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

1. (c) P = 2mE  P  m if E = constant . So 3 1 9 1 2 1 2 1 = = = m m P P . 2. (c) If the kinetic energy would become half, then Potential energy = 2 1 (Initial kinetic energy)  [490 ] 2 1 mgh =  [490 ] 2 1 2  9.8  h =  h = 12.5 m 3. (c) ) ˆ 4 ˆ v = (3i + j   v 3 4 5 m/s 2 2 = + = . So kinetic energy = mv 0.3 (5) 3.75 J 2 1 2 1 2 2 =   = 4. (b) (8 4 400 ) 2 = − = − x − x + dx d dx dU F For the equilibrium condition = − = 0 dx dU F  16 x − 4 = 0  x = 4 / 16  x = 0.25 m . 5. (d) Elastic potential energy of a spring 2 2 1 U = kx  2 U  x So 2 1 2 1 2         = x x U U  2 2 2 10         = cm cm U U  U2 = 25 U 6. (c) Work done to stretch the spring from 1 x to 2 x ( ) 2 1 2 1 2 2 W = k x − x 5 10 75 10 18.75 N.m 2 1 5 10 [(10 10 ) (5 10 ) ] 2 1 3 4 3 2 2 2 2 =     = =   −  − − − 7. . (c) Potential energy of spring k F U 2 2 =  1 2 2 1 k k U U = 2 : 1 1500 3000 = = [If F = constant] 8. (b) When spring is gradually lowered to it's equilibrium position kx = mg  k mg x = . When spring is allowed to fall suddenly it oscillates about it's mean position Let y is the amplitude of vibration then at lower extreme, by the conservation of energy  ky = mgy 2 2 1  k mg y 2 = = 2x. 9. (d) If the spring is stretched by length x, then work done by two equal masses = 2 2 1 kx So work done by each mass on the spring = 2 4 1 kx  Work done by spring on each mass = 2 4 1 − kx . 10. (a) As the force is repulsive in nature between two protons. Therefore potential energy of the system increases. 11. (b) When the rod is lying on a horizontal table, its potential energy = 0 But when we make its stand vertical its centre of mass rises upto high 2 l . So it's potential energy 2 mgl =  Work done = charge in potential energy 2 0 2 l mgl = mg − = . 12. (a) Centre of mass of a stick lies at the mid point and when the stick is displaced through an angle 60o it rises upto height ‘h’ from the initial position. From the figure cos 2 2 l l h = − (1 cos ) 2 = −  l Hence the increment in potential energy of the stick = mgh J l mg o (1 cos 60 ) 1 2 1 (1 cos ) 0.4 10 2 = −  =   − = 13. (a) Potential energy is a relative term but the difference in potential energy is absolute term. If reference level is P Q h  l/2 l/2 cos
fixed once then change in potential energy are same always. 14. (d) As 1/3 part of the chain is hanging from the edge of the table. So by substituting n = 3 in standard expression 2 2n MgL W = 2(3) 18 2 MgL MgL = = . 15. (d) 2 2n MgL W = 2.5 . 2 (4) 4 10 2 2 = J    = 16. (b)       = − 2 1 1 n v gL         =  − 2 (2) 1 10 2 1 = 15 = 3.87 ≃ 4 m/s (approx.) 17. (d) For the given condition potential energy of the two masses will convert into heat and temperature of water will increase W = JQ  2m  g  h = J(mw S t)  2 5 10 10 4.2(2 10 ) 3    =   t  t C C o o 0.119 0.12 8.4 10 1000 3 = =   = . 18. (a) By the conservation of energy Total energy at point A = Total energy at point B  2 1 2 2 1 mgh = mgh + mv  2 2 1 9.8  5 = 9.8  2 + v  58.8 2 v =  v = 7.6 m / s 19. (b) Potential energy of block at starting point = Kinetic energy at point P = Work done against friction in traveling a distance s from point P.  mgh =  mgs  m h s 7.5 0.2 1.5 = = =  i.e. block come to rest at the mid point between P and Q. 20. (c) The engine has to do work against resistive force R as well as car is moving with acceleration a. Power = Force  velocity = (R+ma)v. 21. (c) Force = = (V  ) dt d v dt dm v [A l] dt d = v  dt dl = v A 2 =  Av Power = F  v = Av  v 2  3 =  Av  3 P  v . 22. (d) t mgh P = = time work done  P  m i.e. To obtain twice water from the same pipe in the same time, the power of motor has to be increased to 2 times. 23. (b) 5 60 2.05 10 [25 5 ] 2 1 t m(v v ) 2 1 time Increase in kinetic energy time Work done Power 2 6 2 2 1 2 2     − = − = = = 2.05 10 watt 2.05 MW 6 =  = 24. (a) 100 10 100 Work done Power = = =   t mgh t 10 watt 100 kW 5 = =       = (given ) sec As 100 kg t m 25. (c) P = F⃗. v⃗ = (10î+ 10ĵ+ 20k̂). (5î− 3ĵ+ 6k̂) = 50 − 30 + 120 = 140J ⥂ −⥂ s −1 26. (b) Power = Force  velocity = (Resistive force + Accelerating force)  velocity  30 10 (750 ) 30 3  = + ma   ma = 1000 − 750  2 0.2 1250 250 − a = = ms . 27. (d) Weight of a bus = mass  g = 100 × 100kg × 10m ⥂/⥂ s 2 = 10 5N Total friction force = 10% of weight = 104 N  Power = Force  velocity = 10 4N × 72km/h = 10 4 × 20watt = 2 × 10 5watt = 200kW. 28. (a) Power (P) = t mgh or t m P   1 2 2 1 2 1 t t m m P P = h2 = 2m h1 = 5m A B
11 15 33 45 11 9 3 5  = =            = (g and h are constants) 29. (a) MW t mg h 10 1 2000 10 (550 50) time work done Power =   − =  = = But the system is 80% efficient  Power output = 10  80% = 8 MW. 30. (b) dt mdv F =  F dt = mdv  t m F v = Now P = F  v t m F = F  m F t 2 = If force and mass are constants then P  t. 31. (b) As the ball rebounds with same velocity therefore change in velocity = 2u and the mass colliding with the surface per second = nm Force experienced by the surface dt dv F = m  F = 2 mnu. 32. (a) 1 2 2 2 1 1 2 1 2 1 2 m m m u u m m m m v + +         + − = Substituting m1 = 0, 1 1 2 v = −u + 2u  6 2(4) v1 = − + = 2m/s i.e. the lighter particle will move in original direction with the speed of 2 m/s. 33. (c) Loss of kinetic energy of the colliding body 2 2 2 1 2 1 2 3 1 1 2 2 1 1        = −      + − = −         + − = −  m m m m m m m m K K K K K 9 8 9 1 1  =       = −  Loss of kinetic energy is 9 8 of its initial kinetic energy. 34. (d) Initial velocities of balls are +V and – 2V respectively and we know that for given condition velocities get interchanged after collision. So the velocities of two balls after collision are – 2V and V respectively. 35. (a) From the standard equation u M m M m u m m m m v       + − =         + − = 1 1 2 1 2 1 . 36. (b) By the law of conservation of linear momentum 1 1 2 2 1 1 2 2 m u + m u = m v + m v  4000 2 400  72 + 4000  9 = 400  (−18) +  v  v 2 = 18 km / h . 37. (c) Final velocity of the target 1 2 1 1 2 1 2 2 1 2 2 m m m u u m m m m v + +         + − = As initially target is at rest so by substituting u2 = 0 we get M m u M m Mu v + = + = 1 2 2 2 . 38. (b) Total kinetic energy of the ball will transfer to the bob of simple pendulum. Let it rises to height ‘h’ by the law of conservation of energy. mv = mgh 2 2 1  g v h 2 2 = 39. (c) 1 2 2 2 1 1 2 1 2 1 2 m m m u u m m m m v + +         + − = 1 1 2 1 2 u m m m m         + − = [As u2 = 0 and         = 1.5 1 1 u v given]  1 1 2 1 1 2 1.5 u m m u m m         + − =  1.5( ) m1 + m2 = m1 − m2  5 2 1 = m m . 40. (b) Only this condition satisfies the law of conservation of linear momentum. v h m m m m M u1 = 6 m/s u2 = 4 m/s
41. (c) Kinetic energy retained by projectile 2 1 2 1 2         + − =  m m m m K K  K E 2 8 2 8 2       + −  = = E 0.36 E 25 9 = = . 42. (a) Fraction of kinetic energy retained by projectile 2 1 2 1 2         + − =  m m m m K K Mass of neutron (m1) = 1 and Mass of atom (m2) = A  2 1 1       + − =  A A K K or 2 1 1       + − A A . 43. (b) Kinetic energy transferred to stationary target (carbon nucleus)                 + − = −  2 1 2 1 2 1 m m m m K K               + − = −  2 1 12 1 12 1 K K       = − 169 121 1 169 48 =  (0.6 ) 0.17 . 169 48 K =  MeV = MeV 44. (d) Initial momentum = mv + nm(kv) and final momentum = 0 + nm V By the conservation of momentum, mv + nm(kv) = 0 + nm V  v + nkv = nV  nV = (1 + nk)v  n nk v V (1 + ) = 45. (b) Initial horizontal momentum of the system = m  9 Final horizontal momentum of the system = 2mv cos 300 According to law of conservation of momentum, m  9 = 2mv cos 300  v = 5.2 m/s 46. (c) Let second ball moves with momentum P making an angle  from the horizontal (as shown in the figure). By the conservation of horizontal momentum 1  0.4 = P cos ......(i) By the conservation of vertical momentum 0.3 = P sin ......(ii) From (i) and (ii) we get P = 0.5 kg-m/s 47. (d) In this condition the final resultant momentum makes some angle with x-axis. Which is not possible because initial momentum is along the x-axis and according to law of conservation of momentum initial and final momentum should be equal in magnitude and direction both. 48. (d) From the figure (I) it is clear that before collision initial momentum of the system = 0 After the collision, A becomes stationary, B retraces its path with velocity v. Let C moves with velocity V making an angle  from the horizontal. As the initial momentum of the system is zero, therefore horizontal and vertical momentum after the collision should also be equal to zero. From figure (II) Horizontal momentum cos + cos 30 = 0 o v  v .....(i) Vertical momentum sin − sin 30 = 0 o v  v .....(ii) By solving (i) and (ii) we get o  = −30 and V = v i.e. the C will move with velocity v in the direction of B. 49. (d) Horizontal momentum and vertical momentum both should remain conserve before and after collision. This is possible only for the (d) option. 50. (c) Loss of K.E. in inelastic collision mv mv mv C B 120o 120o 120o A V B C v  A 30o Before collision 1 kg 0.4 m/s  1 kg 0.3 m/s P → m 9 m/s m At rest v v 30o m m 30o