Page 1 53. Minimum force required to start the motion upward = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 4 3 3 3 2 = 7 6 mg 54. Minimum force required to move the block up the incline with constant speed = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 2 3 3 3 2 = 5 6 mg 55. S 1 2 2 8 = × æ è ç ö ø ÷ = (5.22) 9.8 14.3 m Option (c) is correct. 56. a g g ¢ = - 1200 1000 1200 = g 6 a g g = - 1350 1200 1200 = g 8 For accelerated motion v as max 2 2 1 0 2 = + Þ s v a 1 2 2 = max For retarted motion 0 2 2 2 2 = - ¢ v a s max Þ s v a 2 2 2 = ¢ max s s v a a 1 2 2 2 1 1 + = + ¢ é ë ê ù û ú max 25 2 8 6 2 = + é ë ê ù û ú v g g max v max = ´ 50 14 9.8 = - 5.92ms 1 Option (c) is correct. 57. mg sin q = ´ ´ 170 10 8 15 = 906.67 N f 1 170 10 15 17 (max) = ´ ´ ´ 0.2 = 300 N f 2 170 10 15 17 (max) = ´ ´ ´ 0.4 = 600 N The whole system will accelerate as mgsinq is greater than both f 1 (max) and f 2 (max). Total force of friction = + f f 1 2 (max) (max) =900 N Option (a) is correct. 58. mg T ma sin q - - = 300 â€¦(i) and mg T ma sinq + = â€¦(ii) Substituting Eq. (i) by Eq. (ii), 2 300 0 T + = T = - 150 N = 150 N, compressive. Option (a) is correct. Laws of Motion | 101 a 25 m T' = 1000g â€“ a' 1200 g 1200 g T = 1350g Speed = v max Stops q mg sin q A B m = 0.4 f 2 f 1 T T m = 0.2 mg sin q q 8 17 15 Page 2 53. Minimum force required to start the motion upward = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 4 3 3 3 2 = 7 6 mg 54. Minimum force required to move the block up the incline with constant speed = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 2 3 3 3 2 = 5 6 mg 55. S 1 2 2 8 = × æ è ç ö ø ÷ = (5.22) 9.8 14.3 m Option (c) is correct. 56. a g g ¢ = - 1200 1000 1200 = g 6 a g g = - 1350 1200 1200 = g 8 For accelerated motion v as max 2 2 1 0 2 = + Þ s v a 1 2 2 = max For retarted motion 0 2 2 2 2 = - ¢ v a s max Þ s v a 2 2 2 = ¢ max s s v a a 1 2 2 2 1 1 + = + ¢ é ë ê ù û ú max 25 2 8 6 2 = + é ë ê ù û ú v g g max v max = ´ 50 14 9.8 = - 5.92ms 1 Option (c) is correct. 57. mg sin q = ´ ´ 170 10 8 15 = 906.67 N f 1 170 10 15 17 (max) = ´ ´ ´ 0.2 = 300 N f 2 170 10 15 17 (max) = ´ ´ ´ 0.4 = 600 N The whole system will accelerate as mgsinq is greater than both f 1 (max) and f 2 (max). Total force of friction = + f f 1 2 (max) (max) =900 N Option (a) is correct. 58. mg T ma sin q - - = 300 â€¦(i) and mg T ma sinq + = â€¦(ii) Substituting Eq. (i) by Eq. (ii), 2 300 0 T + = T = - 150 N = 150 N, compressive. Option (a) is correct. Laws of Motion | 101 a 25 m T' = 1000g â€“ a' 1200 g 1200 g T = 1350g Speed = v max Stops q mg sin q A B m = 0.4 f 2 f 1 T T m = 0.2 mg sin q q 8 17 15 More than One Cor rect Options 1. (a) Normal force between A and B m g = 2 = ´ = 1 10 10 N \ Force of limiting friction by B on A (or by A on B) = ´ = m 10 2 N Total force opposing applied external force F T = + 2N = + 2 2 N N = 4 N Thus, if F £ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. \ Option (a) is correct. (b) If F > 4 N F T - - = 2 1 a and T a - = 2 1 â€¦(i) Adding above equation F a - = 4 2 â€¦(ii) i.e., F a = + 4 2 For F > 4 N 2 4 4 a + > or a > 0 Þ T - > 2 0 i.e., T > 2 N \ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2 6 4 a = - Þ a = 1 m/s 2 \ Using Eq. (i), T - = 2 1 i.e., T = 3 N Option (d) is correct. 2. At point A T T mg 1 2 cos cos a b = + â€¦(i) and T T 1 2 sin sin a b = â€¦(ii) At point B T mg 2 cos b = â€¦(iii) T F mg 2 sin b = - â€¦(iv) Using Eq. (iii) in Eq. (i), T T 1 2 2 cos cos a b = â€¦(v) Dividing Eq. (ii) by Eq. (v), 2 tan tan a b = â€¦(vi) Option (a) is correct. Squaring and adding Eqs. (iii) and (v), T T T 1 2 2 2 2 2 2 2 4 = + cos sin b b â€¦(vii) Dividing Eq. (iii) by Eq. (iv) tan b = 1 \ cos sin b b = = 1 2 Substituting the values of sinb and cosb in Eq. (vii) T T T 1 2 2 2 2 2 4 1 2 1 2 = æ è ç ö ø ÷ + æ è ç ö ø ÷ = 5 2 2 2 T 102 | Mechanics-1 T T T T 2 N F A B 2 N F = mg B T 2 T 2 b A T 1 a T 1 mg b 1 1 Ö2 Page 3 53. Minimum force required to start the motion upward = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 4 3 3 3 2 = 7 6 mg 54. Minimum force required to move the block up the incline with constant speed = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 2 3 3 3 2 = 5 6 mg 55. S 1 2 2 8 = × æ è ç ö ø ÷ = (5.22) 9.8 14.3 m Option (c) is correct. 56. a g g ¢ = - 1200 1000 1200 = g 6 a g g = - 1350 1200 1200 = g 8 For accelerated motion v as max 2 2 1 0 2 = + Þ s v a 1 2 2 = max For retarted motion 0 2 2 2 2 = - ¢ v a s max Þ s v a 2 2 2 = ¢ max s s v a a 1 2 2 2 1 1 + = + ¢ é ë ê ù û ú max 25 2 8 6 2 = + é ë ê ù û ú v g g max v max = ´ 50 14 9.8 = - 5.92ms 1 Option (c) is correct. 57. mg sin q = ´ ´ 170 10 8 15 = 906.67 N f 1 170 10 15 17 (max) = ´ ´ ´ 0.2 = 300 N f 2 170 10 15 17 (max) = ´ ´ ´ 0.4 = 600 N The whole system will accelerate as mgsinq is greater than both f 1 (max) and f 2 (max). Total force of friction = + f f 1 2 (max) (max) =900 N Option (a) is correct. 58. mg T ma sin q - - = 300 â€¦(i) and mg T ma sinq + = â€¦(ii) Substituting Eq. (i) by Eq. (ii), 2 300 0 T + = T = - 150 N = 150 N, compressive. Option (a) is correct. Laws of Motion | 101 a 25 m T' = 1000g â€“ a' 1200 g 1200 g T = 1350g Speed = v max Stops q mg sin q A B m = 0.4 f 2 f 1 T T m = 0.2 mg sin q q 8 17 15 More than One Cor rect Options 1. (a) Normal force between A and B m g = 2 = ´ = 1 10 10 N \ Force of limiting friction by B on A (or by A on B) = ´ = m 10 2 N Total force opposing applied external force F T = + 2N = + 2 2 N N = 4 N Thus, if F £ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. \ Option (a) is correct. (b) If F > 4 N F T - - = 2 1 a and T a - = 2 1 â€¦(i) Adding above equation F a - = 4 2 â€¦(ii) i.e., F a = + 4 2 For F > 4 N 2 4 4 a + > or a > 0 Þ T - > 2 0 i.e., T > 2 N \ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2 6 4 a = - Þ a = 1 m/s 2 \ Using Eq. (i), T - = 2 1 i.e., T = 3 N Option (d) is correct. 2. At point A T T mg 1 2 cos cos a b = + â€¦(i) and T T 1 2 sin sin a b = â€¦(ii) At point B T mg 2 cos b = â€¦(iii) T F mg 2 sin b = - â€¦(iv) Using Eq. (iii) in Eq. (i), T T 1 2 2 cos cos a b = â€¦(v) Dividing Eq. (ii) by Eq. (v), 2 tan tan a b = â€¦(vi) Option (a) is correct. Squaring and adding Eqs. (iii) and (v), T T T 1 2 2 2 2 2 2 2 4 = + cos sin b b â€¦(vii) Dividing Eq. (iii) by Eq. (iv) tan b = 1 \ cos sin b b = = 1 2 Substituting the values of sinb and cosb in Eq. (vii) T T T 1 2 2 2 2 2 4 1 2 1 2 = æ è ç ö ø ÷ + æ è ç ö ø ÷ = 5 2 2 2 T 102 | Mechanics-1 T T T T 2 N F A B 2 N F = mg B T 2 T 2 b A T 1 a T 1 mg b 1 1 Ö2 Þ 2 5 1 2 T T = Option (c) is correct. 3. Displacement of block in 4 s S = Area under curve = 16 m. DK =Workdone by frictional force 1 2 1 4 1 10 16 2 ´ ´ = ´ ´ ´ m Þ m = 0.1 Option (a) is correct. Option (b) is incorrect. Acceleration, a = f tan = - tan ( ) p q = - tan q = - 1 m/s 2 If half rough retardation = 0.5 m/s 2 \ 16 4 1 2 2 = + - t t ( ) 0.5 i.e., t t 2 16 64 0 - + = or t = 8 s Option (d) is correct. Option (c) is incorrect. 4. Let acceleration of wedge ( ) A a = N ma mg + = sin cos q q N mg ma = - cos sin q q Acceleration of a N M = sin q or Ma mg ma = - ( cos sin ) sin q q q or a M m mg ( sin ) cos sin + = 2 q q q i.e., a mg M m = + cos sin sin q q q 2 = ´ ´ ° ° + ´ ° 0.6 1.7 0.6 g cos sin ( sin ) 45 45 45 2 = + 3 17 3 g = 3 20 g Let a B = Acceleration of block B Net force on B (along inclined plane) ma ma mg B = + cos sin q q Þ a a g B = + cos sin q q Thus, ( ) ( cos sin ) cos a a g B V = + q q q = + a g cos sin cos 2 q q q = + ( ) a g 1 2 = + æ è ç ö ø ÷ 3 20 1 2 g g = 23 40 g ( ) ( cos sin ) sin a a g a B H = + - q q q = - 23 40 3 20 g g = 17 40 g 5. f m g A 1 1 (max.) = m = ´ ´ 0.3 60 10 = 180 N F net on B = + f T 1 (max.) = + 180 125 = 305 N A will remain stationary as T f < 1 (max.) \ f 1 125 = N Force of friction acting between A and B=125 N \ Options (c) and (d) are incorrect. Laws of Motion | 103 O q â€“1 v (ms ) f 4 t (s) q mg N ma (Pseudo force) F N sin q A M a F = mg cos q ma sin q q B f 1 f 2 f 1 A T(pull) = 125 N Page 4 53. Minimum force required to start the motion upward = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 4 3 3 3 2 = 7 6 mg 54. Minimum force required to move the block up the incline with constant speed = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 2 3 3 3 2 = 5 6 mg 55. S 1 2 2 8 = × æ è ç ö ø ÷ = (5.22) 9.8 14.3 m Option (c) is correct. 56. a g g ¢ = - 1200 1000 1200 = g 6 a g g = - 1350 1200 1200 = g 8 For accelerated motion v as max 2 2 1 0 2 = + Þ s v a 1 2 2 = max For retarted motion 0 2 2 2 2 = - ¢ v a s max Þ s v a 2 2 2 = ¢ max s s v a a 1 2 2 2 1 1 + = + ¢ é ë ê ù û ú max 25 2 8 6 2 = + é ë ê ù û ú v g g max v max = ´ 50 14 9.8 = - 5.92ms 1 Option (c) is correct. 57. mg sin q = ´ ´ 170 10 8 15 = 906.67 N f 1 170 10 15 17 (max) = ´ ´ ´ 0.2 = 300 N f 2 170 10 15 17 (max) = ´ ´ ´ 0.4 = 600 N The whole system will accelerate as mgsinq is greater than both f 1 (max) and f 2 (max). Total force of friction = + f f 1 2 (max) (max) =900 N Option (a) is correct. 58. mg T ma sin q - - = 300 â€¦(i) and mg T ma sinq + = â€¦(ii) Substituting Eq. (i) by Eq. (ii), 2 300 0 T + = T = - 150 N = 150 N, compressive. Option (a) is correct. Laws of Motion | 101 a 25 m T' = 1000g â€“ a' 1200 g 1200 g T = 1350g Speed = v max Stops q mg sin q A B m = 0.4 f 2 f 1 T T m = 0.2 mg sin q q 8 17 15 More than One Cor rect Options 1. (a) Normal force between A and B m g = 2 = ´ = 1 10 10 N \ Force of limiting friction by B on A (or by A on B) = ´ = m 10 2 N Total force opposing applied external force F T = + 2N = + 2 2 N N = 4 N Thus, if F £ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. \ Option (a) is correct. (b) If F > 4 N F T - - = 2 1 a and T a - = 2 1 â€¦(i) Adding above equation F a - = 4 2 â€¦(ii) i.e., F a = + 4 2 For F > 4 N 2 4 4 a + > or a > 0 Þ T - > 2 0 i.e., T > 2 N \ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2 6 4 a = - Þ a = 1 m/s 2 \ Using Eq. (i), T - = 2 1 i.e., T = 3 N Option (d) is correct. 2. At point A T T mg 1 2 cos cos a b = + â€¦(i) and T T 1 2 sin sin a b = â€¦(ii) At point B T mg 2 cos b = â€¦(iii) T F mg 2 sin b = - â€¦(iv) Using Eq. (iii) in Eq. (i), T T 1 2 2 cos cos a b = â€¦(v) Dividing Eq. (ii) by Eq. (v), 2 tan tan a b = â€¦(vi) Option (a) is correct. Squaring and adding Eqs. (iii) and (v), T T T 1 2 2 2 2 2 2 2 4 = + cos sin b b â€¦(vii) Dividing Eq. (iii) by Eq. (iv) tan b = 1 \ cos sin b b = = 1 2 Substituting the values of sinb and cosb in Eq. (vii) T T T 1 2 2 2 2 2 4 1 2 1 2 = æ è ç ö ø ÷ + æ è ç ö ø ÷ = 5 2 2 2 T 102 | Mechanics-1 T T T T 2 N F A B 2 N F = mg B T 2 T 2 b A T 1 a T 1 mg b 1 1 Ö2 Þ 2 5 1 2 T T = Option (c) is correct. 3. Displacement of block in 4 s S = Area under curve = 16 m. DK =Workdone by frictional force 1 2 1 4 1 10 16 2 ´ ´ = ´ ´ ´ m Þ m = 0.1 Option (a) is correct. Option (b) is incorrect. Acceleration, a = f tan = - tan ( ) p q = - tan q = - 1 m/s 2 If half rough retardation = 0.5 m/s 2 \ 16 4 1 2 2 = + - t t ( ) 0.5 i.e., t t 2 16 64 0 - + = or t = 8 s Option (d) is correct. Option (c) is incorrect. 4. Let acceleration of wedge ( ) A a = N ma mg + = sin cos q q N mg ma = - cos sin q q Acceleration of a N M = sin q or Ma mg ma = - ( cos sin ) sin q q q or a M m mg ( sin ) cos sin + = 2 q q q i.e., a mg M m = + cos sin sin q q q 2 = ´ ´ ° ° + ´ ° 0.6 1.7 0.6 g cos sin ( sin ) 45 45 45 2 = + 3 17 3 g = 3 20 g Let a B = Acceleration of block B Net force on B (along inclined plane) ma ma mg B = + cos sin q q Þ a a g B = + cos sin q q Thus, ( ) ( cos sin ) cos a a g B V = + q q q = + a g cos sin cos 2 q q q = + ( ) a g 1 2 = + æ è ç ö ø ÷ 3 20 1 2 g g = 23 40 g ( ) ( cos sin ) sin a a g a B H = + - q q q = - 23 40 3 20 g g = 17 40 g 5. f m g A 1 1 (max.) = m = ´ ´ 0.3 60 10 = 180 N F net on B = + f T 1 (max.) = + 180 125 = 305 N A will remain stationary as T f < 1 (max.) \ f 1 125 = N Force of friction acting between A and B=125 N \ Options (c) and (d) are incorrect. Laws of Motion | 103 O q â€“1 v (ms ) f 4 t (s) q mg N ma (Pseudo force) F N sin q A M a F = mg cos q ma sin q q B f 1 f 2 f 1 A T(pull) = 125 N f m m g A B 2 2 (max) ( ) = + m = + 0.2 ( ) 60 40 10 = 200 N f T 1 125 125 250 + = + = N As, f T f 1 2 + > (max. block B /along the A as A is stationary) will move towards right with acceleration. Option (a) is correct. a f T f m m B B A = + - + ( ) (max.) 1 2 = - + 250 200 40 60 =0.5 m/s 2 Option (b) is correct. 6. (See solution to Question no. 4). N mg ma = - cos sin q q Option (c) is correct and option (d) is incorrect. As angle between the directions of a and gsinq will be less than 90°, acceleration of block A will be more than g sin q. Option (a) is correct and option (b) is incorrect. 7. Maximum value of friction. f 1 = between A and B = ´ ´ = 0.25 3 10 7.5 N f 2 = between B and C = ´ ´ = 0.25 17.5 7 10 N f 3 = between C and ground = ´ ´ = 0.25 7.5 15 10 3 N (a) T = + = 17.5 7.5 25 N F T = + + = 37.5 17.5 80 N (c) T a - - = 7.5 17.5 4 â€¦(i) F T a - - - = 37.5 17.5 8 â€¦(ii) F = 200 N â€¦(iii) Solving these equations we get, a = 10 m/s 2 8. Maximum value of friction available to block is less than the maximum value of friction available to man. 9. N ma sin q = = ´ = 1 5 5 â€¦(i) N mg cos q = = ´ = 1 10 10 â€¦(ii) Solving these two equations we get, tanq= 1 2 and N = 5 5 N. 10. Let f 1 = friction between 2 kg and 4 kg f 2 = friction between 4 kg and ground ( ) max f s1 2 10 8 = ´ ´ = 0.4 N ( ) f k1 2 10 4 = ´ ´ = 0.2 N ( ) . max f s2 06 6 10 36 = ´ ´ = N F k2 0 4 6 10 24 = ´ ´ = . N (b) At t = 1 s, F = 2 N < ( ) max f s2 \ Both the blocks are at rest. \ f 1 0 = (c) At t = 4 s, F = 8 N < ( ) max f s2 \ Both the blocks are at rest. f F 2 8 = = N, 11. a = 0, T 1 10 = N, T T 2 1 20 30 = + = N, T 3 20 = N. 12. f max = ´ ´ = 0.3 2 10 6 N (a) At t = 2 s, F N = 2 \ f = 2 N (b) At t = 8 s, F = 8 N > 6 N \ f = 6 N 104 | Mechanics-1 A m a ma q a A a > g sin q net T 17.5 N 37.5 N 7.5 N 17.5 N B F T C Page 5 53. Minimum force required to start the motion upward = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 4 3 3 3 2 = 7 6 mg 54. Minimum force required to move the block up the incline with constant speed = + mg mg k sin cos q m q = + × é ë ê ù û ú mg 1 2 2 3 3 3 2 = 5 6 mg 55. S 1 2 2 8 = × æ è ç ö ø ÷ = (5.22) 9.8 14.3 m Option (c) is correct. 56. a g g ¢ = - 1200 1000 1200 = g 6 a g g = - 1350 1200 1200 = g 8 For accelerated motion v as max 2 2 1 0 2 = + Þ s v a 1 2 2 = max For retarted motion 0 2 2 2 2 = - ¢ v a s max Þ s v a 2 2 2 = ¢ max s s v a a 1 2 2 2 1 1 + = + ¢ é ë ê ù û ú max 25 2 8 6 2 = + é ë ê ù û ú v g g max v max = ´ 50 14 9.8 = - 5.92ms 1 Option (c) is correct. 57. mg sin q = ´ ´ 170 10 8 15 = 906.67 N f 1 170 10 15 17 (max) = ´ ´ ´ 0.2 = 300 N f 2 170 10 15 17 (max) = ´ ´ ´ 0.4 = 600 N The whole system will accelerate as mgsinq is greater than both f 1 (max) and f 2 (max). Total force of friction = + f f 1 2 (max) (max) =900 N Option (a) is correct. 58. mg T ma sin q - - = 300 â€¦(i) and mg T ma sinq + = â€¦(ii) Substituting Eq. (i) by Eq. (ii), 2 300 0 T + = T = - 150 N = 150 N, compressive. Option (a) is correct. Laws of Motion | 101 a 25 m T' = 1000g â€“ a' 1200 g 1200 g T = 1350g Speed = v max Stops q mg sin q A B m = 0.4 f 2 f 1 T T m = 0.2 mg sin q q 8 17 15 More than One Cor rect Options 1. (a) Normal force between A and B m g = 2 = ´ = 1 10 10 N \ Force of limiting friction by B on A (or by A on B) = ´ = m 10 2 N Total force opposing applied external force F T = + 2N = + 2 2 N N = 4 N Thus, if F £ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. \ Option (a) is correct. (b) If F > 4 N F T - - = 2 1 a and T a - = 2 1 â€¦(i) Adding above equation F a - = 4 2 â€¦(ii) i.e., F a = + 4 2 For F > 4 N 2 4 4 a + > or a > 0 Þ T - > 2 0 i.e., T > 2 N \ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2 6 4 a = - Þ a = 1 m/s 2 \ Using Eq. (i), T - = 2 1 i.e., T = 3 N Option (d) is correct. 2. At point A T T mg 1 2 cos cos a b = + â€¦(i) and T T 1 2 sin sin a b = â€¦(ii) At point B T mg 2 cos b = â€¦(iii) T F mg 2 sin b = - â€¦(iv) Using Eq. (iii) in Eq. (i), T T 1 2 2 cos cos a b = â€¦(v) Dividing Eq. (ii) by Eq. (v), 2 tan tan a b = â€¦(vi) Option (a) is correct. Squaring and adding Eqs. (iii) and (v), T T T 1 2 2 2 2 2 2 2 4 = + cos sin b b â€¦(vii) Dividing Eq. (iii) by Eq. (iv) tan b = 1 \ cos sin b b = = 1 2 Substituting the values of sinb and cosb in Eq. (vii) T T T 1 2 2 2 2 2 4 1 2 1 2 = æ è ç ö ø ÷ + æ è ç ö ø ÷ = 5 2 2 2 T 102 | Mechanics-1 T T T T 2 N F A B 2 N F = mg B T 2 T 2 b A T 1 a T 1 mg b 1 1 Ö2 Þ 2 5 1 2 T T = Option (c) is correct. 3. Displacement of block in 4 s S = Area under curve = 16 m. DK =Workdone by frictional force 1 2 1 4 1 10 16 2 ´ ´ = ´ ´ ´ m Þ m = 0.1 Option (a) is correct. Option (b) is incorrect. Acceleration, a = f tan = - tan ( ) p q = - tan q = - 1 m/s 2 If half rough retardation = 0.5 m/s 2 \ 16 4 1 2 2 = + - t t ( ) 0.5 i.e., t t 2 16 64 0 - + = or t = 8 s Option (d) is correct. Option (c) is incorrect. 4. Let acceleration of wedge ( ) A a = N ma mg + = sin cos q q N mg ma = - cos sin q q Acceleration of a N M = sin q or Ma mg ma = - ( cos sin ) sin q q q or a M m mg ( sin ) cos sin + = 2 q q q i.e., a mg M m = + cos sin sin q q q 2 = ´ ´ ° ° + ´ ° 0.6 1.7 0.6 g cos sin ( sin ) 45 45 45 2 = + 3 17 3 g = 3 20 g Let a B = Acceleration of block B Net force on B (along inclined plane) ma ma mg B = + cos sin q q Þ a a g B = + cos sin q q Thus, ( ) ( cos sin ) cos a a g B V = + q q q = + a g cos sin cos 2 q q q = + ( ) a g 1 2 = + æ è ç ö ø ÷ 3 20 1 2 g g = 23 40 g ( ) ( cos sin ) sin a a g a B H = + - q q q = - 23 40 3 20 g g = 17 40 g 5. f m g A 1 1 (max.) = m = ´ ´ 0.3 60 10 = 180 N F net on B = + f T 1 (max.) = + 180 125 = 305 N A will remain stationary as T f < 1 (max.) \ f 1 125 = N Force of friction acting between A and B=125 N \ Options (c) and (d) are incorrect. Laws of Motion | 103 O q â€“1 v (ms ) f 4 t (s) q mg N ma (Pseudo force) F N sin q A M a F = mg cos q ma sin q q B f 1 f 2 f 1 A T(pull) = 125 N f m m g A B 2 2 (max) ( ) = + m = + 0.2 ( ) 60 40 10 = 200 N f T 1 125 125 250 + = + = N As, f T f 1 2 + > (max. block B /along the A as A is stationary) will move towards right with acceleration. Option (a) is correct. a f T f m m B B A = + - + ( ) (max.) 1 2 = - + 250 200 40 60 =0.5 m/s 2 Option (b) is correct. 6. (See solution to Question no. 4). N mg ma = - cos sin q q Option (c) is correct and option (d) is incorrect. As angle between the directions of a and gsinq will be less than 90°, acceleration of block A will be more than g sin q. Option (a) is correct and option (b) is incorrect. 7. Maximum value of friction. f 1 = between A and B = ´ ´ = 0.25 3 10 7.5 N f 2 = between B and C = ´ ´ = 0.25 17.5 7 10 N f 3 = between C and ground = ´ ´ = 0.25 7.5 15 10 3 N (a) T = + = 17.5 7.5 25 N F T = + + = 37.5 17.5 80 N (c) T a - - = 7.5 17.5 4 â€¦(i) F T a - - - = 37.5 17.5 8 â€¦(ii) F = 200 N â€¦(iii) Solving these equations we get, a = 10 m/s 2 8. Maximum value of friction available to block is less than the maximum value of friction available to man. 9. N ma sin q = = ´ = 1 5 5 â€¦(i) N mg cos q = = ´ = 1 10 10 â€¦(ii) Solving these two equations we get, tanq= 1 2 and N = 5 5 N. 10. Let f 1 = friction between 2 kg and 4 kg f 2 = friction between 4 kg and ground ( ) max f s1 2 10 8 = ´ ´ = 0.4 N ( ) f k1 2 10 4 = ´ ´ = 0.2 N ( ) . max f s2 06 6 10 36 = ´ ´ = N F k2 0 4 6 10 24 = ´ ´ = . N (b) At t = 1 s, F = 2 N < ( ) max f s2 \ Both the blocks are at rest. \ f 1 0 = (c) At t = 4 s, F = 8 N < ( ) max f s2 \ Both the blocks are at rest. f F 2 8 = = N, 11. a = 0, T 1 10 = N, T T 2 1 20 30 = + = N, T 3 20 = N. 12. f max = ´ ´ = 0.3 2 10 6 N (a) At t = 2 s, F N = 2 \ f = 2 N (b) At t = 8 s, F = 8 N > 6 N \ f = 6 N 104 | Mechanics-1 A m a ma q a A a > g sin q net T 17.5 N 37.5 N 7.5 N 17.5 N B F T C (c) At t = 10 s, F = 10 N and f = 6 N \ a = - = 10 6 2 2 m/s 2 (d) Block will start at 6 s. After that, net impulse = ´ ´ + + ´ - ´ 1 2 4 6 10 2 10 6 6 ( ) = 16 N-s = mv \ v = = 16 2 8 m/s. 13. f max = ´ ´ = 0.4 2 10 8 N (b) At t = 3 s, F = 6 N \ Common acceleration a = = 6 6 1 m/s 2 \ Pseudo force on 2 kg = ´ = 2 1 2 N (backward) 14. N Mg F = - sin q F N Mg F cos ( sin ) q = m m q = - \ F Mg = + m q m q cos sin For F to be minimum, dF dq = 0 Match the Columns 1. Acceleration after t = 4 s At t = 4 s, F = 8 N \ F max = 8 i.e., m s mg = 8 Þ m s mg = = ´ = 8 8 2 10 0.4 \ (a) ® (r) At t = 4 s, a = - 1 2 ms t = 4 s, F = 8 N F N ma k - = m i.e., m k F ma N = - = - F ma mg = - ´ ´ 8 2 1 2 10 ( ) = 0.3 (b) = (q) At t = 01 . s, F = 0 2 . N \ Force of friction (at t = 01 . s) = 0.2 N \ (c) ® (p) At t = 8 s, F = 16 N \ a F mg m k = - m = - ´ ´ 16 2 10 2 ( ) 0.3 = 5 i.e., a 10 = 0.5 \ (d) ® (s). 2. At q = ° 0 , dragging force = 0 \ Force of force = 0 \ (a) ® (s) At q = ° 90 Normal force on block by plane will be zero. \ Force of friction = 0 \ (b) ® (s) At q = ° 30 Angle of repose = - tan 1 m = - tan ( ) 1 1 = ° 45 As q < angle of repose, the block will not slip and thus, force of friction = mg sin q = ´ ´ ° 2 10 30 sin = 10 N \ (c) ® (p) At q = ° 60 As q > angle of repose Block will accelerate and thus force of friction =m N = ´ ´ ´ ° 1 2 10 60 cos = 10 N \ (d) ® (p). 3. All contact forces (e.g., force of friction and normal reaction) are electromagnetic in nature. \ (a) ® (q), (r) (b) ® (q), (r). Laws of Motion | 105Read More

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