Page 1 8.1 SOLUTIONS TO CONCEPTS CHAPTER – 8 1. M = m c + m b = 90kg u = 6 km/h = 1.666 m/sec v = 12 km/h = 3.333 m/sec Increase in K.E. = ½ Mv 2 – ½ Mu 2 = ½ 90 × (3.333) 2 – ½ × 90 × (1.66) 2 = 494.5 – 124.6 = 374.8 ? 375 J 2. m b = 2 kg. u = 10 m/sec a = 3 m/aec 2 t = 5 sec v = u + at = 10 + 3 I 5 = 25 m/sec. ?F.K.E = ½ mv 2 = ½ × 2 × 625 = 625 J. 3. F = 100 N S = 4m, ? = 0° ? = S . F ? ? ???100 × 4 = 400 J ? 4. m = 5 kg ? = 30° S = 10 m F = mg So, work done by the force of gravity ? = mgh = 5 × 9.8 × 5 = 245 J ? 5. F= 2.50N, S = 2.5m, m =15g = 0.015kg. So, w = F × S ? a = m F = 015 . 0 5 . 2 = 3 500 m/s 2 =F × S cos 0° (acting along the same line) = 2.5 × 2.5 = 6.25J Let the velocity of the body at b = U. Applying work-energy principle ½ mv 2 – 0 = 6.25 ? V = 015 . 0 2 25 . 6 ? = 28.86 m/sec. So, time taken to travel from A to B. ? t = a u v ? = 500 3 86 . 28 ? ? Average power = t W = 3 ) 86 . 28 ( 500 25 . 6 ? ? = 36.1 6. Given j ˆ 3 i ˆ 2 r 1 ? ? ? j ˆ 2 i ˆ 3 r 2 ? ? So, displacement vector is given by, 2 1 r r r ? ? ? ? ? ? j ˆ i ˆ ) j ˆ 3 i ˆ 2 ( ) j ˆ 2 i ˆ 3 ( r ? ? ? ? ? ? ? u=1.66 m/s ? 90kg ? 90kg ? v=3.33 m/s ? u=10 m/s ? 2 kg ? a ? = 3m/s 2 F ? 4m ? R ? 100 N ? mg ? 30° ? 5 ? mg ? F ? 5 log ? 10m ? 30° ? v ? ? A ? B ? Page 2 8.1 SOLUTIONS TO CONCEPTS CHAPTER – 8 1. M = m c + m b = 90kg u = 6 km/h = 1.666 m/sec v = 12 km/h = 3.333 m/sec Increase in K.E. = ½ Mv 2 – ½ Mu 2 = ½ 90 × (3.333) 2 – ½ × 90 × (1.66) 2 = 494.5 – 124.6 = 374.8 ? 375 J 2. m b = 2 kg. u = 10 m/sec a = 3 m/aec 2 t = 5 sec v = u + at = 10 + 3 I 5 = 25 m/sec. ?F.K.E = ½ mv 2 = ½ × 2 × 625 = 625 J. 3. F = 100 N S = 4m, ? = 0° ? = S . F ? ? ???100 × 4 = 400 J ? 4. m = 5 kg ? = 30° S = 10 m F = mg So, work done by the force of gravity ? = mgh = 5 × 9.8 × 5 = 245 J ? 5. F= 2.50N, S = 2.5m, m =15g = 0.015kg. So, w = F × S ? a = m F = 015 . 0 5 . 2 = 3 500 m/s 2 =F × S cos 0° (acting along the same line) = 2.5 × 2.5 = 6.25J Let the velocity of the body at b = U. Applying work-energy principle ½ mv 2 – 0 = 6.25 ? V = 015 . 0 2 25 . 6 ? = 28.86 m/sec. So, time taken to travel from A to B. ? t = a u v ? = 500 3 86 . 28 ? ? Average power = t W = 3 ) 86 . 28 ( 500 25 . 6 ? ? = 36.1 6. Given j ˆ 3 i ˆ 2 r 1 ? ? ? j ˆ 2 i ˆ 3 r 2 ? ? So, displacement vector is given by, 2 1 r r r ? ? ? ? ? ? j ˆ i ˆ ) j ˆ 3 i ˆ 2 ( ) j ˆ 2 i ˆ 3 ( r ? ? ? ? ? ? ? u=1.66 m/s ? 90kg ? 90kg ? v=3.33 m/s ? u=10 m/s ? 2 kg ? a ? = 3m/s 2 F ? 4m ? R ? 100 N ? mg ? 30° ? 5 ? mg ? F ? 5 log ? 10m ? 30° ? v ? ? A ? B ? Chapter 8 8.2 So, work done = s F ? ? ? = 5 × 1 + 5(-1) = 0 7. m b = 2kg, s = 40m, a = 0.5m/sec 2 So, force applied by the man on the box F = m b a = 2 × (0.5) = 1 N ? = FS = 1 × 40 = 40 J ? 8. Given that F= a + bx Where a and b are constants. So, work done by this force during this force during the displacement x = 0 and x = d is given by W = ? ? ? ? d 0 d 0 dx ) bx a ( dx F = ax + (bx 2 /2) = [a + ½ bd] d 9. m b = 250g = .250 kg ? = 37°, S = 1m. Frictional force f = ?R mg sin ? = ? R ..(1) mg cos ?? ..(2) so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J 10. a = ) m M ( 2 F ? (given) a) from fig (1) ma = ? k R 1 and R 1 = mg ? ? = 1 R ma = g ) m M ( 2 F ? b) Frictional force acting on the smaller block f = ?R = ) m M ( 2 F m mg g ) m M ( 2 F ? ? ? ? ? c) Work done w = fs s = d w = d ) m M ( 2 mF ? ? = ) m M ( 2 mFd ? ? 11. Weight = 2000 N, S = 20m, ? = 0.2 a) R + Psin ? - 2000 = 0 ..(1) P cos ? - 0.2 R =0 ..(2) From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0 P = ? ? ? sin 2 . 0 cos 400 ..(3) So, work done by the person, W = PS cos ? = ? ? ? ? sin 2 . 0 cos cos 8000 = ? ? sin 2 . 0 1 8000 = ? ? tan 5 40000 b) For minimum magnitude of force from equn(1) d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2 putting the value in equn (3) W = ? ? tan 5 40000 = ) 2 . 5 ( 40000 = 7690 J 12. w = 100 N, ? = 37°, s = 2m R ? m b g ? m b a ? F ? R ? ?R ? 1 m ? mg ? 37° ? M ? F ? m ? R 1 ? ? k R 1 ? ma ? mg ? ?R 1 ? R 2 ? f ? ma ? mg ? ?R 2 ? ? ? R ? P ? 0.2R ? 2000 N ? Page 3 8.1 SOLUTIONS TO CONCEPTS CHAPTER – 8 1. M = m c + m b = 90kg u = 6 km/h = 1.666 m/sec v = 12 km/h = 3.333 m/sec Increase in K.E. = ½ Mv 2 – ½ Mu 2 = ½ 90 × (3.333) 2 – ½ × 90 × (1.66) 2 = 494.5 – 124.6 = 374.8 ? 375 J 2. m b = 2 kg. u = 10 m/sec a = 3 m/aec 2 t = 5 sec v = u + at = 10 + 3 I 5 = 25 m/sec. ?F.K.E = ½ mv 2 = ½ × 2 × 625 = 625 J. 3. F = 100 N S = 4m, ? = 0° ? = S . F ? ? ???100 × 4 = 400 J ? 4. m = 5 kg ? = 30° S = 10 m F = mg So, work done by the force of gravity ? = mgh = 5 × 9.8 × 5 = 245 J ? 5. F= 2.50N, S = 2.5m, m =15g = 0.015kg. So, w = F × S ? a = m F = 015 . 0 5 . 2 = 3 500 m/s 2 =F × S cos 0° (acting along the same line) = 2.5 × 2.5 = 6.25J Let the velocity of the body at b = U. Applying work-energy principle ½ mv 2 – 0 = 6.25 ? V = 015 . 0 2 25 . 6 ? = 28.86 m/sec. So, time taken to travel from A to B. ? t = a u v ? = 500 3 86 . 28 ? ? Average power = t W = 3 ) 86 . 28 ( 500 25 . 6 ? ? = 36.1 6. Given j ˆ 3 i ˆ 2 r 1 ? ? ? j ˆ 2 i ˆ 3 r 2 ? ? So, displacement vector is given by, 2 1 r r r ? ? ? ? ? ? j ˆ i ˆ ) j ˆ 3 i ˆ 2 ( ) j ˆ 2 i ˆ 3 ( r ? ? ? ? ? ? ? u=1.66 m/s ? 90kg ? 90kg ? v=3.33 m/s ? u=10 m/s ? 2 kg ? a ? = 3m/s 2 F ? 4m ? R ? 100 N ? mg ? 30° ? 5 ? mg ? F ? 5 log ? 10m ? 30° ? v ? ? A ? B ? Chapter 8 8.2 So, work done = s F ? ? ? = 5 × 1 + 5(-1) = 0 7. m b = 2kg, s = 40m, a = 0.5m/sec 2 So, force applied by the man on the box F = m b a = 2 × (0.5) = 1 N ? = FS = 1 × 40 = 40 J ? 8. Given that F= a + bx Where a and b are constants. So, work done by this force during this force during the displacement x = 0 and x = d is given by W = ? ? ? ? d 0 d 0 dx ) bx a ( dx F = ax + (bx 2 /2) = [a + ½ bd] d 9. m b = 250g = .250 kg ? = 37°, S = 1m. Frictional force f = ?R mg sin ? = ? R ..(1) mg cos ?? ..(2) so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J 10. a = ) m M ( 2 F ? (given) a) from fig (1) ma = ? k R 1 and R 1 = mg ? ? = 1 R ma = g ) m M ( 2 F ? b) Frictional force acting on the smaller block f = ?R = ) m M ( 2 F m mg g ) m M ( 2 F ? ? ? ? ? c) Work done w = fs s = d w = d ) m M ( 2 mF ? ? = ) m M ( 2 mFd ? ? 11. Weight = 2000 N, S = 20m, ? = 0.2 a) R + Psin ? - 2000 = 0 ..(1) P cos ? - 0.2 R =0 ..(2) From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0 P = ? ? ? sin 2 . 0 cos 400 ..(3) So, work done by the person, W = PS cos ? = ? ? ? ? sin 2 . 0 cos cos 8000 = ? ? sin 2 . 0 1 8000 = ? ? tan 5 40000 b) For minimum magnitude of force from equn(1) d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2 putting the value in equn (3) W = ? ? tan 5 40000 = ) 2 . 5 ( 40000 = 7690 J 12. w = 100 N, ? = 37°, s = 2m R ? m b g ? m b a ? F ? R ? ?R ? 1 m ? mg ? 37° ? M ? F ? m ? R 1 ? ? k R 1 ? ma ? mg ? ?R 1 ? R 2 ? f ? ma ? mg ? ?R 2 ? ? ? R ? P ? 0.2R ? 2000 N ? Chapter 8 8.3 Force F= mg sin 37° = 100 × 0.60 = 60 N So, work done, when the force is parallel to incline. w = Fs cos ? = 60 × 2 × cos ? = 120 J In ?ABC AB= 2m CB = 37° so, h = C = 1m ?work done when the force in horizontal direction W = mgh = 100 × 1.2 = 120 J ? 13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0 (-a) = S 2 u v 2 2 ? ? a = 50 400 = 8m/sec 2 Frictional force f = ma = 500 × 8 = 4000 N 14. m = 500 kg, u = 0, v = 72 km/h = 20m/s a = s 2 u v 2 2 ? = 50 400 = 8m/sec 2 force needed to accelerate the car F = ma = 500 × 8 = 4000 N 15. Given, v = a x (uniformly accelerated motion) displacement s = d – 0 = d putting x = 0, v 1 = 0 putting x = d, v 2 = a d a = s 2 u v 2 2 2 2 ? = d 2 d a 2 = 2 a 2 force f = ma = 2 ma 2 work done w = FS cos ? = d 2 ma 2 ? = 2 d ma 2 ? 16. a) m = 2kg, ? = 37°, F = 20 N From the free body diagram F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec 2 S = ut + ½ at 2 (u = 0, t = 1s, a = 1.66) = 2m So, work, done w = Fs = 20 × 2 = 40 J b) If W = 40 J S = F W = 20 40 h = 2 sin 37° = 1.2 m So, work done W = –mgh = – 20 × 1.2 = –24 J c) v = u + at = 4 × 10 = 40 m/sec So, K.E. = ½ mv 2 = ½ × 2 × 16 = 16 J 17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec 2 a) t = 1sec So, s= ut + ½ at 2 = 5m 37° ? A ? A ? A ? B ? v=0 ? v=20 m/s m=500 kg ? –a ? 25m ? a ? R mg ? ma f ? 500 kg ? a ? 25m ? R mg ? F F ? ma ? ma ? 2g cos ? ? 20N ? R ? ma 2gsin ? ? 20N ? ma ? mg cos ? ? 20N ? R ? mg sin ? ? ?R ? h ? 37° ? 5m ? C ? A ? B ? 37° ? Page 4 8.1 SOLUTIONS TO CONCEPTS CHAPTER – 8 1. M = m c + m b = 90kg u = 6 km/h = 1.666 m/sec v = 12 km/h = 3.333 m/sec Increase in K.E. = ½ Mv 2 – ½ Mu 2 = ½ 90 × (3.333) 2 – ½ × 90 × (1.66) 2 = 494.5 – 124.6 = 374.8 ? 375 J 2. m b = 2 kg. u = 10 m/sec a = 3 m/aec 2 t = 5 sec v = u + at = 10 + 3 I 5 = 25 m/sec. ?F.K.E = ½ mv 2 = ½ × 2 × 625 = 625 J. 3. F = 100 N S = 4m, ? = 0° ? = S . F ? ? ???100 × 4 = 400 J ? 4. m = 5 kg ? = 30° S = 10 m F = mg So, work done by the force of gravity ? = mgh = 5 × 9.8 × 5 = 245 J ? 5. F= 2.50N, S = 2.5m, m =15g = 0.015kg. So, w = F × S ? a = m F = 015 . 0 5 . 2 = 3 500 m/s 2 =F × S cos 0° (acting along the same line) = 2.5 × 2.5 = 6.25J Let the velocity of the body at b = U. Applying work-energy principle ½ mv 2 – 0 = 6.25 ? V = 015 . 0 2 25 . 6 ? = 28.86 m/sec. So, time taken to travel from A to B. ? t = a u v ? = 500 3 86 . 28 ? ? Average power = t W = 3 ) 86 . 28 ( 500 25 . 6 ? ? = 36.1 6. Given j ˆ 3 i ˆ 2 r 1 ? ? ? j ˆ 2 i ˆ 3 r 2 ? ? So, displacement vector is given by, 2 1 r r r ? ? ? ? ? ? j ˆ i ˆ ) j ˆ 3 i ˆ 2 ( ) j ˆ 2 i ˆ 3 ( r ? ? ? ? ? ? ? u=1.66 m/s ? 90kg ? 90kg ? v=3.33 m/s ? u=10 m/s ? 2 kg ? a ? = 3m/s 2 F ? 4m ? R ? 100 N ? mg ? 30° ? 5 ? mg ? F ? 5 log ? 10m ? 30° ? v ? ? A ? B ? Chapter 8 8.2 So, work done = s F ? ? ? = 5 × 1 + 5(-1) = 0 7. m b = 2kg, s = 40m, a = 0.5m/sec 2 So, force applied by the man on the box F = m b a = 2 × (0.5) = 1 N ? = FS = 1 × 40 = 40 J ? 8. Given that F= a + bx Where a and b are constants. So, work done by this force during this force during the displacement x = 0 and x = d is given by W = ? ? ? ? d 0 d 0 dx ) bx a ( dx F = ax + (bx 2 /2) = [a + ½ bd] d 9. m b = 250g = .250 kg ? = 37°, S = 1m. Frictional force f = ?R mg sin ? = ? R ..(1) mg cos ?? ..(2) so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J 10. a = ) m M ( 2 F ? (given) a) from fig (1) ma = ? k R 1 and R 1 = mg ? ? = 1 R ma = g ) m M ( 2 F ? b) Frictional force acting on the smaller block f = ?R = ) m M ( 2 F m mg g ) m M ( 2 F ? ? ? ? ? c) Work done w = fs s = d w = d ) m M ( 2 mF ? ? = ) m M ( 2 mFd ? ? 11. Weight = 2000 N, S = 20m, ? = 0.2 a) R + Psin ? - 2000 = 0 ..(1) P cos ? - 0.2 R =0 ..(2) From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0 P = ? ? ? sin 2 . 0 cos 400 ..(3) So, work done by the person, W = PS cos ? = ? ? ? ? sin 2 . 0 cos cos 8000 = ? ? sin 2 . 0 1 8000 = ? ? tan 5 40000 b) For minimum magnitude of force from equn(1) d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2 putting the value in equn (3) W = ? ? tan 5 40000 = ) 2 . 5 ( 40000 = 7690 J 12. w = 100 N, ? = 37°, s = 2m R ? m b g ? m b a ? F ? R ? ?R ? 1 m ? mg ? 37° ? M ? F ? m ? R 1 ? ? k R 1 ? ma ? mg ? ?R 1 ? R 2 ? f ? ma ? mg ? ?R 2 ? ? ? R ? P ? 0.2R ? 2000 N ? Chapter 8 8.3 Force F= mg sin 37° = 100 × 0.60 = 60 N So, work done, when the force is parallel to incline. w = Fs cos ? = 60 × 2 × cos ? = 120 J In ?ABC AB= 2m CB = 37° so, h = C = 1m ?work done when the force in horizontal direction W = mgh = 100 × 1.2 = 120 J ? 13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0 (-a) = S 2 u v 2 2 ? ? a = 50 400 = 8m/sec 2 Frictional force f = ma = 500 × 8 = 4000 N 14. m = 500 kg, u = 0, v = 72 km/h = 20m/s a = s 2 u v 2 2 ? = 50 400 = 8m/sec 2 force needed to accelerate the car F = ma = 500 × 8 = 4000 N 15. Given, v = a x (uniformly accelerated motion) displacement s = d – 0 = d putting x = 0, v 1 = 0 putting x = d, v 2 = a d a = s 2 u v 2 2 2 2 ? = d 2 d a 2 = 2 a 2 force f = ma = 2 ma 2 work done w = FS cos ? = d 2 ma 2 ? = 2 d ma 2 ? 16. a) m = 2kg, ? = 37°, F = 20 N From the free body diagram F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec 2 S = ut + ½ at 2 (u = 0, t = 1s, a = 1.66) = 2m So, work, done w = Fs = 20 × 2 = 40 J b) If W = 40 J S = F W = 20 40 h = 2 sin 37° = 1.2 m So, work done W = –mgh = – 20 × 1.2 = –24 J c) v = u + at = 4 × 10 = 40 m/sec So, K.E. = ½ mv 2 = ½ × 2 × 16 = 16 J 17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec 2 a) t = 1sec So, s= ut + ½ at 2 = 5m 37° ? A ? A ? A ? B ? v=0 ? v=20 m/s m=500 kg ? –a ? 25m ? a ? R mg ? ma f ? 500 kg ? a ? 25m ? R mg ? F F ? ma ? ma ? 2g cos ? ? 20N ? R ? ma 2gsin ? ? 20N ? ma ? mg cos ? ? 20N ? R ? mg sin ? ? ?R ? h ? 37° ? 5m ? C ? A ? B ? 37° ? Chapter 8 8.4 Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J b) BC (h) = 5 sin 37° = 3m So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J c) So, frictional force f = mg sin ? work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5 = 60 J ? 18. Given, m = 25o g = 0.250kg, u = 40 cm/sec = 0.4m/sec ? = 0.1, v=0 Here, ? R = ma {where, a = deceleration} a = m R ? = m mg ? = ?g = 0.1 × 9.8 = 0.98 m/sec 2 S = a 2 u v 2 2 ? = 0.082m = 8.2 cm Again, work done against friction is given by – w = ? RS cos ? = 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J ? W = – 0.02 J ? 19. h = 50m, m = 1.8 × 10 5 kg/hr, P = 100 watt, P.E. = mgh = 1.8 × 10 5 × 9.8 × 50 = 882 × 10 5 J/hr Because, half the potential energy is converted into electricity, Electrical energy ½ P.E. = 441 × 10 5 J/hr So, power in watt (J/sec) is given by = 3600 10 441 5 ? ? number of 100 W lamps, that can be lit 100 3600 10 441 5 ? ? = 122.5 ?122 20. m = 6kg, h = 2m P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J P.E. at floor = 0 Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J 21. h = 40m, u = 50 m/sec Let the speed be ‘v’ when it strikes the ground. Applying law of conservation of energy mgh + ½ mu 2 = ½ mv 2 ? 10 × 40 + (1/2) × 2500 = ½ v 2 ? v 2 = 3300 ? v = 57.4 m/sec ?58 m/sec 22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m p = t w , Work w = pt = 460 × 117.56 J Again, W = FS = 200 56 . 117 460 ? = 270.3 N ? 270 N 23. S = 100 m, t = 10.54 sec, m = 50 kg The motion can be assumed to be uniform because the time taken for acceleration is minimum. Page 5 8.1 SOLUTIONS TO CONCEPTS CHAPTER – 8 1. M = m c + m b = 90kg u = 6 km/h = 1.666 m/sec v = 12 km/h = 3.333 m/sec Increase in K.E. = ½ Mv 2 – ½ Mu 2 = ½ 90 × (3.333) 2 – ½ × 90 × (1.66) 2 = 494.5 – 124.6 = 374.8 ? 375 J 2. m b = 2 kg. u = 10 m/sec a = 3 m/aec 2 t = 5 sec v = u + at = 10 + 3 I 5 = 25 m/sec. ?F.K.E = ½ mv 2 = ½ × 2 × 625 = 625 J. 3. F = 100 N S = 4m, ? = 0° ? = S . F ? ? ???100 × 4 = 400 J ? 4. m = 5 kg ? = 30° S = 10 m F = mg So, work done by the force of gravity ? = mgh = 5 × 9.8 × 5 = 245 J ? 5. F= 2.50N, S = 2.5m, m =15g = 0.015kg. So, w = F × S ? a = m F = 015 . 0 5 . 2 = 3 500 m/s 2 =F × S cos 0° (acting along the same line) = 2.5 × 2.5 = 6.25J Let the velocity of the body at b = U. Applying work-energy principle ½ mv 2 – 0 = 6.25 ? V = 015 . 0 2 25 . 6 ? = 28.86 m/sec. So, time taken to travel from A to B. ? t = a u v ? = 500 3 86 . 28 ? ? Average power = t W = 3 ) 86 . 28 ( 500 25 . 6 ? ? = 36.1 6. Given j ˆ 3 i ˆ 2 r 1 ? ? ? j ˆ 2 i ˆ 3 r 2 ? ? So, displacement vector is given by, 2 1 r r r ? ? ? ? ? ? j ˆ i ˆ ) j ˆ 3 i ˆ 2 ( ) j ˆ 2 i ˆ 3 ( r ? ? ? ? ? ? ? u=1.66 m/s ? 90kg ? 90kg ? v=3.33 m/s ? u=10 m/s ? 2 kg ? a ? = 3m/s 2 F ? 4m ? R ? 100 N ? mg ? 30° ? 5 ? mg ? F ? 5 log ? 10m ? 30° ? v ? ? A ? B ? Chapter 8 8.2 So, work done = s F ? ? ? = 5 × 1 + 5(-1) = 0 7. m b = 2kg, s = 40m, a = 0.5m/sec 2 So, force applied by the man on the box F = m b a = 2 × (0.5) = 1 N ? = FS = 1 × 40 = 40 J ? 8. Given that F= a + bx Where a and b are constants. So, work done by this force during this force during the displacement x = 0 and x = d is given by W = ? ? ? ? d 0 d 0 dx ) bx a ( dx F = ax + (bx 2 /2) = [a + ½ bd] d 9. m b = 250g = .250 kg ? = 37°, S = 1m. Frictional force f = ?R mg sin ? = ? R ..(1) mg cos ?? ..(2) so, work done against ?R = ?RS cos 0° = mg sin ? S = 0.250 × 9.8 × 0.60 × 1 = 1.5 J 10. a = ) m M ( 2 F ? (given) a) from fig (1) ma = ? k R 1 and R 1 = mg ? ? = 1 R ma = g ) m M ( 2 F ? b) Frictional force acting on the smaller block f = ?R = ) m M ( 2 F m mg g ) m M ( 2 F ? ? ? ? ? c) Work done w = fs s = d w = d ) m M ( 2 mF ? ? = ) m M ( 2 mFd ? ? 11. Weight = 2000 N, S = 20m, ? = 0.2 a) R + Psin ? - 2000 = 0 ..(1) P cos ? - 0.2 R =0 ..(2) From (1) and (2) P cos ? – 0.2 (2000 – P sin ?)=0 P = ? ? ? sin 2 . 0 cos 400 ..(3) So, work done by the person, W = PS cos ? = ? ? ? ? sin 2 . 0 cos cos 8000 = ? ? sin 2 . 0 1 8000 = ? ? tan 5 40000 b) For minimum magnitude of force from equn(1) d/d ? (cos ??+ 0.2 sin ?) = 0 ? tan ? = 0.2 putting the value in equn (3) W = ? ? tan 5 40000 = ) 2 . 5 ( 40000 = 7690 J 12. w = 100 N, ? = 37°, s = 2m R ? m b g ? m b a ? F ? R ? ?R ? 1 m ? mg ? 37° ? M ? F ? m ? R 1 ? ? k R 1 ? ma ? mg ? ?R 1 ? R 2 ? f ? ma ? mg ? ?R 2 ? ? ? R ? P ? 0.2R ? 2000 N ? Chapter 8 8.3 Force F= mg sin 37° = 100 × 0.60 = 60 N So, work done, when the force is parallel to incline. w = Fs cos ? = 60 × 2 × cos ? = 120 J In ?ABC AB= 2m CB = 37° so, h = C = 1m ?work done when the force in horizontal direction W = mgh = 100 × 1.2 = 120 J ? 13. m = 500 kg, s = 25m, u = 72km/h= 20 m/s, v = 0 (-a) = S 2 u v 2 2 ? ? a = 50 400 = 8m/sec 2 Frictional force f = ma = 500 × 8 = 4000 N 14. m = 500 kg, u = 0, v = 72 km/h = 20m/s a = s 2 u v 2 2 ? = 50 400 = 8m/sec 2 force needed to accelerate the car F = ma = 500 × 8 = 4000 N 15. Given, v = a x (uniformly accelerated motion) displacement s = d – 0 = d putting x = 0, v 1 = 0 putting x = d, v 2 = a d a = s 2 u v 2 2 2 2 ? = d 2 d a 2 = 2 a 2 force f = ma = 2 ma 2 work done w = FS cos ? = d 2 ma 2 ? = 2 d ma 2 ? 16. a) m = 2kg, ? = 37°, F = 20 N From the free body diagram F = (2g sin ?) + ma ? a = (20 – 20 sin ?)/s = 4m/sec 2 S = ut + ½ at 2 (u = 0, t = 1s, a = 1.66) = 2m So, work, done w = Fs = 20 × 2 = 40 J b) If W = 40 J S = F W = 20 40 h = 2 sin 37° = 1.2 m So, work done W = –mgh = – 20 × 1.2 = –24 J c) v = u + at = 4 × 10 = 40 m/sec So, K.E. = ½ mv 2 = ½ × 2 × 16 = 16 J 17. m = 2kg, ? = 37°, F = 20 N, a = 10 m/sec 2 a) t = 1sec So, s= ut + ½ at 2 = 5m 37° ? A ? A ? A ? B ? v=0 ? v=20 m/s m=500 kg ? –a ? 25m ? a ? R mg ? ma f ? 500 kg ? a ? 25m ? R mg ? F F ? ma ? ma ? 2g cos ? ? 20N ? R ? ma 2gsin ? ? 20N ? ma ? mg cos ? ? 20N ? R ? mg sin ? ? ?R ? h ? 37° ? 5m ? C ? A ? B ? 37° ? Chapter 8 8.4 Work done by the applied force w = FS cos 0° = 20 × 5 = 100 J b) BC (h) = 5 sin 37° = 3m So, work done by the weight W = mgh = 2 × 10 × 3 = 60 J c) So, frictional force f = mg sin ? work done by the frictional forces w = fs cos0° = (mg sin ?) s = 20 × 0.60 × 5 = 60 J ? 18. Given, m = 25o g = 0.250kg, u = 40 cm/sec = 0.4m/sec ? = 0.1, v=0 Here, ? R = ma {where, a = deceleration} a = m R ? = m mg ? = ?g = 0.1 × 9.8 = 0.98 m/sec 2 S = a 2 u v 2 2 ? = 0.082m = 8.2 cm Again, work done against friction is given by – w = ? RS cos ? = 0.1 × 2.5 × 0.082 × 1 ( ? = 0°) = 0.02 J ? W = – 0.02 J ? 19. h = 50m, m = 1.8 × 10 5 kg/hr, P = 100 watt, P.E. = mgh = 1.8 × 10 5 × 9.8 × 50 = 882 × 10 5 J/hr Because, half the potential energy is converted into electricity, Electrical energy ½ P.E. = 441 × 10 5 J/hr So, power in watt (J/sec) is given by = 3600 10 441 5 ? ? number of 100 W lamps, that can be lit 100 3600 10 441 5 ? ? = 122.5 ?122 20. m = 6kg, h = 2m P.E. at a height ‘2m’ = mgh = 6 × (9.8) × 2 = 117.6 J P.E. at floor = 0 Loss in P.E. = 117.6 – 0 = 117. 6 J ? 118 J 21. h = 40m, u = 50 m/sec Let the speed be ‘v’ when it strikes the ground. Applying law of conservation of energy mgh + ½ mu 2 = ½ mv 2 ? 10 × 40 + (1/2) × 2500 = ½ v 2 ? v 2 = 3300 ? v = 57.4 m/sec ?58 m/sec 22. t = 1 min 57.56 sec = 11.56 sec, p= 400 W, s =200 m p = t w , Work w = pt = 460 × 117.56 J Again, W = FS = 200 56 . 117 460 ? = 270.3 N ? 270 N 23. S = 100 m, t = 10.54 sec, m = 50 kg The motion can be assumed to be uniform because the time taken for acceleration is minimum. Chapter 8 8.5 a) Speed v = S/t = 9.487 e/s So, K.E. = ½ mv 2 = 2250 J b) Weight = mg = 490 J given R = mg /10 = 49 J so, work done against resistance W F = – RS = – 49 × 100 = – 4900 J c) To maintain her uniform speed, she has to exert 4900 j of energy to over come friction P = t W = 4900 / 10.54 = 465 W 24. h = 10 m flow rate = (m/t) = 30 kg/min = 0.5 kg/sec power P = t mgh = (0.5) × 9.8 × 10 = 49 W So, horse power (h.p) P/746 = 49/746 = 6.6 × 10 –2 hp 25. m = 200g = 0.2kg, h = 150cm = 1.5m, v = 3m/sec, t = 1 sec Total work done = ½ mv 2 + mgh = (1/2) × (0.2) ×9 + (0.2) × (9.8) × (1.5) = 3.84 J h.p. used = 746 84 . 3 = 5.14 × 10 –3 26. m = 200 kg, s = 12m, t = 1 min = 60 sec So, work W = F cos ? = mgs cos0° [ ? = 0°, for minimum work] = 2000 × 10 × 12 = 240000 J So, power p = t W = 60 240000 = 4000 watt h.p = 746 4000 = 5.3 hp. ? 27. The specification given by the company are U = 0, m = 95 kg, P m = 3.5 hp V m = 60 km/h = 50/3 m/sec t m = 5 sec So, the maximum acceleration that can be produced is given by, a = 5 0 ) 3 / 50 ( ? = 3 10 So, the driving force is given by F = ma = 95 × 3 10 = 3 950 N So, the velocity that can be attained by maximum h.p. white supplying 3 950 will be v = F p ? v = 950 5 746 5 . 3 ? ? = 8.2 m/sec. Because, the scooter can reach a maximum of 8.s m/sec while producing a force of 950/3 N, the specifications given are some what over claimed. 28. Given m = 30kg, v = 40 cm/sec = 0.4 m/sec s = 2m From the free body diagram, the force given by the chain is, F = (ma – mg) = m(a – g) [where a = acceleration of the block] a = 2s u2) (v2 = 4 . 0 16 . 0 = 0.04 m/sec 2 mg ? F ? ma ? mg ? F ?Read More

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