Title 06. FRICTION Med Ans.pdf ✅

1. (d) μ does not depend on normal reaction. 2. (d)a = gsin 45o – g cos 45o a = 1 1 – 2 2 2 2 g g     =   3. (b) 4. (a) 5. (d)For equilibrium, normal to pane N = mg cos  ... (a) Net force along the plane downward F = mg sin  + f k ...(b) where f k is kinetic fricton but f k = N =  mg cos  ...(c) from eq. (a), (b), and (c) we get  F = mg sin  +  mg cos  According to Newton’s IInd law F = ma  ma = mg sin  +  mg cos   Retardation a = g sin  +  g cos  From equation v = u + at, (we have) o = u – (g sin  +  g cos  )t  g sin  + m g cos  = u t  10 × sin 30° + × 10 cos 30° = 5 0.5  10 × 1 2 + 10 × 3 2 = 10  5 3  = 5 or  = 1 3 6. (c)a = g sin 45o + g cos45o = g 1 1 2 2   +     7. (d) 8. (a) fmax = 0.6 × 1 × g = 6N fmax > 5 so f = 5 N 9. (b) a = 100 – mg m  a = 1 100 – 10 10 2 10   = 5 m/s2 10. (a)Block B will come to rest, if force applied to it will vanish due to frictional force acting between block B and surface, i.e, foce applied = frictional force 11. (c)F = ma   mg = ma   = a g Now vc, v = u + at or ;k 0 = 6 + 10a or ;k 0.6 10 = a = – 0.6 so vr%  = 0.6 0.06 10 a g = = 12. (d)s = 2 100 100 2 2 0.5 10 k v  g  =   = 100 100 5 2   = 1000 m  f = mg = 0.98 N 13. (a) Let acceleration in Ist case is a1 and that in second case is a2
Now vc, 1 2 a1 t 2 = 1 2 a2 (2t)2  a2 = 1 4 a ............(i) Clearly a1 = mg sin m  = g sin ............(ii) and a2 = mg mg sin cos m    − = g sin – g cos ............(iii) From (i), (ii) and (iii), we get  = 0.75. 14. (c)The normal reaction on the block is N = mg – F sin  Net force on block is Fcos – μN = Fcos – μ mg + μFsin or acceleration of the block is a = F μ μmg (cos sin ) m   + − = F m (cos + μsin) – μg 15. (a) N = 50 – 40 sin30 = 30 a = 40cos30o 0.2 30 5 −  = 5.73 m/sec2 16. (c)Let velocity of projection be V and velocity of the block when it returns back = V' then V > V' (since some K.E. is lost to friction) Hence average velocity during ascent > average velocity during descent  ta < td 17. (a) 18. (a) 19. (d)tan   for sliding not depends on mass 20. (b) 21. (a) 22. (b)tan  =    = tan 60o   = 3 = 1.732 23. (d) s mg  75 N  s  75 20g = 0.35 24. (a)ma = mg  a = g 25. (c) Acceleration of train will be from right to left.  Pseudo force will act on the box from left to right therefore friction will act from right to left. 26. (b) 27. (a) since  > tan the block will not slide therefore f = mg sin = 2 × 9.8 × 1 2 = 9.8 N. 28. (a)Friction force depends only on normal reaction. 29. (d)
a = s k f f m − = ( ) s k mg m   − = (S – k ) g = (0.5 – 0.4)10 = 1 m/sec2 30. (a) T = mg  T   × 10 g  mg  0.20 × 10  m  2 kg 31. (c)While the horse pulling a cart, the horse exerts a force on the ground, therefore from the third law of newton, the ground will also exerts a force on the horse that causes the horse to move forward. 32. (c) The length of the rope which can overhang from the edge of the table without sliding down is given by 1 = 1         +  33. (a)a = 2 m/sec F = 20 N. 34. (d) 35. (c) 36. (b) 37. (d)F  mgsin + g cos Fmin = mgsin + g cos 38. (c) 39. (c) Tcos45° = 450 Tsin45° = W W = 450 N 40. (d)consider the equilibrium of the block for minimum value of force we have F F mgsin60 external s + = so     − =     3 1 1 Mg Mg . 2 2 3 3 ext = 200 F 120 1.732 consider the equilibrium of the block for maximum value of force we have F F mgsin60 external s − = so    = =  = =   system maximum 100 a 5 5 * 0.2 1 20 =     + =     3 1 1 Mg . Mg 2 2 3 =200 N 41. (a) N = mg + F sin 60 = 3 × 10 + 3 2 F ...... (i) F cos 60 = N ...... (ii)  2 F = 1 2 3 × (10 3 + 3 2 F )  2 F = 5 + 4 F  4 F = 5  F = 20 N 42. (d) Key Idea : The tension in the string is equal to static frictional force between block A and the surface. Let the mass of the block B is M. 450N 45° W T
In equilibrium T = Mg = 0  T = Mg .........(i) If blocks do not move then T = ƒs where ƒs = frictional force = sR = smg = sR = smg  T = smg ..........(ii) Thus, from Eqs. (i) and (ii), we have Mg = smg or M = sm Given : s = 0.2, m = 2 kg  M = 0.2 × 2 = 0.4 kg 43. (d)Force cy, F = R = Mg weight of block = R = 0.2 ×10 = 2N 44. (a)Let the mass of block be m. Frictional force in rest position F = mg sin 30o 10 = m × 10 × 1 2  m = 2 10 10  = 2 kg 45. (c) (Easy) Max. frictional force fmax = N = (mg + F sin53°) = 0.2 (20 × 10 + 30 × 4 5 ) = 44.8 N As applied horizontal force is Fcos53° = 18N < fmax , friction force will also be 18 N. = 18N < fmax , 46. (a)N = mg + Qcos frectional froce f = (mg + Qcos) P + Qsin = (mg + Qcos)  = +  +  P Q sin mg Qcos 47. (b)For min. m, 5 kg block will have a tendency to move left. so 48. (a)Apply Newton’s law for system along the string mB g = (mA + mC ) × g  mC = mB  – mA = 9 0.2 – 10 = 15 kg 49. (c)The free body diagram of the block is as shown in the figure. N is the normal reaction exerted by wedge on the block. The wedge moves towards left with acceleration ‘a’, then the component of acceleration of block normal to the plane is  Applying Newtons second law to the block normal to plane. mg cos  – N = ma sin  For N to be zero a = g cot . Hence the friction shall be zero when a = g cot . mg cos  – N = ma sin  N a = g cot . a = g cot . mg 30o mg sin 30o mg cos30o R F mg Fsin53° F Fcos53° 53° N N