Tiêu đề 06. FRICTION EASY Ans.pdf ✅

1. (c) 2. (d) 0.1 10 1 100 9.8 98 = =  = = = mg F R F  3. (c) Here applied horizontal force F acts as normal reaction. For holding the block Force of friction = Weight of block f = W   R = W   F = W   W F = As   1  F  W 4. (b) 5. (c) 6. (c) Fl = sR = 0.4 mg = 0.4 10 = 4N i.e. minimum 4N force is required to start the motion of a body. But applied force is only 3N. So the block will not move. 7. (a) For limiting condition A C B m m m +  =  + mC = 10 5 0.2  2 + 0.2mC = 5  mC = 15kg 8. (a) 9. (d) Ball and bearing produce rolling motion for which force of friction is low. Lubrication and polishing reduce roughness of surface. 10. (c) For given condition we can apply direct formula l l         + = 1 1   11. (c) Sliding friction is greater than rolling friction. 12. (b) N W F 49 0.2 1 9.8 =  = =  13. (a) 20% 0.25 1 5 0.25 1 '  = =      + =         + = l l l l   of l. 14. (a) m k g m m m B B A B 2 10  =  0.2 =  = 15. (d) Work done by friction can be positive, negative and zero depending upon the situation. 16. (c) Lenght of chain lying on the table Lenght of chain hanging from the table  = L l l − = 17. (b) Surfaces always slide over each other. 18. (a) Coefficient of friction 20 9.8 75 75  = = = R mg Fl  s = 0.38 19. (c) AB BG F = f + f =  ABma g +  BG(mA + mB )g = 0.2100 10 +0.3(300 )10 = 200 + 900 = 1100 N 20. (c) 21. (b)  = tan (Angle of repose) = tan 60 = 1.732 22. (a) Applied force = 2.5 N Limiting friction = mg = 0.4  2 9.8 = 7.84 N For the given condition applied force is very smaller than limiting friction.  Static friction on a body = Applied force = 2.5 N 23. (c) Sand is used to increase the friction. 24. (a) F = R = 0.3  250 = 75 N 25. (b) For the given condition, Static friction = Applied force = Weight of body = 2 10 = 20 N 26. (a)  W F =  W = F = 0.2 10 = 2N 27. (d) A B s m m  =  2 0.2 mB =  mB = 0.4 kg 28. (a) m k g m m m B B A B s 2 10  =  0.2 =  = f W R F B A fBG F fAB Ground
29. (d) Length of the chain lying on the table Lenght of the chain hanging from the table  s = 2 1 2 / 3 / 3 / 3 / 3 = = − = l l l l l 30. (d) 31. (a) 32. (b) 33. (d) In the given condition the required centripetal force is provided by frictional force between the road and tyre. mg R mv =  2  v =  Rg 34. (a) Retarding force F = ma =  R =  mg  a = g Now from equation of motion v u 2as 2 2 = − 0 u 2as 2  = −  g u a u s 2 2 2 2 = =  g v 2 2 0 = 35. (d) Net force = Applied force – Friction force ma = 24 −  mg = 24 − 0.4 5 9.8 = 24 −19.6  2 0.88 / 5 4.4 a = = m s 36. (a) Work done = Force × Displacement =  mg (v  t) W = (0.2) 2  9.8  2  5 joule Heat generated Q J W = 4.2 0.2  2  9.8  2  5 = = 9.33 cal 37. (c) For given condition 2 1 m s   2 2 2 1 1 2 300 200       =         = m m s s  s s 16 m 9 4 36 9 4 2 = 1  =  = 38. (a) There is no friction between the body B and surface of the table. If the body B is pulled with force F then F = (mA + mB )a Due to this force upper body A will feel the pseudo force in a backward direction. f = m A  a But due to friction between A and B, body will not move. The body A will start moving when pseudo force is more than friction force. i.e. for slipping, mAa =  mAg  a =  g 39. (d) Limiting friction = sR = smg = 0.5  60 10 = 300 N Kinetic friction = kR = kmg = 0.4 60 10 = 240 N Force applied on the body = 300 N and if the body is moving then, Net accelerating force =Applied force – Kinetic friction  ma = 300 − 240 = 60  2 1 / 60 60 a = = m s 40. (a) 20 0.5 10 100 2 =  =  = = g v v g r r   41. (b) 42. (b) m g u S 25 2 0.2 10 (10) 2 2 2 =   = =  43. (d) For limiting condition f = R F sin 30  = (mg − F cos 30 ) , By solving F = 294 .3 N . 44. (c) Net force on the body = Applied force – Friction ma = F − kmg  0.3 10 9.8 129 .4 10 10 =  −  = − = mg F ma k 45. (c) v = gr = 0.5 9.8  40 = 196 = 14 m/s 46. (b) m g u s 40 2 0.5 10 (20) 2 2 2 =   = =  47. (d) Net force in forward direction = Accelerating force + Friction = ma +  mg = m(a +  g) = (1500 + 500 )(1 + 0.2 10) = 2000  3 = 6000 N 48. (b) v = rg = 0.4  30  9.8 = 10.84 m /s 49. (a) W = mgS = 0.250 9.8 1 = 98 J 50. (a) Fl = mg = 0.6 19.8 = 5.88 N Pseudo force on the block = ma = 15 = 5 N Pseudo is less then limiting friction hence static force of friction = 5 N. 51. (d) m g P gm m u g u S 2 2 2 2 2 2 2 2 2 = = = B A f R F F F sin 30° F cos 30° f mg  R
52. (d) Weight of the body = 64N so mass of the body m = 6.4 kg , = 0.6 s , k = 0.4 Net acceleration Massof the body Applied force - Kinetic friction = g g g m mg mg s k s k = ( − ) = (0.6 − 0.4) = 0.2 − =     53. (b) 54. (b) mass Applied force – Kinetic friction a = 2 5 / 10 100 0.5 10 10 = m s −   = 55. (b) 56. (d) v = u − at  u − gt = 0  0.06 10 10 6 =  = = gt u  57. (b) From the relation F − mg = ma 2 10 / 10 129 .4 0.3 10 9.8 m s m F mg a = −   = − =  58. (b) Let body is dragged with force P, making an angle 60° with the horizontal. Fk = Kinetic friction in the motion = kR From the figure F = Pcos60 k and R = mg − P sin 60   Pcos60 = (mg − Psin60) k          =  − 2 3 0.5 60 10 2 P P  P = 315 .1 N  Fk P N 2 315 .1 = cos60 = Work done F s Joule k 2 315 2 315 .1 =  =  = 59. (d) a u v = u − at  t = [As v = 0 ] F u m t  = 6 sec 5000 30 1000 =  = 60. (c) 61. (b) Kinetic energy acquired by body = (Total work done on the body) – (work against friction) = F  S − mgS = 25 10 − 0.2  5 10 10 = 250 −100 = 150 Joule 62. (a) v = rg = 0.5  500 10 = 50 m /s 63. (d) Net downward acceleration Mass Weight- Friction force = m (mg −  R) = 60 60 10 − 0.5 600 = 2 5 / 60 300 = = m s 64. (a) Kinetic friction = kR = 0.2(mg − F sin 30 )       =  −  2 1 0.2 5 10 40 = 0.2(50 − 20) = 6 N Acceleration of the block Mass cos 30 − Kinetic friction = F 2 5.73 / 5 6 2 3 40 = m s  − = 65. (b) We know g u s 2 2 =  0.2 2 10 9 (6) 2 2 2 =   = = gs u  66. (d) m g u s 1000 2 0.5 10 (100 ) 2 2 2 =   = =  67. (d) Kinetic energy of the cylinder will go against friction  2 2 1 mv =  mgs  m g u s 10 2 (0.5) 10 (10) 2 2 2 =   = =  68. (b) When the body is at rest then static friction works on it, which is less than limiting friction ( R). 69. (b) 70. (c) Coefficient of friction = Tangent of angle of repose   = tan  71. (a)       = − 2 1 tan 1 n   2 1 1 n = − [As  = 45 ] 72. (a) Retardation in upward motion = g(sin  +  cos ) P P cos 60° P sin 60° Fk mg 60° R W R F 600N F F cos 30° F sin 30° Fk mg 30° R
 Force required just to move up F = mg (sin +  cos) up Similarly for down ward motion a = g(sin  −  cos )  Force required just to prevent the body sliding down F = mg(sin −  cos) dn According to problem Fup = 2Fdn  mg (sin  +  cos ) = 2mg (sin  −  cos )  sin +  cos = 2 sin − 2 cos  3 cos  = sin   tan  = 3  tan (3 ) tan (3 0.25) tan (0.75) −1 −1 −1  =  =  = = 36.8 73. (c)       = − 2 1 tan 1 n    = 45 and n = 2 (Given)  0.75 4 3 4 1 1 2 1 tan 45 1 2  = − = =       =  − 74. (a) (sin cos ) 9.8(sin 5 0.5 cos 45 ) o o a = g  −   =  − 2 /sec 2 4.9 = m 75. (d) Because if the angle of inclination is equal to or more than angle of repose then box will automatically slides down the plane. 76. (d) Net force along the plane = P − mg sin = 750 − 500 = 250 N Limiting friction = Fl =  sR =  smg cos = 0.4 × 102 × 9.8 × cos 30 = 346 N As net external force is less than limiting friction therefore friction on the body will be 250 N. 77. (c) a = g(sin  −  cos ) = 10(sin 60  − 0.25 cos 60 ) 2 a = 7.4 m / s 78. (b) Fk = kR = kmg cos = 1.7  0.110  cos 30 Fk N 2 3 = 1.7  79. (a) 4 3 2 1 tan 30 1 1 tan 1 2 2  =       = −      = − n   80. (a) For angle of repose, Friction =Component of weight along the plane = mg sin  o = 2  9.8  sin 45 o = 19 .6 sin 45 81. (d) For upper half 2 / 2 2( sin) / 2 sin 2 2 v = u + al = g l = gl For lower half  2 0 2 (sin cos ) 2 l = u + g  −    −gl sin = gl(sin −  cos )   cos  = 2 sin   = 2 tan  82. (c) Resultant downward force along the incline = mg (sin −  cos ) Normal reaction = mg cos  Given : mg cos  = 2mg (sin −  cos ) By solving o  = 45 . 83. (b) F = mg (sin +  cos ) = 10 9.8(sin30 + 0.5 cos30) = 91.4 N . 84. (c) W mg S 1 2.45 J 2 1 =  cos = 0.5 19.8   = 85. (d) F = mg sin 30 = 50 N = 5kg -wt . 86. (a) 3 1  = tan 30 = . 87. (a) Work done against gravity = mgh = 210 10 = 200 J Work done against friction = (Total work done – work done against gravity) = 300 − 200 =100 J 88. (a) 2 0 2 sin 30 2 2 v = u + as = +  g 2  v = 20 R mg cos    F P mg