1000W
         After collision
         M_Au_A + M_B U_b   =  M_A U_A + M_BU_B
 1000 x 10+ 1000 x 0  =  1000 x 0+ 1000 x VB
                         V^-_B   =  10 mS^-₁.

31. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms-1 by applying a force. The trolley comes to rest after traversing a distance of 16m: (a) How much work is done on the trolley (b) How much work is done by the girl?
                                               U = 4mS-1, V = 0, S = 16 m
                                               
                                          Force  = ma = 40 x (-¹/₂) = -20N
                                   Work done on the trolley
                                                             = 20 N x 16 m = 320 J
                                                  Work done by the girl = OJ.

32. Four men lift a 250 kg or lowering if . (a) How much work is done by the men in lifting the box? (b) How much work do they do in just holding it? (c) why do they get tired while holditing it? (g=10 ms-2).
                                     (a) F  = 250 kg x g
                                              = 2500 N
                                           S = 1m
                                          W = F.S
                                              = 2500 NM = 2500J
(b) Zero; as the box does not move at all, while holding if.
(c) In order to hold the box, men are applying a force which a opposite and equal to the gravitational force acting on the box. While applying the force, muscular effort is involved. So they feel tired.

33. What is power? How do you differentiate kilowatt from kilowatt hour? The Jog falls in Karnataka state are nearly 20 m high. 2000 tones of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 ms-2)
Power is the rate of doing work. Kilowatt is the unit of power and kilo watt hour is the unit of energy.
                                             H = 20m,
                                 And mass = 2000 x 103 kg = 2x 106 kg
                                       

34. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100w, be able to lift at constant speed of 1 ms-1 vertically? (g = 10ms-2)
                                                

35. Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 w for each kg. What force does it exert in moving the car at a speed of 25 ms^-1?
                              One watt is the power of an agent which does work at rate of 1JS-1.
                                            1 kilowatt = 1000 J S^-1
                                                       
                                                           = 3.75 x 10³N
                                                 Force = 3750 N.

36.  Certain Force acting on a 20 kg mass changes its velocity from 5ms^-1  to 2ms^-1. Caculate the work done by the force.           

Work done by the force when velocity is 5 ms-1

             = 1/2 mv2

             = 1/2 × 20kg × (5ms-1)2 = 250J

 Work done by the force when velocity is 2 ms^-1

             = 1/2 mv²

             = 1/2 × 20kg × (2 ms^-1) = 40 J

therefore Resultant work done by the force

                 = 250 J - 40 J - 40 J = 210J.

37. An Object of mass 40 kg is raised to a of 5 m above the ground. What is its fall, find energy? if the object is allowed down.  Take  g  =  10 ms^-2.

                      i)  Potential energy  = mgh

                                                     = 40 kg × 10 ms^-2 × 5m

                                                     = 2000 J

                    ii) According to the law of conservation of energy:

                                              K.E. When it is half - way   down  =

                                             P.E. When it is half- way down

38. An electroc heater is rated 1500 w. How much energy does it use in to hours?

                         Energy  used by heater

                                = Power  × Time

                               = 1500 × 10h

40.Compare the power at which each of the following is moving upwards against the force of gravity? (given g=10ms^-2).
i) a butterfly of mass 1.0g that flies upward at a rate of 0.5ms^-1
ii) a 250g squirrel climbing up on a tree at a rate of 0.5ms-1
 i)  Power   =  mg ×  velocity

              g   = 10mS^-2
  

Hence, the power with which the squirrel is climbing is much higher than that of a butterfly flying.

41. What type of energy does a flying airplane possess with respect to the earth?
A flying airplane possesses both the potential energy ( height above ground) and kinetic energy (due to motion) with respect to the earth.

42. A labourer, with a load on his head, is going along a horizontal road. Is he doing some work?
No, the lobourer is not doing any work because force and direction of motion are perpendicular to each other.

43. Give one example each of:
a)positive work done by a force
b) Negative work done by a force
a) positive work done by a force:
When we hit a ball, the ball starts moving in the direction of force; it is an example of positive work done by a force.
b) Negative work done by a force:
The frictional force always acts in the direction opposite to the movement of the body thus , it is an example of negative work done by a force.

44. The kinetic energy of an object of mass ‘m’ moving with velocity of 5m/s is 25J What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
                   From question, mass of the object  =m,
                   Velocity of the object(v) = 5ms-1
                   Kinetic energy (k.E) of the object =25J
                   K.E = ½ mv^-2=25=m(5)2
                       ⇒   M= 2kg
When velocity of the object is double , ie, 2V

45. What change should be affected in the velocity of a body to maintain the same kinetic energy if its mass is increased four times?
                 K.E =1/2mv2. If mass is increased four times, velocity must be reduced to half to maintain the same K.E.

46. Calculate the work required to be done to stop a car of 1500kg moving at a velocity of 60 km/h?
                          Mass of the car, m=1500kg
           Velocity of car, V = 60Kmh-1=16.67ms-1
The work required to be done to stop the car
                            W  = Ek =¹/₂ mv²
                                 = ¹/₂ ×1500 × (16.67)²
                                 = 208416.67 J = 208.4KJ

47. What is the amount of work done in the following cases? Justify your answer by giving the appropriate reason.
a) by an electron revolving in a circular orbit of radius ‘r’ around a nucleus
b) by the force of gravity, when a stone of mass ‘m’ is dropped from the top of a multi-storyed building of height ‘h’
       a) work done by an electron revolving in a circular orbit of radius r around a nucleus is zero because the direction of force is perpendicular in the direction of displacement.
        b) Work done = Decrease in potential energy = mgh

48. Find the expression for the gravitational potential energy of a body of mass ‘m’ at a height ‘h’
If g is the acceleration due to gravity, then force acting on the body,
                                                   F = mg                               …….(i)
Now amount of work done in lifting the body through a height h,
                                                  N  = F × Displacement
                                                      = mg × h = mgh
But the total amount of work done in lifting the stone is equal to the energy spent .
                          Thus, energy spent = mgh
This energy spent in lifting the body is stored within the body in the form of potential energy
                      Thus, potential energy = mgh

49. Identify the energy transformation in the following;
a) steam engine b) solar cell
(a) (i) The chemical energy of the coil changes in to the heat energy
(ii) The heat energy changes in to the kinetic energy of the steam.
(iii) The kinetic energy of the steam changes in to the mechanical energy of the engine.
(b)(i) The heat energy of the sun changes in to the chemical energy of the battery
(ii) The chemical energy of the battery changes in to electric energy
(iii) The electric energy changes into heat and light energy.

50. A boy lifts a suitcase of 20kg from the ground to a height of 1.2m. calculate the work done by him on the suitcase (given : g=10ms-2)
                             Mass, M  = 10kg
             Distance (height), h  = 1.2m,
there fore Work done ,      W = Force × Distance
                                            = mg × h
                                            = 20 × 10 × 1.2 = 240 J.

51. An electric bulb of 100 w works for 4 hrs a day. Calculate the units of energy consumed in 15 days.
             Power of electric bulb =100 w  = 0.1 kw,
                                              Time (t) = 4h,
                  Energy consumed in 1 day = p × t
                                            = 0.1 × 2 = .04kwh,
Therefore energy consumed in 15 days = 15 × 0.4= 6 kwh

52. Define work state two factors on which the magnitude of work depends.
                                                     When a force acts on a body, the amount of work done by the force is equal to the product of force and the distance moved in the direction of force.
Magnitude of work depends on :
Force
Displacement of the body in the direction of force.

53. A body of 5kg initially at rest is subjected to a force of 20N Calculate the kinetic energy acquired by the body at the end of 10 seconds.
 Here
      M= 5Kg, u = 0, F= 29 N ,  t = 10s
Acceleration,
                   a = F/m = ²⁰/₅ =4 ms^-2
Velocity after 10 second = V = u + at = 0 + 4 × 10= 40ms^-1
Thus, Kinetic energy = ½ mv-²
                               = 1/2× 5× (40)2 = 4000J.

54. The potential energy of a freely falling object decreases progressively What happens
(i) Kinetic energy (ii) total mechanical energy?
State the law on which your answer is based
b) A house hold consumes 1kwh of energy per day How much energy is this in joules?
            a) (i) The potential energy of a freely falling object decreases progressively , because a part of the potential energy changes in to kinetic energy continuously . Thus kinetic energy increases progressively.
ii)Total mechanical energy = potential energy + Kinetic energy = Constant
This relation is based on the law of conservation of energy
b) 1kwh    = 1000w× 1h
               = 1000w × 60× 60s
               = 3600000 Joules
               = 3.6× 106 Joules
               = 3.6 m Joules.

55. What is meant by term power? A boy pulls a bucket of water of mass 10kg from a 5 m deep well in 10 seconds . Calculate the power developed by the boy (g =10 m s^-2).
The rate of doing work is called power mass of water (m) = 10kg
Displacement (D) =5m
therefore  Force exerted by the boy (F) =mg
                               = 10×10=100N
therefore  Work done by the boy (w) = F× D
                = 100 × 5 = 500 J
Time for doing work (t) =10s
therefore Power of the boy (p)
                          

56. Define the term potential energy . write an expression for potential energy of an object of mass ‘m’ at a height ‘h’ above the ground what is the potential energy possessed by an object of mass 1 kg kept at a height of 1m (g =10ms^-2)
Potential energy : The energy possessed by a body on account of its position or configuration is called potential energy.
                  Potential energy  = mgh
                              Here, m  = 1 kg
                                         g = 10m s-2
                   Potential energy = mgh
                                            = 1× 1 × 10 = 10 J.

57. Two women shanty and kamala each of mass 50 kg and 60 kg respectively climb up through a height of 10m. Shanti takes 20 s while kamala takes 40 s to reach . Calculate the difference in the power expanded by shanthi and kamala (Assuming g =10mg^-2)
                             

58. (a) state the law of conservation of energy
(b) Name the commercial unit of energy
( c) Water stored in Bhakra Dam is used to Produce electric power . Explain various types of energy transformation taking place in the process.
                 (a) The energy in a system can neither be created not destroyed. If may be transformed from one form to another, but total energy of the system remains constant.
                 (b) Commercial unit of electrical energy is kilowatt hour (kwh)
                  (c) As the water falls from a height , the potential energy of water stored in the dam is converted in to a kinetic energy of running water . Which get converted in to rotational kinetic energy of the turbine that ultimately changes into electrical energy of the generator.

59. a) Define power
b) In an office a tube light of 40 w, a fan of 75 w and a cooler of 150w have been installed . If all these appliances are used for 8 hours a day . calculate the energy consumed per day in commercial unit of energy.
                       a) power is defined as the rate of doing work
                                      P = ^w/_t
                                  b) E = (40 + 75+ 150) w × 8h per day
                                         
                                         = 2.12 kwh per day.

60. a) The potential energy of a freely falling object decreases progressively . What happens to its :
i) Kinetic energy
ii) total mechanical energy ? state the law on which your answer is based:
(b) A house hold consumes 1 kwh of energy per day how much energy is this in joules ?
a) The potential energy of a freely falling object decreases progressively since its height decreases continuously . similarly its kinetic energy increases since its velocity increases continuously . Thus its total mechanical energy remains constant, which is based on the law of conservation of energy. According to this law, the energy in a system can neither be created nor destroyed . It may be transformed from one form to another , but total energy of the system remains constant
b) Energy consumed
                = 1kwh =1kw× 1h
                = 1000w× 1× 60 × 60s
                
                = 3600000J =3.6× 10⁶ J.

61. a) State and define SI unit of power
b) A person carrying 10 bricks each of man 2.5 kg on his head moves to a height 20meters in 50 seconds calculate the power spent in carrying bricks of the person
(g =10m/S²
           (a) S.I unit of power is watt (W)
                Amount of work done by 1 J of energy in 1 second is called 1 watt
                Total mass (m) =10× 2.5 kg = 25kg
                Time (t) = 50s
                g = 10ms-1
                Forced exerted by person = mg
                        = 12 kg × 10ms-2= 250N
                Work done = Force × Displacement
                                 = 250 N × 20m = 5000J

62. a) Derive the formula of kinetic energy of an object of mass ‘m’ moving with a uniform velocity ‘V^-’
b) A force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2ms-1 , calculate the work done by the force
a) Let U and V be the initial and final velocities and also a and S be the uniform acceleration and displacement respectively then from equation of motion.
                    V² =
                     
  Work done by the force , f is
                    W= F× S
            
           ⇒  W = 1/0 m
When the body is starting from the rest position , then U = 0
           T hus W= ½ m
                    W=  ½ mv2
But work done is equal to the change in kinetic energy of an object.
So, the kinetic energy acquired by a body of mass, m and moving with a uniform velocity
V is given by Ek =1/2 mv²
b) Given m= 20kg, u =5ms-1
and V  = 2ms^-1
       w = ½ m
          = 1/2 × 20
          = -10 × 21 =-210 J
(-ve sign indicates that force opposes the motion of the body)

63. a) State the law of conservation of energy
b) Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum balb to one side and allow it to oscillate.
c)Why does a saw become warmwhen it is used to cut a log of wood?
     (a) law of conservation of energy
                                 According to this law, energy can only be transformed from one form to another form It can neither be created nor destroyed . The total energy before and after the transformation always remains constant.
b) At mean position ie, at X potential energy P.E is zero . when it is displaced from X to Y it gains P.E. if Y is at a distance h vertically above ‘X’ the bob has a P.E of mgh When it is left and it returns from Y to say Y it loses some P.E and has a corresponding gain of K.E on reaching X its whole P.E at Y gets converted into K.E at X.
   i.e,   =  1/2  mv²  = mgh   ⇒  v-  = √2gh
Where 'v' is the speed of the bob at X.

Due to this speed, the bob overshoots X and reaches Z and at Z its energy is partly kinetic and partly potential but the sum of K.E and P.E is the same as K.E at X and P.E at Y When it reaches Z whole of its energy become P.E
The bob comes to rest after same time because of the resistance offered by air to its motion and work done against friction at the point of oscillation O
There is no violation of the law of conservation of energy . the energy initially possessed by the bob is mostly converted to heat energy due to resistance of air and friction of energy is also converted into sound energy. However , the law of conservation of energy is not violated at all.
C) A saw become warm when it is used to cut a log of wood because it possess kinetic energy due to its motion . This kinetic energy is converted in to heat energy due to friction at the point of contact on the log of wood.

64. a) Derive the formula of kinetic energy of an object of mass ‘m’ moving with a uniform velocity ‘V’
b) A force acting on a 20 kg mass changes its velocity from 5m-1 to 2m5-1 calculate the work done by the force.
             Let a body of mass m, whose initial velocity U = 0, after covering a distances attains a velocity V- . If a is the acceleration then  V^{-2  =  u}2  ^{+  2as}

  ⇒  V^-2   =   O²   +  2as

  ⇒   a   =  v²/2s

therefore  Accelerating force acting on the body = mass × Acceleration
             = m × v ²/₂ S
Now work done by the accelerating force in covering the distance
S= mv²/₂ s × s = ½ mv2
But work done by the accelerating force in covering the distance
S= Kinetic energy of the body
Thus kinetic energy of the body =1/2 mv2
b) m=20kg
u=5ms-1
            v =  2ms^-1
work done by the force = change in kinetic energy
         =  ½ mv2-1/2 mu2
         =  1/2×20×22-1/2×20×52
         = 40-250 = -210J
-ve sgn shows that force is acting in a direction opposite to the displacement of the body.

65. a) define potential energy Give two examples for it
b) Define power state its S.I unit what is the power of a lamp which consumes 600 J of energy is one minute?
a) The potential energy possessed by a body is the energy present in it due to its position or configuration,
Potential energy = mghTwo examples of potential energy
An object lying on some height from the ground level
An arrow and the stretched string on the bow
b) The rate of doing work or rate of transfer of energy is called power
                power =  ^work/_^Time
S.I unit of power is watt
               Here, t   = 1 minute 60S
                      W  = 600J
                       P  = w/t
                           = 600/60 = 10W

66. State the energy transformation taking place in the following cases
a) When brakes are applied to a speeding vehicle.
b) When an arrow is released from a stretched bow.
c) When some one speaks in front of a microphone
d) When a tuning for is hit against rubber pad
e) When the coil of a motor moves in magnetic field.
              a) When brakes are applied to a speeding vehicle then kinetic energy of the vehicle is converted into heat energy.
              b) When an arrow is released from a stretched bow, then potential energy of the bow is converted in to kinetic energy of the arrow.
              c) When some one speaks in front of a microphone then sound energy is converted in to electrical energy.
              d) When tuning fork is hit against a rubber pad then kinetic energy is converted in to sound energy
              e) When the coil of a motor moves in magnetic field then mechanical energy converted into electric energy.

67. Two identical pointed objects made from iron and wood . They allowed to fall on a heap of sand from the same height . The iron object penetrates more in sand than the wooden object. Which of the objects has more potential energy?
Since iron object penetrates the sand more than the wooden object, so work done by the iron object is more than the work done by the wooden object since , potential energy of the object = work done by the object, so potential energy of iron object is more than the potential energy of the wooden block.

68. A girl having mass of 35 kg sits on a trolley of mass 5 kg . the trolling is given an initial velocity of 4ms-1 by applying a force . The trolley comes to rest after traversing a distance of 16m (a) How much work is done on the trolley? (b) How much work is done by the girl?
                              U  = 4ms-1
                               V = 0,
                              S  = 16m
                   

                                Force = ma=40×
  Work done on the trolley
                                          = 20N ×16m = 320J
          Work done by the girl = 0J

69. Moon goes round the earth in circular orbit write the gravitational force between them and also prove that the work done is zero in the process of maintaining the circular path.

               Me = mass of earth , mm= mass of moon,
               ∂    = distance between moon an earth
Gravitational force acts towards earth as experienced by moon and the displacement is tangential in the circular path since work done is F d cos and work done will be zero in maintaining the circular path.

70. A road going up a steep hill is always winding and has a smaller slope . Give reason.
                              It is difficult for a vehicle to climb up steep hill because work is done against gravity and k.E. gained by a vehicle decreases . The K.E of vehicle is converted in toP.E , While going up a steep hill. Hence, road for going up a steep hill is made winding so that vehicle can gain P.E sufficient to run up a steep hill.
The slope is made smaller, otherwise K.E of vehicle will become zero and vehicle will start rolling back.

71. A bus of mass 10,000kg is moving with avelocity of 60kmh-1 calculate the work done to stop this bus.
                     M  = 10000kg
                      V  = 60kmh^-1
                          = 60× ⁵/₁₈ ms^-1
                          = ³⁰⁰/₁₈ = 16.67 ms^-1
                       V = 0
         Work done = ¹/₂ mv² -1/2 mv²
                          = 0-¹/₂ × 10000kg ×(16.67ms-1)²
                          = -1389444.5 J

72. Does a book placed at the top of a table possess potential energy? Which has more potential energy , a book placed on table or a book placed at the top of an almirah placed in the room? Why?
                Yes, a book placed on the table possesses some gravitational potential energy because some work has been done on the book. in order to raise it from floor to top of table
The book placed at the top of an almirah has more potential energy because now the height of book is more than that at the table top.
73. You lift a heavily packed carton of mass m in vertically upward direction through a height h what is the work done (i) by you on the carton, (ii) by force of gravity on the carton?
   i) Work done by me is positive having a value W= mgh . It is because I am applying a force (f=mg) in vertically upward direction on the carton to hold it and displacement is also in the same direction.
   ii) Work done by force of gravity on carton is W =-mgh . Because force of gravity is vertically downwards but motion is vertically upward.

74. Does the work done by gravity on an object so as to take it to a heoght h, depends upon the path followed ? Explain giving an example
                 The work done by force of gravity on an object depends only on the difference in vertical heights of the initial and final positions of the object . In no case it depends on the path along which the object is moved
Figure shows two different levels A and B and the difference in their height levels is h. Then work done by gravity for taking an object of mass m from level A to level B is W=-mgh In four possible paths (a) ,(b), (c) and (d) having different path lengths have been shown but work done along all these paths is same. As an example consider a person going to top of a building from ground level . work done is same whether the person goes there either through a staircase or a ramp or climbing along a pipe etc.

75. Identify the kind of energy transformation that takes place when you start your car by applying the key in it.
When we want to start our car by applying key, the electric circuit of car battery is completed. In the process chemical energy stored in car battery is transformed into electric energy.
                          The electric energy so generated cause spark in the spark plug. In the process electrical energy is transformed in heat and light energy
Due to spark taking place fuel begins to ignite . In the process chemical energy of fuel is transformed in to heat energy.
The hot gases push the pison of car engine and heat energy is transformed into mechanical energy and the car starts.

76. State in each of the following cases, if work is done /not done and why?
a) A girl climbing a staircase
b) A man standing and holding a briefcase in hand
c) A porter carrying a heavy load and going down the stairs.
d) A boy preparing for examination
e) A planet going around the sun
a) When a girl is climbing a staircase clearly work is done against gravity, which the applied forces is equal to the weight of the girl and is responsible for the certain dis-placement.
b)In this cause there is no displacement of the briefcase in the direction of applied force and hence no work is done.
c) A porter carrying a heavy load and going down the stairs, clearly does work, because there is a certain displacement in the direction of the heavy load.
d) No, work is done as mental work is not taken as work in physics.
e) When a planet goes around the sun in a circular path, then the force of gravity acts on it and is directed towards the centre of the circular path. Thus force of gravity and the direction of motion of the planet are perpendicular to each other . so work done =F scos 900 =0 . Thus is this case there is no work done .