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Page 1 82  Mechanics1 6. T W 2 2 = and T T 2 1 2 = 7. N = 40 …(i) f = 20 …(ii) and f N x ´ = ´ 10 …(iii) 8. T f N + ° = ° cos sin 30 30 …(i) N f W cos sin 30 30 °+ ° = …(ii) T R fR ´ = …(iii) 9. V T W + = 3 2 …(i) H T = 2 …(ii) Net moment about O = zero \ . W l T l ´ = ´ 2 3 2 …(iii) 10. (a) a =  + + = 100 40 6 4 10 3 2 m/s (b) Net force = ma \ F F 6 4 18 12 = = N, N and F 10 = 30 N (c) N F  = = 40 30 10 \ N = 70 N. 11. a F m m m = + + = = 1 2 3 60 60 1 m/s 2 (a) T m a 1 1 10 = = N T T m a 2 1 2  = \ T 2 10 20  = \ T 2 30 = N (b) T 1 0 = . New acceleration a¢ = = 60 50 1.2 m s 2 / T m a 2 2 24 = ¢ = N 12. (a) T g a 1 2 2  = (b) T g a 2 5 5  = x f 10 cm W = 20 N N 40 N N T 30° 30° f W W V O T/2 H 3 T 2 1.9 kg 0.1 kg T 1 2 g a 45° W T 1 T 2 1.9 kg 0.2 kg T 2 5 g a 2.9 kg Page 2 82  Mechanics1 6. T W 2 2 = and T T 2 1 2 = 7. N = 40 …(i) f = 20 …(ii) and f N x ´ = ´ 10 …(iii) 8. T f N + ° = ° cos sin 30 30 …(i) N f W cos sin 30 30 °+ ° = …(ii) T R fR ´ = …(iii) 9. V T W + = 3 2 …(i) H T = 2 …(ii) Net moment about O = zero \ . W l T l ´ = ´ 2 3 2 …(iii) 10. (a) a =  + + = 100 40 6 4 10 3 2 m/s (b) Net force = ma \ F F 6 4 18 12 = = N, N and F 10 = 30 N (c) N F  = = 40 30 10 \ N = 70 N. 11. a F m m m = + + = = 1 2 3 60 60 1 m/s 2 (a) T m a 1 1 10 = = N T T m a 2 1 2  = \ T 2 10 20  = \ T 2 30 = N (b) T 1 0 = . New acceleration a¢ = = 60 50 1.2 m s 2 / T m a 2 2 24 = ¢ = N 12. (a) T g a 1 2 2  = (b) T g a 2 5 5  = x f 10 cm W = 20 N N 40 N N T 30° 30° f W W V O T/2 H 3 T 2 1.9 kg 0.1 kg T 1 2 g a 45° W T 1 T 2 1.9 kg 0.2 kg T 2 5 g a 2.9 kg 13. (a) a g =  200 16 16 (b) T g a 1 11 11  = (c) T g a 2 9 9  = 14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity. 15. Tension on B T = Tension on A T = 3 Now in these situations a T µ 1 16. x x x A C B + + = 2 constant. Differentiating twice w.r.t. time we get the acceleration relation. 17. a a A B = sin q \ a a A B = sin q 18. x y + = 6 …(i) y x  = 4 …(ii) Solving, we get x = 1 m/s 2 19. a g g =  = 7 3 10 4 m/s 2 40 4 1  = T a 30 3 1 2 +  = T T a T a 3 10 1  = 20. T a 1 1 = …(i) T a a r 1 20 2 2  =  ( / ) …(ii) 30 3 2 1  = + T a a r ( / ) …(iii) 21. T Mg Ma  ° = sin 30 …(i) 2 2 2 2 Mg T M a  = × …(ii) 22. T a = 1 …(i) 10 1  = T a …(ii) 23. 50 2 5  = T a …(i) T a  = 40 4 2 ( ) …(ii) 24. (a) N = 40 N, m s N = 24 N F N s < m \ f = 20 N and a = 0 (b) N = 20 N, m s N = 12 N and m k N = 8 N Laws of Motion 83 q q a B a A x y x y 1 2 3 a r 2 3 1 a a r T 1 T 1 2T 1 T 1 a 2 T 1 T = T 1 2 T a 2M 2T 2 Mg a 2 M Mg sin 30° B 2T 50 a A 2T 40 2a Page 3 82  Mechanics1 6. T W 2 2 = and T T 2 1 2 = 7. N = 40 …(i) f = 20 …(ii) and f N x ´ = ´ 10 …(iii) 8. T f N + ° = ° cos sin 30 30 …(i) N f W cos sin 30 30 °+ ° = …(ii) T R fR ´ = …(iii) 9. V T W + = 3 2 …(i) H T = 2 …(ii) Net moment about O = zero \ . W l T l ´ = ´ 2 3 2 …(iii) 10. (a) a =  + + = 100 40 6 4 10 3 2 m/s (b) Net force = ma \ F F 6 4 18 12 = = N, N and F 10 = 30 N (c) N F  = = 40 30 10 \ N = 70 N. 11. a F m m m = + + = = 1 2 3 60 60 1 m/s 2 (a) T m a 1 1 10 = = N T T m a 2 1 2  = \ T 2 10 20  = \ T 2 30 = N (b) T 1 0 = . New acceleration a¢ = = 60 50 1.2 m s 2 / T m a 2 2 24 = ¢ = N 12. (a) T g a 1 2 2  = (b) T g a 2 5 5  = x f 10 cm W = 20 N N 40 N N T 30° 30° f W W V O T/2 H 3 T 2 1.9 kg 0.1 kg T 1 2 g a 45° W T 1 T 2 1.9 kg 0.2 kg T 2 5 g a 2.9 kg 13. (a) a g =  200 16 16 (b) T g a 1 11 11  = (c) T g a 2 9 9  = 14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity. 15. Tension on B T = Tension on A T = 3 Now in these situations a T µ 1 16. x x x A C B + + = 2 constant. Differentiating twice w.r.t. time we get the acceleration relation. 17. a a A B = sin q \ a a A B = sin q 18. x y + = 6 …(i) y x  = 4 …(ii) Solving, we get x = 1 m/s 2 19. a g g =  = 7 3 10 4 m/s 2 40 4 1  = T a 30 3 1 2 +  = T T a T a 3 10 1  = 20. T a 1 1 = …(i) T a a r 1 20 2 2  =  ( / ) …(ii) 30 3 2 1  = + T a a r ( / ) …(iii) 21. T Mg Ma  ° = sin 30 …(i) 2 2 2 2 Mg T M a  = × …(ii) 22. T a = 1 …(i) 10 1  = T a …(ii) 23. 50 2 5  = T a …(i) T a  = 40 4 2 ( ) …(ii) 24. (a) N = 40 N, m s N = 24 N F N s < m \ f = 20 N and a = 0 (b) N = 20 N, m s N = 12 N and m k N = 8 N Laws of Motion 83 q q a B a A x y x y 1 2 3 a r 2 3 1 a a r T 1 T 1 2T 1 T 1 a 2 T 1 T = T 1 2 T a 2M 2T 2 Mg a 2 M Mg sin 30° B 2T 50 a A 2T 40 2a F N s > m \ f N k = = m 8 N and a =  = 20 8 2 6 m/s 2 (c) N =  = 60 20 40 N m s N = 8 N and m k N = 4 N Since, F N s cos 45° > m \ f N k = = m 4 N and a =  20 4 6 = 8 3 m/s 2 25. a g = = m 3 m/s 2 (a) v at = \ 6 3 = t or t = 2 s (b) s at = 1 2 2 \ s = ´ ´ = 1 2 3 4 6 m. 26. f = ´ ´ = 0 4 1 10 4 . N a f 1 1 4 = = m/s 2 a f 2 2 2 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or 2 4 8 2 + =  t t \ t = 1 s (b) v v 1 2 2 4 1 6 = = + ´ = m/s (c) s u t a t 1 1 1 2 1 2 = + s u t a t 2 2 2 2 1 2 =  27. f = = ( . )( )( ) 06 2 10 12 N a 2 12 2 6 = = m/s 2 and a 1 12 1 12 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or u a t u a t 1 1 2 2 + = + or 3 6 18 12  =  + t t \ t = 7 6 s (b) Common velocity at this instant is v 1 or v 2 . (c) s u t a t 1 1 1 2 1 2 = + and s u t a t 2 2 2 2 1 2 = + 28. N = 20 N m s N = 16 N and m k N = 12 N Since, W = 20 N > m s N, friction m k N will act. \ a =  = 20 12 2 4 m/s 2 29. N = 20 N mN = 16 N Block will start moving when F N = m or 2 16 t = or t = 8 s. After 8 s a t t =  =  2 16 2 8 i.e., at graph is a straight line with positive slope and negative intercept. 30. N = 60 N, m s N = 36 N, m k N = 24 N Block will start moving when F N s = m 84  Mechanics1 a 1 1 kg 2 kg f f a 2 a 2 2 kg f a 1 f 1 kg –ve +ve Page 4 82  Mechanics1 6. T W 2 2 = and T T 2 1 2 = 7. N = 40 …(i) f = 20 …(ii) and f N x ´ = ´ 10 …(iii) 8. T f N + ° = ° cos sin 30 30 …(i) N f W cos sin 30 30 °+ ° = …(ii) T R fR ´ = …(iii) 9. V T W + = 3 2 …(i) H T = 2 …(ii) Net moment about O = zero \ . W l T l ´ = ´ 2 3 2 …(iii) 10. (a) a =  + + = 100 40 6 4 10 3 2 m/s (b) Net force = ma \ F F 6 4 18 12 = = N, N and F 10 = 30 N (c) N F  = = 40 30 10 \ N = 70 N. 11. a F m m m = + + = = 1 2 3 60 60 1 m/s 2 (a) T m a 1 1 10 = = N T T m a 2 1 2  = \ T 2 10 20  = \ T 2 30 = N (b) T 1 0 = . New acceleration a¢ = = 60 50 1.2 m s 2 / T m a 2 2 24 = ¢ = N 12. (a) T g a 1 2 2  = (b) T g a 2 5 5  = x f 10 cm W = 20 N N 40 N N T 30° 30° f W W V O T/2 H 3 T 2 1.9 kg 0.1 kg T 1 2 g a 45° W T 1 T 2 1.9 kg 0.2 kg T 2 5 g a 2.9 kg 13. (a) a g =  200 16 16 (b) T g a 1 11 11  = (c) T g a 2 9 9  = 14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity. 15. Tension on B T = Tension on A T = 3 Now in these situations a T µ 1 16. x x x A C B + + = 2 constant. Differentiating twice w.r.t. time we get the acceleration relation. 17. a a A B = sin q \ a a A B = sin q 18. x y + = 6 …(i) y x  = 4 …(ii) Solving, we get x = 1 m/s 2 19. a g g =  = 7 3 10 4 m/s 2 40 4 1  = T a 30 3 1 2 +  = T T a T a 3 10 1  = 20. T a 1 1 = …(i) T a a r 1 20 2 2  =  ( / ) …(ii) 30 3 2 1  = + T a a r ( / ) …(iii) 21. T Mg Ma  ° = sin 30 …(i) 2 2 2 2 Mg T M a  = × …(ii) 22. T a = 1 …(i) 10 1  = T a …(ii) 23. 50 2 5  = T a …(i) T a  = 40 4 2 ( ) …(ii) 24. (a) N = 40 N, m s N = 24 N F N s < m \ f = 20 N and a = 0 (b) N = 20 N, m s N = 12 N and m k N = 8 N Laws of Motion 83 q q a B a A x y x y 1 2 3 a r 2 3 1 a a r T 1 T 1 2T 1 T 1 a 2 T 1 T = T 1 2 T a 2M 2T 2 Mg a 2 M Mg sin 30° B 2T 50 a A 2T 40 2a F N s > m \ f N k = = m 8 N and a =  = 20 8 2 6 m/s 2 (c) N =  = 60 20 40 N m s N = 8 N and m k N = 4 N Since, F N s cos 45° > m \ f N k = = m 4 N and a =  20 4 6 = 8 3 m/s 2 25. a g = = m 3 m/s 2 (a) v at = \ 6 3 = t or t = 2 s (b) s at = 1 2 2 \ s = ´ ´ = 1 2 3 4 6 m. 26. f = ´ ´ = 0 4 1 10 4 . N a f 1 1 4 = = m/s 2 a f 2 2 2 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or 2 4 8 2 + =  t t \ t = 1 s (b) v v 1 2 2 4 1 6 = = + ´ = m/s (c) s u t a t 1 1 1 2 1 2 = + s u t a t 2 2 2 2 1 2 =  27. f = = ( . )( )( ) 06 2 10 12 N a 2 12 2 6 = = m/s 2 and a 1 12 1 12 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or u a t u a t 1 1 2 2 + = + or 3 6 18 12  =  + t t \ t = 7 6 s (b) Common velocity at this instant is v 1 or v 2 . (c) s u t a t 1 1 1 2 1 2 = + and s u t a t 2 2 2 2 1 2 = + 28. N = 20 N m s N = 16 N and m k N = 12 N Since, W = 20 N > m s N, friction m k N will act. \ a =  = 20 12 2 4 m/s 2 29. N = 20 N mN = 16 N Block will start moving when F N = m or 2 16 t = or t = 8 s. After 8 s a t t =  =  2 16 2 8 i.e., at graph is a straight line with positive slope and negative intercept. 30. N = 60 N, m s N = 36 N, m k N = 24 N Block will start moving when F N s = m 84  Mechanics1 a 1 1 kg 2 kg f f a 2 a 2 2 kg f a 1 f 1 kg –ve +ve or 4 36 t = \ t = 9 s After 9 s a t t =  =  4 24 6 2 3 4 31. N mg = = cos q 30 N mg sin 30 30 3 ° = N » 52 N. m s N = 18 N and m k N = 12 N (a) F =  = 52 18 34 N. (b) F + = 12 52 \ F = 40 N (c) F   = ´ 52 12 6 4 \ F = 88 N Objective Questions (Level1) Single Correct Option 1. a mg F m g F m =  =  m m A B > \ a a A B > or ball A reaches earlier. 2. a g g g =  = 4 2 6 3 Now, 2 2 3 g T g  = \ T g = = 4 3 13 N 3. 3 2 100 1 T = …(i) T T 1 2 2 = …(ii) 4. mg T ma  = \ a g T m min max =  =  = g mg m g 2 3 3 5. a g g g =  = 10 5 15 3 10 10 3 g T g  = ´ \ T g = = 20 3 Reading of spring balance. 6. a mg mg m = ° = 2 30 3 0 sin \ T mg = 7. a g g g g 1 45 45 =  = ° ° sin cos sin cos q m q m a g g 2 45 = ° sin sin q = Now t s a = 2 or t a µ 1 \ t t a a 1 2 2 1 = or 2 45 45 45 = ° ° ° g g g sin sin cos m Solving, we get m = 3 4 Laws of Motion 85 F 52 N 12 N v F 52 N a 12 N 30° T 1 T 2 100 N mg T a F 52 N 18 N Page 5 82  Mechanics1 6. T W 2 2 = and T T 2 1 2 = 7. N = 40 …(i) f = 20 …(ii) and f N x ´ = ´ 10 …(iii) 8. T f N + ° = ° cos sin 30 30 …(i) N f W cos sin 30 30 °+ ° = …(ii) T R fR ´ = …(iii) 9. V T W + = 3 2 …(i) H T = 2 …(ii) Net moment about O = zero \ . W l T l ´ = ´ 2 3 2 …(iii) 10. (a) a =  + + = 100 40 6 4 10 3 2 m/s (b) Net force = ma \ F F 6 4 18 12 = = N, N and F 10 = 30 N (c) N F  = = 40 30 10 \ N = 70 N. 11. a F m m m = + + = = 1 2 3 60 60 1 m/s 2 (a) T m a 1 1 10 = = N T T m a 2 1 2  = \ T 2 10 20  = \ T 2 30 = N (b) T 1 0 = . New acceleration a¢ = = 60 50 1.2 m s 2 / T m a 2 2 24 = ¢ = N 12. (a) T g a 1 2 2  = (b) T g a 2 5 5  = x f 10 cm W = 20 N N 40 N N T 30° 30° f W W V O T/2 H 3 T 2 1.9 kg 0.1 kg T 1 2 g a 45° W T 1 T 2 1.9 kg 0.2 kg T 2 5 g a 2.9 kg 13. (a) a g =  200 16 16 (b) T g a 1 11 11  = (c) T g a 2 9 9  = 14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity. 15. Tension on B T = Tension on A T = 3 Now in these situations a T µ 1 16. x x x A C B + + = 2 constant. Differentiating twice w.r.t. time we get the acceleration relation. 17. a a A B = sin q \ a a A B = sin q 18. x y + = 6 …(i) y x  = 4 …(ii) Solving, we get x = 1 m/s 2 19. a g g =  = 7 3 10 4 m/s 2 40 4 1  = T a 30 3 1 2 +  = T T a T a 3 10 1  = 20. T a 1 1 = …(i) T a a r 1 20 2 2  =  ( / ) …(ii) 30 3 2 1  = + T a a r ( / ) …(iii) 21. T Mg Ma  ° = sin 30 …(i) 2 2 2 2 Mg T M a  = × …(ii) 22. T a = 1 …(i) 10 1  = T a …(ii) 23. 50 2 5  = T a …(i) T a  = 40 4 2 ( ) …(ii) 24. (a) N = 40 N, m s N = 24 N F N s < m \ f = 20 N and a = 0 (b) N = 20 N, m s N = 12 N and m k N = 8 N Laws of Motion 83 q q a B a A x y x y 1 2 3 a r 2 3 1 a a r T 1 T 1 2T 1 T 1 a 2 T 1 T = T 1 2 T a 2M 2T 2 Mg a 2 M Mg sin 30° B 2T 50 a A 2T 40 2a F N s > m \ f N k = = m 8 N and a =  = 20 8 2 6 m/s 2 (c) N =  = 60 20 40 N m s N = 8 N and m k N = 4 N Since, F N s cos 45° > m \ f N k = = m 4 N and a =  20 4 6 = 8 3 m/s 2 25. a g = = m 3 m/s 2 (a) v at = \ 6 3 = t or t = 2 s (b) s at = 1 2 2 \ s = ´ ´ = 1 2 3 4 6 m. 26. f = ´ ´ = 0 4 1 10 4 . N a f 1 1 4 = = m/s 2 a f 2 2 2 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or 2 4 8 2 + =  t t \ t = 1 s (b) v v 1 2 2 4 1 6 = = + ´ = m/s (c) s u t a t 1 1 1 2 1 2 = + s u t a t 2 2 2 2 1 2 =  27. f = = ( . )( )( ) 06 2 10 12 N a 2 12 2 6 = = m/s 2 and a 1 12 1 12 = = m/s 2 (a) Relative motion will stop when v v 1 2 = or u a t u a t 1 1 2 2 + = + or 3 6 18 12  =  + t t \ t = 7 6 s (b) Common velocity at this instant is v 1 or v 2 . (c) s u t a t 1 1 1 2 1 2 = + and s u t a t 2 2 2 2 1 2 = + 28. N = 20 N m s N = 16 N and m k N = 12 N Since, W = 20 N > m s N, friction m k N will act. \ a =  = 20 12 2 4 m/s 2 29. N = 20 N mN = 16 N Block will start moving when F N = m or 2 16 t = or t = 8 s. After 8 s a t t =  =  2 16 2 8 i.e., at graph is a straight line with positive slope and negative intercept. 30. N = 60 N, m s N = 36 N, m k N = 24 N Block will start moving when F N s = m 84  Mechanics1 a 1 1 kg 2 kg f f a 2 a 2 2 kg f a 1 f 1 kg –ve +ve or 4 36 t = \ t = 9 s After 9 s a t t =  =  4 24 6 2 3 4 31. N mg = = cos q 30 N mg sin 30 30 3 ° = N » 52 N. m s N = 18 N and m k N = 12 N (a) F =  = 52 18 34 N. (b) F + = 12 52 \ F = 40 N (c) F   = ´ 52 12 6 4 \ F = 88 N Objective Questions (Level1) Single Correct Option 1. a mg F m g F m =  =  m m A B > \ a a A B > or ball A reaches earlier. 2. a g g g =  = 4 2 6 3 Now, 2 2 3 g T g  = \ T g = = 4 3 13 N 3. 3 2 100 1 T = …(i) T T 1 2 2 = …(ii) 4. mg T ma  = \ a g T m min max =  =  = g mg m g 2 3 3 5. a g g g =  = 10 5 15 3 10 10 3 g T g  = ´ \ T g = = 20 3 Reading of spring balance. 6. a mg mg m = ° = 2 30 3 0 sin \ T mg = 7. a g g g g 1 45 45 =  = ° ° sin cos sin cos q m q m a g g 2 45 = ° sin sin q = Now t s a = 2 or t a µ 1 \ t t a a 1 2 2 1 = or 2 45 45 45 = ° ° ° g g g sin sin cos m Solving, we get m = 3 4 Laws of Motion 85 F 52 N 12 N v F 52 N a 12 N 30° T 1 T 2 100 N mg T a F 52 N 18 N 8. F mg mg 1 = sin cos q + m q F mg mg 2 = sin cos q  m q Given that F F 1 2 2 = 9. For equilibrium of block, net force from plane should be equal and opposite of weight. 10. No solution is required. 11. Angle of repose q m = = °  tan ( ) 1 30 \ h R R R =  =  æ è ç ö ø ÷ cos q 1 3 2 12. Net pulling force =  = = 15 5 10 g g g F Net retarding force = = = ( )( ) 0.2 5g g f \ a F f g =  = 25 9 25 T g a g 1 5 5 9 5  = = \ T g 1 34 5 = 15 15 27 5 2 g T a g  = = \ T 2 48 5 = g \ T T 1 2 17 24 = 13. Relative to lift, a g a r = + ( ) sin q along the plane. Now, t s a r = 2 = + 2L g a ( ) sin q 14. f mg = sin q (if block is at rest) 15. 2 30 T F cos ° = \ T F = 3 a T m = ° cos 60 = F m 2 3 16. N mg F =  sin q m f q N N mg F = = f  (tan ) (tan )( sin ) F N cos q m = 17. mmg = ´ ´ = 0.2 4 10 8 N At t = 2 s, F = 4 N Since F mg < m \ Force of friction f F = = 4 N 18. a g 1 = sin q a g g 2 = sin cos q  m q t a = 25 or t a µ 1 t t a a 1 2 2 1 = 1 2 =  g g g sin cos sin q m q q 19. Net accelaration of man relative to ground = + = a a a 2 T mg m a  = ( ) 2 \ T m g a = + ( ) 2 20. a = 0 3 50 25 75 F g g = + = ( ) \ F = 25 g = 250 N 86  Mechanics1 q a v q F h q T 60° 30° F aRead More

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