Page 1 6.1 SOLUTIONS TO CONCEPTS CHAPTER 6 1. Let m = mass of the block From the freebody diagram, R – mg = 0 ? R = mg ...(1) Again ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4 The co-efficient of kinetic friction between the block and the plane is 0.4 ? 2. Due to friction the body will decelerate Let the deceleration be ‘a’ R – mg = 0 ? R = mg ...(1) ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g = 0.1 × 10 = 1m/s 2. Initial velocity u = 10 m/s Final velocity v = 0 m/s a = –1m/s 2 (deceleration) S = a 2 u v 2 2 ? = ) 1 ( 2 10 0 2 ? ? = 2 100 = 50m It will travel 50m before coming to rest. 3. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f ? frictional force F ? Applied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, R – mg cos ? = 0 ? R = mg cos ? ..(1) For the block U = 0, s = 8m, t = 2sec. ?s = ut + ½ at 2 ? 8 = 0 + ½ a 2 2 ? a = 4m/s 2 Again, ?R + ma – mg sin ? = 0 ? ? mg cos ? + ma – mg sin ? = 0 [from (1)] ? m( ?g cos ? + a – g sin ?) = 0 ? ? × 10 × cos 30° = g sin 30° – a ??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4 ? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11 ? Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram 4 – 4a – ?R + 4g sin 30° = 0 …(1) R – 4g cos 30° = 0 ...(2) ? R = 4g cos 30° Putting the values of R is & in equn. (1) 4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0 ? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0 ? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s 2 For the block u =0, t = 2sec, a = 5m/s 2 Distance s = ut + ½ at 2 ? s = 0 + (1/2) 5 × 2 2 = 10m The block will move 10m. R mg ma ?R a R mg ma ?R a velocity F p o mg R 30° R ??? ?R mg ma 4kg 30° 4N R ??? ?R mg ma a velocity Page 2 6.1 SOLUTIONS TO CONCEPTS CHAPTER 6 1. Let m = mass of the block From the freebody diagram, R – mg = 0 ? R = mg ...(1) Again ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4 The co-efficient of kinetic friction between the block and the plane is 0.4 ? 2. Due to friction the body will decelerate Let the deceleration be ‘a’ R – mg = 0 ? R = mg ...(1) ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g = 0.1 × 10 = 1m/s 2. Initial velocity u = 10 m/s Final velocity v = 0 m/s a = –1m/s 2 (deceleration) S = a 2 u v 2 2 ? = ) 1 ( 2 10 0 2 ? ? = 2 100 = 50m It will travel 50m before coming to rest. 3. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f ? frictional force F ? Applied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, R – mg cos ? = 0 ? R = mg cos ? ..(1) For the block U = 0, s = 8m, t = 2sec. ?s = ut + ½ at 2 ? 8 = 0 + ½ a 2 2 ? a = 4m/s 2 Again, ?R + ma – mg sin ? = 0 ? ? mg cos ? + ma – mg sin ? = 0 [from (1)] ? m( ?g cos ? + a – g sin ?) = 0 ? ? × 10 × cos 30° = g sin 30° – a ??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4 ? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11 ? Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram 4 – 4a – ?R + 4g sin 30° = 0 …(1) R – 4g cos 30° = 0 ...(2) ? R = 4g cos 30° Putting the values of R is & in equn. (1) 4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0 ? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0 ? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s 2 For the block u =0, t = 2sec, a = 5m/s 2 Distance s = ut + ½ at 2 ? s = 0 + (1/2) 5 × 2 2 = 10m The block will move 10m. R mg ma ?R a R mg ma ?R a velocity F p o mg R 30° R ??? ?R mg ma 4kg 30° 4N R ??? ?R mg ma a velocity Chapter 6 6.2 6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline = ? R + 2 g sin 30° = 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)] = 3.39 + 9.8 = 13N With this minimum force the body move up the incline with a constant velocity as net force on it is zero. b) Net force acting down the incline is given by, F = 2 g sin 30° – ?R = 2 × 9.8 × (1/2) – 3.39 = 6.41N Due to F = 6.41N the body will move down the incline with acceleration. No external force is required. ? Force required is zero. 7. From the free body diagram g = 10m/s 2 , m = 2kg, ? = 30°, ? = 0.2 R – mg cos ? - F sin ? = 0 ? R = mg cos ? + F sin ?? ...(1) And mg sin ? + ?R – F cos ? = 0 ? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0 ? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0 ? F = ) cos sin ( ) cos mg sin mg ( ? ? ? ? ? ? ? ? ? F = ) 2 / 3 ( ) 2 / 1 ( 2 . 0 ) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2 ? ? ? ? ? ? ? ? = 76 . 0 464 . 13 = 17.7N ? 17.5N ? 8. m ? mass of child R – mg cos 45° = 0 ? R = mg cos 45° = mg /v 2 ...(1) Net force acting on the boy due to which it slides down is mg sin 45° - ?R = mg sin 45° - ? mg cos 45° = m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 ) = m [(5/ 2 ) – 0.6 × (5 / 2 )] = m(2 2 ) acceleration = mass Force = m ) 2 2 ( m = 2 2 m/s 2 ? 9. Suppose, the body is accelerating down with acceleration ‘a’. From the free body diagram R – mg cos ? = 0 ? R = mg cos ? ...(1) ma + mg sin ? – ? R = 0 ? a = m ) cos (sin mg ? ? ? ? = g (sin ? – ? cos ?) For the first half mt. u = 0, s = 0.5m, t = 0.5 sec. So, v = u + at = 0 + (0.5)4 = 2 m/s S = ut + ½ at 2 ? 0.5 = 0 + ½ a (0/5) 2 ? a = 4m/s 2 ...(2) For the next half metre u` = 2m/s, a = 4m/s 2 , s= 0.5. ? 0.5 = 2t + (1/2) 4 t 2 ? 2 t 2 + 2 t – 0.5 =0 (body moving us) ?R R ??? F mg (body moving down) ?R R ??? mg R 45° ?R mg F 30° ?R R 30° mg mg ma ?R R Page 3 6.1 SOLUTIONS TO CONCEPTS CHAPTER 6 1. Let m = mass of the block From the freebody diagram, R – mg = 0 ? R = mg ...(1) Again ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4 The co-efficient of kinetic friction between the block and the plane is 0.4 ? 2. Due to friction the body will decelerate Let the deceleration be ‘a’ R – mg = 0 ? R = mg ...(1) ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g = 0.1 × 10 = 1m/s 2. Initial velocity u = 10 m/s Final velocity v = 0 m/s a = –1m/s 2 (deceleration) S = a 2 u v 2 2 ? = ) 1 ( 2 10 0 2 ? ? = 2 100 = 50m It will travel 50m before coming to rest. 3. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f ? frictional force F ? Applied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, R – mg cos ? = 0 ? R = mg cos ? ..(1) For the block U = 0, s = 8m, t = 2sec. ?s = ut + ½ at 2 ? 8 = 0 + ½ a 2 2 ? a = 4m/s 2 Again, ?R + ma – mg sin ? = 0 ? ? mg cos ? + ma – mg sin ? = 0 [from (1)] ? m( ?g cos ? + a – g sin ?) = 0 ? ? × 10 × cos 30° = g sin 30° – a ??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4 ? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11 ? Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram 4 – 4a – ?R + 4g sin 30° = 0 …(1) R – 4g cos 30° = 0 ...(2) ? R = 4g cos 30° Putting the values of R is & in equn. (1) 4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0 ? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0 ? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s 2 For the block u =0, t = 2sec, a = 5m/s 2 Distance s = ut + ½ at 2 ? s = 0 + (1/2) 5 × 2 2 = 10m The block will move 10m. R mg ma ?R a R mg ma ?R a velocity F p o mg R 30° R ??? ?R mg ma 4kg 30° 4N R ??? ?R mg ma a velocity Chapter 6 6.2 6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline = ? R + 2 g sin 30° = 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)] = 3.39 + 9.8 = 13N With this minimum force the body move up the incline with a constant velocity as net force on it is zero. b) Net force acting down the incline is given by, F = 2 g sin 30° – ?R = 2 × 9.8 × (1/2) – 3.39 = 6.41N Due to F = 6.41N the body will move down the incline with acceleration. No external force is required. ? Force required is zero. 7. From the free body diagram g = 10m/s 2 , m = 2kg, ? = 30°, ? = 0.2 R – mg cos ? - F sin ? = 0 ? R = mg cos ? + F sin ?? ...(1) And mg sin ? + ?R – F cos ? = 0 ? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0 ? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0 ? F = ) cos sin ( ) cos mg sin mg ( ? ? ? ? ? ? ? ? ? F = ) 2 / 3 ( ) 2 / 1 ( 2 . 0 ) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2 ? ? ? ? ? ? ? ? = 76 . 0 464 . 13 = 17.7N ? 17.5N ? 8. m ? mass of child R – mg cos 45° = 0 ? R = mg cos 45° = mg /v 2 ...(1) Net force acting on the boy due to which it slides down is mg sin 45° - ?R = mg sin 45° - ? mg cos 45° = m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 ) = m [(5/ 2 ) – 0.6 × (5 / 2 )] = m(2 2 ) acceleration = mass Force = m ) 2 2 ( m = 2 2 m/s 2 ? 9. Suppose, the body is accelerating down with acceleration ‘a’. From the free body diagram R – mg cos ? = 0 ? R = mg cos ? ...(1) ma + mg sin ? – ? R = 0 ? a = m ) cos (sin mg ? ? ? ? = g (sin ? – ? cos ?) For the first half mt. u = 0, s = 0.5m, t = 0.5 sec. So, v = u + at = 0 + (0.5)4 = 2 m/s S = ut + ½ at 2 ? 0.5 = 0 + ½ a (0/5) 2 ? a = 4m/s 2 ...(2) For the next half metre u` = 2m/s, a = 4m/s 2 , s= 0.5. ? 0.5 = 2t + (1/2) 4 t 2 ? 2 t 2 + 2 t – 0.5 =0 (body moving us) ?R R ??? F mg (body moving down) ?R R ??? mg R 45° ?R mg F 30° ?R R 30° mg mg ma ?R R Chapter 6 6.3 ? 4 t 2 + 4 t – 1 = 0 ? = 4 2 16 16 4 ? ? ? ? = 8 656 . 1 = 0.207sec Time taken to cover next half meter is 0.21sec. 10. f ? applied force F i ? contact force F ? frictional force R ? normal reaction ? = tan ? = F/R When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction ( ?R) Before reaching limiting friction F < ?R ? tan ? = R R R F ? ? ? tan ? ? ? ? ??? tan –1 ? ? 11. From the free body diagram T + 0.5a – 0.5 g = 0 ...(1) ?R + 1a + T 1 – T = 0 ...(2) ?R + 1a – T 1 = 0 ?R + 1a = T 1 ...(3) From (2) & (3) ? ?R + a = T – T 1 ? T – T 1 = T 1 ? T = 2T 1 Equation (2) becomes ?R + a + T 1 – 2T 1 = 0 ? ?R + a – T 1 = 0 ? T 1 = ?R + a = 0.2g + a ...(4) Equation (1) becomes 2T 1 + 0/5a – 0.5g = 0 ? T 1 = 2 a 5 . 0 g 5 . 0 ? = 0.25g – 0.25a ...(5) From (4) & (5) 0.2g + a = 0.25g – 0.25a ? a = 25 . 1 05 . 0 × 10 = 0.04 I 10 = 0.4m/s 2 a) Accln of 1kg blocks each is 0.4m/s 2 b) Tension T 1 = 0.2g + a + 0.4 = 2.4N c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N 12. From the free body diagram ? 1 R + 1 – 16 = 0 ? ? 1 (2g) + (–15) = 0 ? ? 1 = 15/20 = 0.75 ? 2 R 1 + 4 × 0.5 + 16 – 4g sin 30° = 0 ?? 2 (20 3 ) + 2 + 16 – 20 = 0 ? ? 2 = 3 20 2 = 32 . 17 1 = 0.057 ? 0.06 ?Co-efficient of friction ? 1 = 0.75 & ? 2 = 0.06 ? F f ? ? Fi R Limiting Friction ?R T 1 R A 1g 1a ? ?R R 1g 1a ? 0.5g 0.5g a B A ?=0.2 0.5kg 1kg 1kg ?=0.2 2×0.5 16N ?R 1 4g 16N=T ? 2R 4×0.5 30° a ? 1 0.5 m/s 2 2kg 4kg ? 2 Page 4 6.1 SOLUTIONS TO CONCEPTS CHAPTER 6 1. Let m = mass of the block From the freebody diagram, R – mg = 0 ? R = mg ...(1) Again ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4 The co-efficient of kinetic friction between the block and the plane is 0.4 ? 2. Due to friction the body will decelerate Let the deceleration be ‘a’ R – mg = 0 ? R = mg ...(1) ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g = 0.1 × 10 = 1m/s 2. Initial velocity u = 10 m/s Final velocity v = 0 m/s a = –1m/s 2 (deceleration) S = a 2 u v 2 2 ? = ) 1 ( 2 10 0 2 ? ? = 2 100 = 50m It will travel 50m before coming to rest. 3. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f ? frictional force F ? Applied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, R – mg cos ? = 0 ? R = mg cos ? ..(1) For the block U = 0, s = 8m, t = 2sec. ?s = ut + ½ at 2 ? 8 = 0 + ½ a 2 2 ? a = 4m/s 2 Again, ?R + ma – mg sin ? = 0 ? ? mg cos ? + ma – mg sin ? = 0 [from (1)] ? m( ?g cos ? + a – g sin ?) = 0 ? ? × 10 × cos 30° = g sin 30° – a ??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4 ? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11 ? Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram 4 – 4a – ?R + 4g sin 30° = 0 …(1) R – 4g cos 30° = 0 ...(2) ? R = 4g cos 30° Putting the values of R is & in equn. (1) 4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0 ? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0 ? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s 2 For the block u =0, t = 2sec, a = 5m/s 2 Distance s = ut + ½ at 2 ? s = 0 + (1/2) 5 × 2 2 = 10m The block will move 10m. R mg ma ?R a R mg ma ?R a velocity F p o mg R 30° R ??? ?R mg ma 4kg 30° 4N R ??? ?R mg ma a velocity Chapter 6 6.2 6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline = ? R + 2 g sin 30° = 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)] = 3.39 + 9.8 = 13N With this minimum force the body move up the incline with a constant velocity as net force on it is zero. b) Net force acting down the incline is given by, F = 2 g sin 30° – ?R = 2 × 9.8 × (1/2) – 3.39 = 6.41N Due to F = 6.41N the body will move down the incline with acceleration. No external force is required. ? Force required is zero. 7. From the free body diagram g = 10m/s 2 , m = 2kg, ? = 30°, ? = 0.2 R – mg cos ? - F sin ? = 0 ? R = mg cos ? + F sin ?? ...(1) And mg sin ? + ?R – F cos ? = 0 ? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0 ? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0 ? F = ) cos sin ( ) cos mg sin mg ( ? ? ? ? ? ? ? ? ? F = ) 2 / 3 ( ) 2 / 1 ( 2 . 0 ) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2 ? ? ? ? ? ? ? ? = 76 . 0 464 . 13 = 17.7N ? 17.5N ? 8. m ? mass of child R – mg cos 45° = 0 ? R = mg cos 45° = mg /v 2 ...(1) Net force acting on the boy due to which it slides down is mg sin 45° - ?R = mg sin 45° - ? mg cos 45° = m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 ) = m [(5/ 2 ) – 0.6 × (5 / 2 )] = m(2 2 ) acceleration = mass Force = m ) 2 2 ( m = 2 2 m/s 2 ? 9. Suppose, the body is accelerating down with acceleration ‘a’. From the free body diagram R – mg cos ? = 0 ? R = mg cos ? ...(1) ma + mg sin ? – ? R = 0 ? a = m ) cos (sin mg ? ? ? ? = g (sin ? – ? cos ?) For the first half mt. u = 0, s = 0.5m, t = 0.5 sec. So, v = u + at = 0 + (0.5)4 = 2 m/s S = ut + ½ at 2 ? 0.5 = 0 + ½ a (0/5) 2 ? a = 4m/s 2 ...(2) For the next half metre u` = 2m/s, a = 4m/s 2 , s= 0.5. ? 0.5 = 2t + (1/2) 4 t 2 ? 2 t 2 + 2 t – 0.5 =0 (body moving us) ?R R ??? F mg (body moving down) ?R R ??? mg R 45° ?R mg F 30° ?R R 30° mg mg ma ?R R Chapter 6 6.3 ? 4 t 2 + 4 t – 1 = 0 ? = 4 2 16 16 4 ? ? ? ? = 8 656 . 1 = 0.207sec Time taken to cover next half meter is 0.21sec. 10. f ? applied force F i ? contact force F ? frictional force R ? normal reaction ? = tan ? = F/R When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction ( ?R) Before reaching limiting friction F < ?R ? tan ? = R R R F ? ? ? tan ? ? ? ? ??? tan –1 ? ? 11. From the free body diagram T + 0.5a – 0.5 g = 0 ...(1) ?R + 1a + T 1 – T = 0 ...(2) ?R + 1a – T 1 = 0 ?R + 1a = T 1 ...(3) From (2) & (3) ? ?R + a = T – T 1 ? T – T 1 = T 1 ? T = 2T 1 Equation (2) becomes ?R + a + T 1 – 2T 1 = 0 ? ?R + a – T 1 = 0 ? T 1 = ?R + a = 0.2g + a ...(4) Equation (1) becomes 2T 1 + 0/5a – 0.5g = 0 ? T 1 = 2 a 5 . 0 g 5 . 0 ? = 0.25g – 0.25a ...(5) From (4) & (5) 0.2g + a = 0.25g – 0.25a ? a = 25 . 1 05 . 0 × 10 = 0.04 I 10 = 0.4m/s 2 a) Accln of 1kg blocks each is 0.4m/s 2 b) Tension T 1 = 0.2g + a + 0.4 = 2.4N c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N 12. From the free body diagram ? 1 R + 1 – 16 = 0 ? ? 1 (2g) + (–15) = 0 ? ? 1 = 15/20 = 0.75 ? 2 R 1 + 4 × 0.5 + 16 – 4g sin 30° = 0 ?? 2 (20 3 ) + 2 + 16 – 20 = 0 ? ? 2 = 3 20 2 = 32 . 17 1 = 0.057 ? 0.06 ?Co-efficient of friction ? 1 = 0.75 & ? 2 = 0.06 ? F f ? ? Fi R Limiting Friction ?R T 1 R A 1g 1a ? ?R R 1g 1a ? 0.5g 0.5g a B A ?=0.2 0.5kg 1kg 1kg ?=0.2 2×0.5 16N ?R 1 4g 16N=T ? 2R 4×0.5 30° a ? 1 0.5 m/s 2 2kg 4kg ? 2 Chapter 6 6.4 13. From the free body diagram T + 15a – 15g = 0 T – (T 1 + 5a+ ?R)= 0 T 1 – 5g – 5a = 0 ? ? T = 15g – 15 a ...(i) ? T – (5g + 5a + 5a + ?R) = 0 ?T 1 =5g + 5a …(iii) ? T = 5g + 10a + ?R …(ii) ? From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g) ? 25a = 90 ? a = 3.6m/s 2 Equation (ii) ? T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 ? 96N in the left string Equation (iii) T 1 = 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string. 14. s = 5m, ? = 4/3, g = 10m/s 2 u = 36km/h = 10m/s, v = 0, a = s 2 u v 2 2 ? = 5 2 10 0 2 ? ? = –10m/s 2 From the freebody diagrams, R – mg cos ? = 0 ; g = 10m/s 2 ? R = mg cos ??….(i) ; ? = 4/3. Again, ma + mg sin ? - ? R = 0 ? ma + mg sin ? – ? mg cos ? = 0 ??a + g sin ? – mg cos ? = 0 ? 10 + 10 sin ? - (4/3) × 10 cos ? = 0 ? 30 + 30 sin ? – 40 cos ? =0 ? 3 + 3 sin ? – 4 cos ? = 0 ? 4 cos ? - 3 sin ? = 3 ? 4 ? ? 2 sin 1 = 3 + 3 sin ? ? 16 (1 – sin 2 ?) = 9 + 9 sin 2 ??+ 18 sin ? sin ? = 25 2 ) 7 )( 25 ( 4 18 18 2 ? ? ? ? ? = 50 32 18 ? ? = 50 14 = 0.28 [Taking +ve sign only] ? ? = sin –1 (0.28) = 16° Maximum incline is ? = 16° ? 15. to reach in minimum time, he has to move with maximum possible acceleration. Let, the maximum acceleration is ‘a’ ? ma – ?R = 0 ? ma = ? mg ? a = ? g = 0.9 × 10 = 9m/s 2 a) Initial velocity u = 0, t = ? a = 9m/s 2 , s = 50m s = ut + ½ at 2 ? 50 = 0 + (1/2) 9 t 2 ? t = 9 100 = 3 10 sec. b) After overing 50m, velocity of the athelete is V = u + at = 0 + 9 × (10/3) = 30m/s He has to stop in minimum time. So deceleration ia –a = –9m/s 2 (max) T 1 5g 5a T 15g 15a A B R 5g T 1 T r=5g ?R C B A a 15kg 15kg a a R ma mg ?R a ? ? velocity ??? the max. angle ?R a R mg ma ?R a R mg ma Page 5 6.1 SOLUTIONS TO CONCEPTS CHAPTER 6 1. Let m = mass of the block From the freebody diagram, R – mg = 0 ? R = mg ...(1) Again ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g ? 4 = ?g ? ? = 4/g = 4/10 = 0.4 The co-efficient of kinetic friction between the block and the plane is 0.4 ? 2. Due to friction the body will decelerate Let the deceleration be ‘a’ R – mg = 0 ? R = mg ...(1) ma – ? R = 0 ? ma = ? R = ? mg (from (1)) ? a = ?g = 0.1 × 10 = 1m/s 2. Initial velocity u = 10 m/s Final velocity v = 0 m/s a = –1m/s 2 (deceleration) S = a 2 u v 2 2 ? = ) 1 ( 2 10 0 2 ? ? = 2 100 = 50m It will travel 50m before coming to rest. 3. Body is kept on the horizontal table. If no force is applied, no frictional force will be there f ? frictional force F ? Applied force From grap it can be seen that when applied force is zero, frictional force is zero. 4. From the free body diagram, R – mg cos ? = 0 ? R = mg cos ? ..(1) For the block U = 0, s = 8m, t = 2sec. ?s = ut + ½ at 2 ? 8 = 0 + ½ a 2 2 ? a = 4m/s 2 Again, ?R + ma – mg sin ? = 0 ? ? mg cos ? + ma – mg sin ? = 0 [from (1)] ? m( ?g cos ? + a – g sin ?) = 0 ? ? × 10 × cos 30° = g sin 30° – a ??? × 10 × ) 3 / 3 ( = 10 × (1/2) – 4 ? ) 3 / 5 ( ? =1 ? ? = 1/ ) 3 / 5 ( = 0.11 ? Co-efficient of kinetic friction between the two is 0.11. 5. From the free body diagram 4 – 4a – ?R + 4g sin 30° = 0 …(1) R – 4g cos 30° = 0 ...(2) ? R = 4g cos 30° Putting the values of R is & in equn. (1) 4 – 4a – 0.11 × 4g cos 30° + 4g sin 30° = 0 ? 4 – 4a – 0.11 × 4 × 10 × ( 2 / 3 ) + 4 × 10 × (1/2) = 0 ? 4 – 4a – 3.81 + 20 = 0 ? a ? 5 m/s 2 For the block u =0, t = 2sec, a = 5m/s 2 Distance s = ut + ½ at 2 ? s = 0 + (1/2) 5 × 2 2 = 10m The block will move 10m. R mg ma ?R a R mg ma ?R a velocity F p o mg R 30° R ??? ?R mg ma 4kg 30° 4N R ??? ?R mg ma a velocity Chapter 6 6.2 6. To make the block move up the incline, the force should be equal and opposite to the net force acting down the incline = ? R + 2 g sin 30° = 0.2 × (9.8) 3 + 2 I 9.8 × (1/2) [from (1)] = 3.39 + 9.8 = 13N With this minimum force the body move up the incline with a constant velocity as net force on it is zero. b) Net force acting down the incline is given by, F = 2 g sin 30° – ?R = 2 × 9.8 × (1/2) – 3.39 = 6.41N Due to F = 6.41N the body will move down the incline with acceleration. No external force is required. ? Force required is zero. 7. From the free body diagram g = 10m/s 2 , m = 2kg, ? = 30°, ? = 0.2 R – mg cos ? - F sin ? = 0 ? R = mg cos ? + F sin ?? ...(1) And mg sin ? + ?R – F cos ? = 0 ? mg sin ? + ?(mg cos ? + F sin ?) – F cos ? = 0 ? mg sin ? + ? mg cos ? + ? F sin ? – F cos ? = 0 ? F = ) cos sin ( ) cos mg sin mg ( ? ? ? ? ? ? ? ? ? F = ) 2 / 3 ( ) 2 / 1 ( 2 . 0 ) 2 / 3 ( 10 2 2 . 0 ) 2 / 1 ( 10 2 ? ? ? ? ? ? ? ? = 76 . 0 464 . 13 = 17.7N ? 17.5N ? 8. m ? mass of child R – mg cos 45° = 0 ? R = mg cos 45° = mg /v 2 ...(1) Net force acting on the boy due to which it slides down is mg sin 45° - ?R = mg sin 45° - ? mg cos 45° = m × 10 (1/ 2 ) – 0.6 × m × 10 × (1/ 2 ) = m [(5/ 2 ) – 0.6 × (5 / 2 )] = m(2 2 ) acceleration = mass Force = m ) 2 2 ( m = 2 2 m/s 2 ? 9. Suppose, the body is accelerating down with acceleration ‘a’. From the free body diagram R – mg cos ? = 0 ? R = mg cos ? ...(1) ma + mg sin ? – ? R = 0 ? a = m ) cos (sin mg ? ? ? ? = g (sin ? – ? cos ?) For the first half mt. u = 0, s = 0.5m, t = 0.5 sec. So, v = u + at = 0 + (0.5)4 = 2 m/s S = ut + ½ at 2 ? 0.5 = 0 + ½ a (0/5) 2 ? a = 4m/s 2 ...(2) For the next half metre u` = 2m/s, a = 4m/s 2 , s= 0.5. ? 0.5 = 2t + (1/2) 4 t 2 ? 2 t 2 + 2 t – 0.5 =0 (body moving us) ?R R ??? F mg (body moving down) ?R R ??? mg R 45° ?R mg F 30° ?R R 30° mg mg ma ?R R Chapter 6 6.3 ? 4 t 2 + 4 t – 1 = 0 ? = 4 2 16 16 4 ? ? ? ? = 8 656 . 1 = 0.207sec Time taken to cover next half meter is 0.21sec. 10. f ? applied force F i ? contact force F ? frictional force R ? normal reaction ? = tan ? = F/R When F = ?R, F is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction ( ?R) Before reaching limiting friction F < ?R ? tan ? = R R R F ? ? ? tan ? ? ? ? ??? tan –1 ? ? 11. From the free body diagram T + 0.5a – 0.5 g = 0 ...(1) ?R + 1a + T 1 – T = 0 ...(2) ?R + 1a – T 1 = 0 ?R + 1a = T 1 ...(3) From (2) & (3) ? ?R + a = T – T 1 ? T – T 1 = T 1 ? T = 2T 1 Equation (2) becomes ?R + a + T 1 – 2T 1 = 0 ? ?R + a – T 1 = 0 ? T 1 = ?R + a = 0.2g + a ...(4) Equation (1) becomes 2T 1 + 0/5a – 0.5g = 0 ? T 1 = 2 a 5 . 0 g 5 . 0 ? = 0.25g – 0.25a ...(5) From (4) & (5) 0.2g + a = 0.25g – 0.25a ? a = 25 . 1 05 . 0 × 10 = 0.04 I 10 = 0.4m/s 2 a) Accln of 1kg blocks each is 0.4m/s 2 b) Tension T 1 = 0.2g + a + 0.4 = 2.4N c) T = 0.5g – 0.5a = 0.5 × 10 – 0.5 × 0.4 = 4.8N 12. From the free body diagram ? 1 R + 1 – 16 = 0 ? ? 1 (2g) + (–15) = 0 ? ? 1 = 15/20 = 0.75 ? 2 R 1 + 4 × 0.5 + 16 – 4g sin 30° = 0 ?? 2 (20 3 ) + 2 + 16 – 20 = 0 ? ? 2 = 3 20 2 = 32 . 17 1 = 0.057 ? 0.06 ?Co-efficient of friction ? 1 = 0.75 & ? 2 = 0.06 ? F f ? ? Fi R Limiting Friction ?R T 1 R A 1g 1a ? ?R R 1g 1a ? 0.5g 0.5g a B A ?=0.2 0.5kg 1kg 1kg ?=0.2 2×0.5 16N ?R 1 4g 16N=T ? 2R 4×0.5 30° a ? 1 0.5 m/s 2 2kg 4kg ? 2 Chapter 6 6.4 13. From the free body diagram T + 15a – 15g = 0 T – (T 1 + 5a+ ?R)= 0 T 1 – 5g – 5a = 0 ? ? T = 15g – 15 a ...(i) ? T – (5g + 5a + 5a + ?R) = 0 ?T 1 =5g + 5a …(iii) ? T = 5g + 10a + ?R …(ii) ? From (i) & (ii) 15g – 15a = 5g + 10a + 0.2 (5g) ? 25a = 90 ? a = 3.6m/s 2 Equation (ii) ? T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 ? 96N in the left string Equation (iii) T 1 = 5g + 5a = 5 × 10 + 5 × 3.6 =68N in the right string. 14. s = 5m, ? = 4/3, g = 10m/s 2 u = 36km/h = 10m/s, v = 0, a = s 2 u v 2 2 ? = 5 2 10 0 2 ? ? = –10m/s 2 From the freebody diagrams, R – mg cos ? = 0 ; g = 10m/s 2 ? R = mg cos ??….(i) ; ? = 4/3. Again, ma + mg sin ? - ? R = 0 ? ma + mg sin ? – ? mg cos ? = 0 ??a + g sin ? – mg cos ? = 0 ? 10 + 10 sin ? - (4/3) × 10 cos ? = 0 ? 30 + 30 sin ? – 40 cos ? =0 ? 3 + 3 sin ? – 4 cos ? = 0 ? 4 cos ? - 3 sin ? = 3 ? 4 ? ? 2 sin 1 = 3 + 3 sin ? ? 16 (1 – sin 2 ?) = 9 + 9 sin 2 ??+ 18 sin ? sin ? = 25 2 ) 7 )( 25 ( 4 18 18 2 ? ? ? ? ? = 50 32 18 ? ? = 50 14 = 0.28 [Taking +ve sign only] ? ? = sin –1 (0.28) = 16° Maximum incline is ? = 16° ? 15. to reach in minimum time, he has to move with maximum possible acceleration. Let, the maximum acceleration is ‘a’ ? ma – ?R = 0 ? ma = ? mg ? a = ? g = 0.9 × 10 = 9m/s 2 a) Initial velocity u = 0, t = ? a = 9m/s 2 , s = 50m s = ut + ½ at 2 ? 50 = 0 + (1/2) 9 t 2 ? t = 9 100 = 3 10 sec. b) After overing 50m, velocity of the athelete is V = u + at = 0 + 9 × (10/3) = 30m/s He has to stop in minimum time. So deceleration ia –a = –9m/s 2 (max) T 1 5g 5a T 15g 15a A B R 5g T 1 T r=5g ?R C B A a 15kg 15kg a a R ma mg ?R a ? ? velocity ??? the max. angle ?R a R mg ma ?R a R mg ma Chapter 6 6.5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) on Decelerati ( s / m 9 g a ) force frictional (max R ma ma R 2 ? u 1 = 30m/s, v 1 = 0 t = a u v 1 1 ? = a 30 0 ? ? = a 30 ? ? = 3 10 sec. 16. Hardest brake means maximum force of friction is developed between car’s type & road. Max frictional force = ?R From the free body diagram R – mg cos ? =0 ? R = mg cos ? ...(i) and ?R + ma – mg sin ) = 0 …(ii) ? ?mg cos ? + ma – mg sin ? = 0 ? ?g cos ? + a – 10 × (1/2) = 0 ??a = 5 – {1 – (2 3 )} × 10 ( 2 / 3 ) = 2.5 m/s 2 When, hardest brake is applied the car move with acceleration 2.5m/s 2 S = 12.8m, u = 6m/s S0, velocity at the end of incline V = as 2 u 2 ? = ) 8 . 12 )( 5 . 2 ( 2 6 2 ? = 64 36 ? = 10m/s = 36km/h Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36km/h. 17. Let, , a maximum acceleration produced in car. ? ma = ?R [For more acceleration, the tyres will slip] ? ma = ? mg ? a = ?g = 1 × 10 = 10m/s 2 For crossing the bridge in minimum time, it has to travel with maximum acceleration u = 0, s = 500m, a = 10m/s 2 s = ut + ½ at 2 ? 500 = 0 + (1/2) 10 t 2 ? t = 10 sec. If acceleration is less than 10m/s 2 , time will be more than 10sec. So one can’t drive through the bridge in less than 10sec. ? 18. From the free body diagram R = 4g cos 30° = 4 × 10 × 2 / 3 = 20 3 ...(i) ? 2 R + 4a – P – 4g sin 30° = 0 ? 0.3 (40) cos 30° + 4a – P – 40 sin 20° = 0 …(ii) P + 2a + ? 1 R 1 – 2g sin 30° = 0 …(iii) R 1 = 2g cos 30° = 2 × 10 × 2 / 3 = 10 3 ...(iv) Equn. (ii) 6 3 + 4a – P – 20 = 0 Equn (iv) P + 2a + 2 3 – 10 = 0 From Equn (ii) & (iv) 6 3 + 6a – 30 + 2 3 = 0 ? 6a = 30 – 8 3 = 30 – 13.85 = 16.15 ? a = 6 15 . 16 = 2.69 = 2.7m/s 2 b) can be solved. In this case, the 4 kg block will travel with more acceleration because, coefficient of friction is less than that of 2kg. So, they will move separately. Drawing the free body diagram of 2kg mass only, it can be found that, a = 2.4m/s 2 . ? ma R a mg ?R ?R a mg R 4kg 30° 2kg a ?a R 4g ? ?R P 2a R 2g P ? 1 R 1 ?Read More

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