Title 07. Work, Energy and Power Easy Ans.pdf ✅

1. (b) Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular. W = F.s = Fscos = Fscos(90) = 0 2. (a) Work = Force × Displacement (length) If unit of force and length be increased by four times then the unit of energy will increase by 16 times. 3. (c) No displacement is there. 4. (d) Stopping distance 2 S  u . If the speed is doubled then the stopping distance will be four times. 5. (c) W = Fscos  2 1 50 25 cos = = = Fs W    = 60 6. (b) Work done = Force × displacement = Weight of the book × Height of the book shelf 7. (b) Work done does not depend on time. 8. (c) ) ˆ ˆ ).(2 ˆ 3 ˆ W = F.s = (5i + j i − j = 10 – 3 = 7 J 9. (a) 2 3 8t 3t dt dx v = = − +  v 3 m / s 0 = and v 19 m/s 4 = ( ) 2 1 2 0 2 4 W = m v −v (According to work energy theorem) 0.03 (19 3 ) 5.28 J 2 1 2 2 =   − = 10. (d) As the body moves in the direction of force therefore work done by gravitational force will be positive. W = Fs = mgh = 10  9.8 10 = 980 J 11. (d) 12. (b) W = mg sin  s 2 10 sin 15 10 3 =     = 5.17 kJ 13. (d) ) ˆ 5 ˆ ).(6 ˆ 4 ˆ 6 ˆ W = F.s = (5i + j − k i + k = 30 − 20 = 10 units 14. (b) 2 2 1 W = Fs = F at       = + 2 2 1 from s ut at              = 2 2 1 t m F W F J m F t 6 5 30 25 2 15 25 (1) 2 2 2 2 = =   = = 15. (b) Work done on the body = K.E. gained by the body Fscos = 1  Fcos N s 2.5 0.4 1 1 = = = 16. (b) Work done = mgh = 10 9.8 1 = 98 J 17. (b) 18. (d) 4 2 t s =  dt t ds 2 = N t dt d dt md s F ma 3 4 6 2 2 2 2 2 =         = = = Now   J t d t t W F d s (2) (0) 3 4 3 2 2 3 2 3 2 2 2 0 2 2 0 2 0 = − =         = = =   19. (d) Net force on body 4 3 5N 2 2 = + =  2 a = F / m = 5 /10 = 1 / 2m/s Kinetic energy = 2 2 1 mv m(at) 125 J 2 1 2 = = 20. (d) m g u s 10 2 0.5 10 10 10 2 2 =    = =  21. (d) W = F.s i j i j) 9 16 25 J ˆ 4 ˆ ).(3 ˆ 4 ˆ = (3 + + = + = 22. (d) Total mass = (50 + 20) = 70 kg Total height = 20 × 0.25 = 5m  Work done = mgh = 70 × 9.8 × 5 = 3430 J 23. (d) ) 0 ˆ ˆ 3 ˆ ).(2 ˆ 3 ˆ 2 ˆ W = F.s = (6i + j − k i − j + xk = 12 − 6 − 3x = 0  x = 2 24. (a) ) ˆ 15 ˆ 11 ˆ )(11 ˆ 3 ˆ ˆ W = F.(r2 −r1 ) = (4i + j + k i + j + k W = 44 +11 + 45 = 100 Joule 25. (c) W i cj k i j k) 6 J ˆ 3 ˆ 2 ˆ ).( 4 ˆ 2 ˆ ˆ = (3 + + − + + = W = −12 + 2c +6 = 6  c = 6 26. (a) Both part will have numerically equal momentum and lighter part will have more velocity. 27. (d) Watt and Horsepower are the unit of power 28. (b) Work = Force × Displacement If force and displacement both are doubled then work would be four times. 29. (d) W = FS cos = 10  4  cos60 = 20 Joule 30. (a) ) ˆ 3 ˆ 5 ˆ ).(6 ˆ 4 ˆ W = F.s = (5i + j i − j + k = 30 − 20 = 10 J 31. (b) Fraction of length of the chain hanging from the table
10 3 200 1 60 = = = cm cm n  3 10 n = Work done in pulling the chain on the table 2 2n mgL W = 3.6 J 2 (10/3) 4 10 2 2 =    = 32. (c) When a force of constant magnitude which is perpendicular to the velocity of particle acts on a particle, work done is zero and hence change in kinetic energy is zero. 33. (a) The ball rebounds with the same speed. So change in it's Kinetic energy will be zero i.e. work done by the ball on the wall is zero. 34. (b) W F r i j k i j) 10 3 7 J ˆ ˆ ).(2 ˆ 2 ˆ 3 ˆ = . = (5 + + − = − = 35. (a) K.E. acquired by the body = work done on the body K E = mv = Fs 2 2 1 . . i.e. it does not depend upon the mass of the body although velocity depends upon the mass m v 2 1  [If F and s are constant] 36. (d) ) ˆ 6 ˆ 0 ˆ ).(3 ˆ 0 ˆ 5 ˆ W = F.s = (4i + j + k i + j + k = 4  3 units 37. (a) As surface is smooth so work done against friction is zero. Also the displacement and force of gravity are perpendicular so work done against gravity is zero. 38. (c) Opposing force in vertical pulling = mg But opposing force on an inclined plane is mg sin, which is less than mg. 39. (c) Velocity of fall is independent of the mass of the falling body. 40. (a) Work done = F.s i j i j) 6 3 2 1 18 2 16 J ˆ ˆ ) (3 ˆ 2 ˆ = (6 +  − =  −  = − = 41. (c) When the ball is released from the top of tower then ratio of distances covered by the ball in first, second and third second : : = 1 : 3 : 5 : hI hII hIII [because h  (2n −1)] n  Ratio of work done mghI mghII mghIII : : = 1:3:5 42. (b) 2 1 0 2 0 0 2 1 2 . 1 1 1 Cx x W F dx Cx dx C x x x =         = =   43. (c) When the block moves vertically downward with acceleration 4 g then tension in the cord Mg g T M g 4 3 4  =      = − Work done by the cord = F.s = Fscos = Td cos(180 ) d Mg        = − 4 3 4 3 d = − Mg 44. (c) k F W 2 2 = If both springs are stretched by same force then k W 1  As 1 2 k  k therefore W1  W2 i.e. more work is done in case of second spring. 45. (a) P.E. 10[(0.25) (0.20) ] 2 1 ( ) 2 1 2 2 2 1 2 = k x2 − x =  − = 5 0.45 0.05 = 0.1 J 46. (a) kS 10 J 2 1 2 = (given in the problem)   2 2 2 2 1 (2 ) ( ) 3 2 1 k S − S =  kS = 3 × 10 = 30 J 47. (c) k F U 2 2 =  1 2 2 1 k k U U = (if force are same)  1 2 1500 3000 2 1 = = U U 48. (d) Here N m x F k 10 / 1 10 10 4 3 =  = = − W k x 10 (40 10 ) 8 J 2 1 2 1 2 4 3 2 = =    = − 49. (d)   = = − + 5 0 2 5 0 W Fdx (7 2x 3x ) dx = 5 0 2 3 [7x − x + x ] = 35 – 25 + 125 = 135 J 50. (d) 3 3 t S =  dS t dt 2 = 2 3 2 2 2 2 2 / 3 t m s t dt d dt d S a =         = = Now work done by the force   = = 2 0 2 0 W F.dS ma.dS    2 0 2 3 2t t dt  = 2 0 3 6t dt =   2 0 4 2 3 t = 24 J 51. (b) 2 2 1 W = kx If both wires are stretched through same distance then W  k . As 2 2 1 k = k so W2 = 2W1 L/n T d
52. (b) 2 2 2 1 2 1 mv = kx  m k m x v 0.1 1000 0.1 = = 10 = 53. (c) Force constant of a spring 2 2 10 1 10 −   = = = x mg x F k  k = 500 N/m Increment in the length = 60 – 50 = 10 cm U k x 500 (10 10 ) 2.5 J 2 1 2 1 2 2 2 = =  = − 54. (b) 2 2 2 4 1 2 2 800 (15 5 ) 10 2 1 ( ) 2 1 − W = k x − x =   −  = 8 J 55. (c) 2 2 1 100 = kx (given) [(2 ) ] 2 1 ( ) 2 1 2 2 2 1 2 2 W = k x − x = k x − x k x 3 100 300 J 2 1 3 2  =  =      =  56. (d) 2 2 1 U = kx if x becomes 5 times then energy will become 25 times i.e. 4  25 = 100 J 57. (c) 2 3 2 2 4 1 2 2 5 10 (10 5 ) 10 2 1 ( ) 2 1 − W = k x − x =   −  = 18.75 J 58. (a) The kinetic energy of mass is converted into potential energy of a spring 2 2 2 1 2 1 mv = kx  m k mv x 0.15 50 0.5 (1.5) 2 2 =  = = 59. (a) This condition is applicable for simple harmonic motion. As particle moves from mean position to extreme position its potential energy increases according to expression 2 2 1 U = kx and accordingly kinetic energy decreases. 60. (c) Potential energy 2 2 1 U = kx  2 U  x [if k = constant] If elongation made 4 times then potential energy will become 16 times. 61. (b) 62. (d) 2 U  x  25 0.02 0.1 2 2 1 2 1 2  =      =         = x x U U  U2 = 25U 63. (a) If x is the extension produced in spring. F = kx  k F x = = cm k mg 4.9 4000 20 9.8 =  = 64. (a) k T k F U 2 2 2 2 = = 65. (b) 2 U = A − Bx  Bx dx dU F = − = 2  F  x 66. (d) Condition for stable equilibrium = − = 0 dx dU F  0 12 6 =       − − x b x a dx d  12 6 0 13 7 − + = − − ax bx  13 7 12 6 x b x a =  2 6 x b a =  6 2 b a x = 67. (d) Friction is a non-conservative force. 68. (c) P = 2mE  P  m (if E = const.)  2 1 2 1 m m P P = 69. (c) Work in raising a box = (weight of the box) × (height by which it is raised) 70. (a) m P E 2 2 = if P = constant then m E 1  71. (a) Body at rest may possess potential energy. 72. (b) Due to theory of relativity. 73. (d) m E 2 P 2 =  2 E  P i.e. if P is increased n times then E will increase n 2 times. 74. (c) 75. (c) P.E. of bob at point A = mgl This amount of energy will be converted into kinetic energy  K.E. of bob at point B = mgl and as the collision between bob and block (of same mass) is elastic so after collision bob will come to rest and total Kinetic energy will be transferred to block. So kinetic energy of block = mgl 76. (b) According to conservation of momentum Momentum of tank = Momentum of shell 125000 × vtank = 25 × 1000  vtank = 0.2 ft/sec. A B m m m
77. (d) As the initial momentum of bomb was zero, therefore after explosion two parts should possess numerically equal momentum i.e. A A B B m v = m v  4 v A = 8 6  v m s A = 12 /  Kinetic energy of other mass A, = 2 2 1 A A m v = 2 4 (12) 2 1   = 288 J. 78. (c) Let the thickness of one plank is s if bullet enters with velocity u then it leaves with velocity u u v u 20 19 20  =      = − from v u 2as 2 2 = −  u u 2as 20 19 2 2  = −       as u 39 2 400 2 = Now if the n planks are arranged just to stop the bullet then again from v u 2as 2 2 = − 0 u 2ans 2 = −  39 400 2 2 = = as u n  n =10.25 As the planks are more than 10 so we can consider n = 11 79. (b) Let h is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy  mgh = 2 490  2 490 2  9.8  h =  h = 12.5m. 80. (c) P = 2mE . If E are same then P  m  2 1 4 1 2 1 2 1 = = = m m P P 81. (a) Let initial kinetic energy, E1 = E Final kinetic energy, E2 = E + 300 % of E = 4E As P  E  2 4 1 2 1 2 = = = E E E E P P  P2 = 2P1  P2 = P1 + 100 % of P1 i.e. Momentum will increase by 100%. 82. (b) P = 2mE if E are equal then P  m i.e. heavier body will possess greater momentum. 83. (c) Let P1 = P , P2 = P1 + 50% of P1 = 2 3 2 1 1 1 P P P + = 2 E  P  4 3 /2 9 2 1 1 2 1 2 1 2 =         =         = P P P P E E  E2 = 2.25 E 1 25 1 = E +1. E  E2 = E1 +125 % of E1 i.e. kinetic energy will increase by 125%. 84. (b) As the body splits into two equal parts due to internal explosion therefore momentum of system remains conserved i.e. 8 2 4 1 4 2  = v + v  v1 + v2 = 4 ...(i) By the law of conservation of energy Initial kinetic energy + Energy released due to explosion = Final kinetic energy of the system  2 2 2 1 2 4 2 1 4 2 1 8 (2) 16 2 1   + = v + v  16 2 2 2 v1 + v = ...(ii) By solving eq. (i) and (ii) we get v1 = 4 and v2 = 0 i.e. one part comes to rest and other moves in the same direction as that of original body. 85. (d) P = 2mE  P  E i.e. if kinetic energy of a particle is doubled the its momentum will becomes 2 times. 86. (b) Potential energy = mgh Potential energy is maximum when h is maximum 87. (c) If particle is projected vertically upward with velocity of 2m/s then it returns with the same velocity. So its kinetic energy mv 2 (2) 4 J 2 1 2 1 2 2 = =   = 88. (b) 89. (c) m P E 2 2 = if bodies possess equal linear momenta then m E 1  i.e. 1 2 2 1 m m E E = 90. (d) 2 s  u i.e. if speed becomes double then stopping distance will become four times i.e. 8 4 = 32m 91. (c) 2 s  u i.e. if speed becomes three times then distance needed for stopping will be nine times. 4kg 8kg A B vA vB 4kg v1 4kg v2 After explosion 8kg 2m/s Before explosion