Page 1 12. Decrease in KE =  1 2 0 2 mv = 1 2 2 mv Change in mechanical energy PE + KE = 1 2 2 mv Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. \ Change ME =  1 2 2 mv mgh Thus, assertion is true. As explained above the reason is false. (m does not change with the increase in angle of inclination) \ Correct option is (c). Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by gravity + work done by force F \ 1 2 2 mv mgh FR = + (Displacement of force F is R) = ´ ´ + ´ 1 2 10 5 5 5 = 50 \ u m = 100 = 200 =  14.14 ms 1 \ Correct option is (a). 2. P Fv = = mav = æ è ç ö ø ÷ m v dv dx v a dv dt dv ds ds dt v dv ds = = × = æ è ç ö ø ÷ \ ds m p v dv = 2 i.e., ds m p v dv v v ò ò = 2 2 or s m p v v v = é ë ê ù û ú 3 2 3 =  m p v v 3 2 3 3 [( ) ( ) ] = 7 3 3 mv p \ Correct option is (a). 3. Loss of PE of block = Gain in PE of spring mg d k ( ) sin + ° = ´ 2 30 1 2 2 2 or 10 10 2 1 2 1 2 100 2 2 ´ ´ + ´ = ´ ´ ( ) d d = 2 m Let, v A = velocity of block when it just touches spring \ 1 2 30 2 mv mgd A = ° sin v A 2 10 2 2 1 2 = ´ ´ ´ v A =  20 1 ms Correct option is (a). 128  Mechanics1 30° f' v' Stops PE = mgh h v Slope CaseI Rough f d A k 2m A 30° d m Page 2 12. Decrease in KE =  1 2 0 2 mv = 1 2 2 mv Change in mechanical energy PE + KE = 1 2 2 mv Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. \ Change ME =  1 2 2 mv mgh Thus, assertion is true. As explained above the reason is false. (m does not change with the increase in angle of inclination) \ Correct option is (c). Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by gravity + work done by force F \ 1 2 2 mv mgh FR = + (Displacement of force F is R) = ´ ´ + ´ 1 2 10 5 5 5 = 50 \ u m = 100 = 200 =  14.14 ms 1 \ Correct option is (a). 2. P Fv = = mav = æ è ç ö ø ÷ m v dv dx v a dv dt dv ds ds dt v dv ds = = × = æ è ç ö ø ÷ \ ds m p v dv = 2 i.e., ds m p v dv v v ò ò = 2 2 or s m p v v v = é ë ê ù û ú 3 2 3 =  m p v v 3 2 3 3 [( ) ( ) ] = 7 3 3 mv p \ Correct option is (a). 3. Loss of PE of block = Gain in PE of spring mg d k ( ) sin + ° = ´ 2 30 1 2 2 2 or 10 10 2 1 2 1 2 100 2 2 ´ ´ + ´ = ´ ´ ( ) d d = 2 m Let, v A = velocity of block when it just touches spring \ 1 2 30 2 mv mgd A = ° sin v A 2 10 2 2 1 2 = ´ ´ ´ v A =  20 1 ms Correct option is (a). 128  Mechanics1 30° f' v' Stops PE = mgh h v Slope CaseI Rough f d A k 2m A 30° d m Work, Energy and Power  129 4. Mass per unit length of chain = m R p / 2 dm Rd m R = q p /2 y ydm dm CM = ò ò =  ò R m R Rd m 0 2 1 2 p q p q / ( cos ) / =  ò 2 1 0 2 R d p q q p ( cos ) / =  é ë ê ù û ú 2 2 1 R p p =  é ë ê ù û ú R 1 2 p Now, 1 2 2 mv mg y + CM or v g y 2 2 = COM v gR =  æ è ç ö ø ÷ 2 1 2 p \ Correct option is (c). 5. The moment string is cut Net force on A mg = 2 (downward) \ a g 1 2 = Net force on B = 0 \ a 2 0 = \ Correct option is (b). 6. Let x be the expansion in the spring. Increase in PE of spring = Decrease in PE of block C 1 2 2 1 kx M gx = i.e., kx M g = 2 1 For block A to remain at rest kx Mg = m min or 2 1 M g Mg = m min \ m min = 2 1 M M \ Correct option is (c). 7. T mg i = KX mg mg i = = 2 2 When one spring is cut. It means KX i becomes zero. Downward acceleration, a KX m mg m g i = = = 2 2 2 Now drawing FBD of lower mass : mg T m a mg f  = = . 2 \ T mg f = 2 or DT T T mg f i =  = 2 8. a mg mg m =  sin cos q m q =  g g sin cos q m q For v to be maximum dv dx = 0 or v dv dx = 0 F (= mg) mg mg A B F = mg T = mg T F = kx = mg mg (weight) F mg Mg 1 A B kx kx T T C T T R (1 – cos q) CM y ( ) R 1 – 2 p q q a x mmg cos q mg sin q Page 3 12. Decrease in KE =  1 2 0 2 mv = 1 2 2 mv Change in mechanical energy PE + KE = 1 2 2 mv Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. \ Change ME =  1 2 2 mv mgh Thus, assertion is true. As explained above the reason is false. (m does not change with the increase in angle of inclination) \ Correct option is (c). Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by gravity + work done by force F \ 1 2 2 mv mgh FR = + (Displacement of force F is R) = ´ ´ + ´ 1 2 10 5 5 5 = 50 \ u m = 100 = 200 =  14.14 ms 1 \ Correct option is (a). 2. P Fv = = mav = æ è ç ö ø ÷ m v dv dx v a dv dt dv ds ds dt v dv ds = = × = æ è ç ö ø ÷ \ ds m p v dv = 2 i.e., ds m p v dv v v ò ò = 2 2 or s m p v v v = é ë ê ù û ú 3 2 3 =  m p v v 3 2 3 3 [( ) ( ) ] = 7 3 3 mv p \ Correct option is (a). 3. Loss of PE of block = Gain in PE of spring mg d k ( ) sin + ° = ´ 2 30 1 2 2 2 or 10 10 2 1 2 1 2 100 2 2 ´ ´ + ´ = ´ ´ ( ) d d = 2 m Let, v A = velocity of block when it just touches spring \ 1 2 30 2 mv mgd A = ° sin v A 2 10 2 2 1 2 = ´ ´ ´ v A =  20 1 ms Correct option is (a). 128  Mechanics1 30° f' v' Stops PE = mgh h v Slope CaseI Rough f d A k 2m A 30° d m Work, Energy and Power  129 4. Mass per unit length of chain = m R p / 2 dm Rd m R = q p /2 y ydm dm CM = ò ò =  ò R m R Rd m 0 2 1 2 p q p q / ( cos ) / =  ò 2 1 0 2 R d p q q p ( cos ) / =  é ë ê ù û ú 2 2 1 R p p =  é ë ê ù û ú R 1 2 p Now, 1 2 2 mv mg y + CM or v g y 2 2 = COM v gR =  æ è ç ö ø ÷ 2 1 2 p \ Correct option is (c). 5. The moment string is cut Net force on A mg = 2 (downward) \ a g 1 2 = Net force on B = 0 \ a 2 0 = \ Correct option is (b). 6. Let x be the expansion in the spring. Increase in PE of spring = Decrease in PE of block C 1 2 2 1 kx M gx = i.e., kx M g = 2 1 For block A to remain at rest kx Mg = m min or 2 1 M g Mg = m min \ m min = 2 1 M M \ Correct option is (c). 7. T mg i = KX mg mg i = = 2 2 When one spring is cut. It means KX i becomes zero. Downward acceleration, a KX m mg m g i = = = 2 2 2 Now drawing FBD of lower mass : mg T m a mg f  = = . 2 \ T mg f = 2 or DT T T mg f i =  = 2 8. a mg mg m =  sin cos q m q =  g g sin cos q m q For v to be maximum dv dx = 0 or v dv dx = 0 F (= mg) mg mg A B F = mg T = mg T F = kx = mg mg (weight) F mg Mg 1 A B kx kx T T C T T R (1 – cos q) CM y ( ) R 1 – 2 p q q a x mmg cos q mg sin q or dx dt dv dx × = 0 or dv dt = 0 or a = 0 i.e., g g sin cos q m q = or sin cos q m q = or 3 5 3 10 4 5 = × x Þ x = = 10 4 2.5 m \ Correct option is (d). 9. (a) Between points E and F, dU dr is  ive. Now, F dU dr =  , the force between E and F will be + ive i.e., repulsive. (b) At point C the potential energy is minimum. Thus, C is point of stable equilibrium. \ Correct option is (c). 10. Power = × ® ® F v = × + ® ® ® m t g u g ( ) = ° + + × ® ® mg u m t cos ( ) 90 q g g i.e., P mg u mg t =  + sin q 2 [   ,  u g ® ® = = u g] Therefore, the graph between P and t will be as shown in option (c). 11. PE of spring due to its compression by x =  Work done by frictional force when displaced by 2x i.e., 1 2 2 2 kx mg x = m Þ x mg x = 4 m \ Correct option is (c). 12. Power delivered by man = × ® ® T v = T vcos q   T ® = T and   v ® = T 13. f = + 3 4 x y \ F x x =  ¶f ¶ =  3 N and F y y =  ¶f ¶ =  4 N PR PQ = 5 4 Þ PR PQ = ´ 5 4 = 10 m \ Work done by the conservative force on the particle = ´ F PR net = ´ 5 10 N m = 50 Nm = 50 J \ Correct option is (c). 14. Both at x x = 1 and x x = 2 the force acting on the body is zero i.e., it is in equilibrium. Now, if the body (when at x x = 1 ) is moved towards right (i.e., x x > 1 ) the force acting on it is + ive i.e., the body will not come back and if the body (when at x x = 2 ) is moved toward rght (i.e., x x > 2 ) the force acting on it is  ive i.e., the body will return back. Then,x x = 2 is the position of stable equilibrium. \ Correct option is (b). 130  Mechanics1 v mg u mg ® ® ® ® q x 2x mmg m v T ® ® P (6m, 8m) 4N 3N F net = 5N x R y 6 m Q Page 4 12. Decrease in KE =  1 2 0 2 mv = 1 2 2 mv Change in mechanical energy PE + KE = 1 2 2 mv Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. \ Change ME =  1 2 2 mv mgh Thus, assertion is true. As explained above the reason is false. (m does not change with the increase in angle of inclination) \ Correct option is (c). Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by gravity + work done by force F \ 1 2 2 mv mgh FR = + (Displacement of force F is R) = ´ ´ + ´ 1 2 10 5 5 5 = 50 \ u m = 100 = 200 =  14.14 ms 1 \ Correct option is (a). 2. P Fv = = mav = æ è ç ö ø ÷ m v dv dx v a dv dt dv ds ds dt v dv ds = = × = æ è ç ö ø ÷ \ ds m p v dv = 2 i.e., ds m p v dv v v ò ò = 2 2 or s m p v v v = é ë ê ù û ú 3 2 3 =  m p v v 3 2 3 3 [( ) ( ) ] = 7 3 3 mv p \ Correct option is (a). 3. Loss of PE of block = Gain in PE of spring mg d k ( ) sin + ° = ´ 2 30 1 2 2 2 or 10 10 2 1 2 1 2 100 2 2 ´ ´ + ´ = ´ ´ ( ) d d = 2 m Let, v A = velocity of block when it just touches spring \ 1 2 30 2 mv mgd A = ° sin v A 2 10 2 2 1 2 = ´ ´ ´ v A =  20 1 ms Correct option is (a). 128  Mechanics1 30° f' v' Stops PE = mgh h v Slope CaseI Rough f d A k 2m A 30° d m Work, Energy and Power  129 4. Mass per unit length of chain = m R p / 2 dm Rd m R = q p /2 y ydm dm CM = ò ò =  ò R m R Rd m 0 2 1 2 p q p q / ( cos ) / =  ò 2 1 0 2 R d p q q p ( cos ) / =  é ë ê ù û ú 2 2 1 R p p =  é ë ê ù û ú R 1 2 p Now, 1 2 2 mv mg y + CM or v g y 2 2 = COM v gR =  æ è ç ö ø ÷ 2 1 2 p \ Correct option is (c). 5. The moment string is cut Net force on A mg = 2 (downward) \ a g 1 2 = Net force on B = 0 \ a 2 0 = \ Correct option is (b). 6. Let x be the expansion in the spring. Increase in PE of spring = Decrease in PE of block C 1 2 2 1 kx M gx = i.e., kx M g = 2 1 For block A to remain at rest kx Mg = m min or 2 1 M g Mg = m min \ m min = 2 1 M M \ Correct option is (c). 7. T mg i = KX mg mg i = = 2 2 When one spring is cut. It means KX i becomes zero. Downward acceleration, a KX m mg m g i = = = 2 2 2 Now drawing FBD of lower mass : mg T m a mg f  = = . 2 \ T mg f = 2 or DT T T mg f i =  = 2 8. a mg mg m =  sin cos q m q =  g g sin cos q m q For v to be maximum dv dx = 0 or v dv dx = 0 F (= mg) mg mg A B F = mg T = mg T F = kx = mg mg (weight) F mg Mg 1 A B kx kx T T C T T R (1 – cos q) CM y ( ) R 1 – 2 p q q a x mmg cos q mg sin q or dx dt dv dx × = 0 or dv dt = 0 or a = 0 i.e., g g sin cos q m q = or sin cos q m q = or 3 5 3 10 4 5 = × x Þ x = = 10 4 2.5 m \ Correct option is (d). 9. (a) Between points E and F, dU dr is  ive. Now, F dU dr =  , the force between E and F will be + ive i.e., repulsive. (b) At point C the potential energy is minimum. Thus, C is point of stable equilibrium. \ Correct option is (c). 10. Power = × ® ® F v = × + ® ® ® m t g u g ( ) = ° + + × ® ® mg u m t cos ( ) 90 q g g i.e., P mg u mg t =  + sin q 2 [   ,  u g ® ® = = u g] Therefore, the graph between P and t will be as shown in option (c). 11. PE of spring due to its compression by x =  Work done by frictional force when displaced by 2x i.e., 1 2 2 2 kx mg x = m Þ x mg x = 4 m \ Correct option is (c). 12. Power delivered by man = × ® ® T v = T vcos q   T ® = T and   v ® = T 13. f = + 3 4 x y \ F x x =  ¶f ¶ =  3 N and F y y =  ¶f ¶ =  4 N PR PQ = 5 4 Þ PR PQ = ´ 5 4 = 10 m \ Work done by the conservative force on the particle = ´ F PR net = ´ 5 10 N m = 50 Nm = 50 J \ Correct option is (c). 14. Both at x x = 1 and x x = 2 the force acting on the body is zero i.e., it is in equilibrium. Now, if the body (when at x x = 1 ) is moved towards right (i.e., x x > 1 ) the force acting on it is + ive i.e., the body will not come back and if the body (when at x x = 2 ) is moved toward rght (i.e., x x > 2 ) the force acting on it is  ive i.e., the body will return back. Then,x x = 2 is the position of stable equilibrium. \ Correct option is (b). 130  Mechanics1 v mg u mg ® ® ® ® q x 2x mmg m v T ® ® P (6m, 8m) 4N 3N F net = 5N x R y 6 m Q 15. The man will stop when m q q mg mg cos sin = or ( ) cos sin m q q 0 x mg mg = or x = tan q m 0 16. Taking moment about A AC F k x l × = ( ) 2 \ AC k x l F = ( ) 2 = + ( ) ( ) k x l k k x 2 1 2 = + k k k l 2 1 2 \ Correct option is (d). 17. mgh + work done by the force of friction = 1 2 2 mv \ Work done by the force of friction =  1 2 2 mv mgh = ´ ´ æ è ç ö ø ÷  ´ ´ 1 2 1 2 1 10 1 2 ( ) =  8 J \ Correct option is (c). 18. U a x b x =  12 6 U ax bx =    12 6 For stable equilibrium, dU dx = 0 i.e., a x b x ( ) ( )    =   12 6 0 13 7 i.e., 6 12 7 12 b x a x   = i.e., x a b = æ è ç ö ø ÷ 2 1 6 \ Correct option is (a). 19. For the rotational equilibrium of the rod m g l k x l × = 2 i.e., x mg k = 2 \ PE stored in the spring = 1 2 2 kx = æ è ç ö ø ÷ 1 2 2 2 k mg k = ( ) mg k 2 8 \ Correct option is (c). 20. a m m m m g =  + 1 2 1 2 Speed (v) with which mass m 1 strikes the floor = + 0 2 2 gh =  + æ è ç ö ø ÷ 2 1 2 1 2 m m m m gh \ Correct option is (a). 21. F ax bx =  + 2 \  =  + dU dx ax bx 2 or dU ax bx dx =  ( ) 2 or U ax bx dx =  ò ( ) 2 or U ax bx c =  + 2 3 2 3 At x = 0, F = 0 \ U = 0 and so, c = 0 Thus, U ax bx =  2 3 2 3 U = 0, when ax bx 2 3 2 3 = i.e., at x a b = 3 2 Work, Energy and Power  131 C B A kx 1 kx 2 F = (k + k )x 12 q x mmg cos q mg sin q l kx l/2 mg Page 5 12. Decrease in KE =  1 2 0 2 mv = 1 2 2 mv Change in mechanical energy PE + KE = 1 2 2 mv Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. \ Change ME =  1 2 2 mv mgh Thus, assertion is true. As explained above the reason is false. (m does not change with the increase in angle of inclination) \ Correct option is (c). Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by gravity + work done by force F \ 1 2 2 mv mgh FR = + (Displacement of force F is R) = ´ ´ + ´ 1 2 10 5 5 5 = 50 \ u m = 100 = 200 =  14.14 ms 1 \ Correct option is (a). 2. P Fv = = mav = æ è ç ö ø ÷ m v dv dx v a dv dt dv ds ds dt v dv ds = = × = æ è ç ö ø ÷ \ ds m p v dv = 2 i.e., ds m p v dv v v ò ò = 2 2 or s m p v v v = é ë ê ù û ú 3 2 3 =  m p v v 3 2 3 3 [( ) ( ) ] = 7 3 3 mv p \ Correct option is (a). 3. Loss of PE of block = Gain in PE of spring mg d k ( ) sin + ° = ´ 2 30 1 2 2 2 or 10 10 2 1 2 1 2 100 2 2 ´ ´ + ´ = ´ ´ ( ) d d = 2 m Let, v A = velocity of block when it just touches spring \ 1 2 30 2 mv mgd A = ° sin v A 2 10 2 2 1 2 = ´ ´ ´ v A =  20 1 ms Correct option is (a). 128  Mechanics1 30° f' v' Stops PE = mgh h v Slope CaseI Rough f d A k 2m A 30° d m Work, Energy and Power  129 4. Mass per unit length of chain = m R p / 2 dm Rd m R = q p /2 y ydm dm CM = ò ò =  ò R m R Rd m 0 2 1 2 p q p q / ( cos ) / =  ò 2 1 0 2 R d p q q p ( cos ) / =  é ë ê ù û ú 2 2 1 R p p =  é ë ê ù û ú R 1 2 p Now, 1 2 2 mv mg y + CM or v g y 2 2 = COM v gR =  æ è ç ö ø ÷ 2 1 2 p \ Correct option is (c). 5. The moment string is cut Net force on A mg = 2 (downward) \ a g 1 2 = Net force on B = 0 \ a 2 0 = \ Correct option is (b). 6. Let x be the expansion in the spring. Increase in PE of spring = Decrease in PE of block C 1 2 2 1 kx M gx = i.e., kx M g = 2 1 For block A to remain at rest kx Mg = m min or 2 1 M g Mg = m min \ m min = 2 1 M M \ Correct option is (c). 7. T mg i = KX mg mg i = = 2 2 When one spring is cut. It means KX i becomes zero. Downward acceleration, a KX m mg m g i = = = 2 2 2 Now drawing FBD of lower mass : mg T m a mg f  = = . 2 \ T mg f = 2 or DT T T mg f i =  = 2 8. a mg mg m =  sin cos q m q =  g g sin cos q m q For v to be maximum dv dx = 0 or v dv dx = 0 F (= mg) mg mg A B F = mg T = mg T F = kx = mg mg (weight) F mg Mg 1 A B kx kx T T C T T R (1 – cos q) CM y ( ) R 1 – 2 p q q a x mmg cos q mg sin q or dx dt dv dx × = 0 or dv dt = 0 or a = 0 i.e., g g sin cos q m q = or sin cos q m q = or 3 5 3 10 4 5 = × x Þ x = = 10 4 2.5 m \ Correct option is (d). 9. (a) Between points E and F, dU dr is  ive. Now, F dU dr =  , the force between E and F will be + ive i.e., repulsive. (b) At point C the potential energy is minimum. Thus, C is point of stable equilibrium. \ Correct option is (c). 10. Power = × ® ® F v = × + ® ® ® m t g u g ( ) = ° + + × ® ® mg u m t cos ( ) 90 q g g i.e., P mg u mg t =  + sin q 2 [   ,  u g ® ® = = u g] Therefore, the graph between P and t will be as shown in option (c). 11. PE of spring due to its compression by x =  Work done by frictional force when displaced by 2x i.e., 1 2 2 2 kx mg x = m Þ x mg x = 4 m \ Correct option is (c). 12. Power delivered by man = × ® ® T v = T vcos q   T ® = T and   v ® = T 13. f = + 3 4 x y \ F x x =  ¶f ¶ =  3 N and F y y =  ¶f ¶ =  4 N PR PQ = 5 4 Þ PR PQ = ´ 5 4 = 10 m \ Work done by the conservative force on the particle = ´ F PR net = ´ 5 10 N m = 50 Nm = 50 J \ Correct option is (c). 14. Both at x x = 1 and x x = 2 the force acting on the body is zero i.e., it is in equilibrium. Now, if the body (when at x x = 1 ) is moved towards right (i.e., x x > 1 ) the force acting on it is + ive i.e., the body will not come back and if the body (when at x x = 2 ) is moved toward rght (i.e., x x > 2 ) the force acting on it is  ive i.e., the body will return back. Then,x x = 2 is the position of stable equilibrium. \ Correct option is (b). 130  Mechanics1 v mg u mg ® ® ® ® q x 2x mmg m v T ® ® P (6m, 8m) 4N 3N F net = 5N x R y 6 m Q 15. The man will stop when m q q mg mg cos sin = or ( ) cos sin m q q 0 x mg mg = or x = tan q m 0 16. Taking moment about A AC F k x l × = ( ) 2 \ AC k x l F = ( ) 2 = + ( ) ( ) k x l k k x 2 1 2 = + k k k l 2 1 2 \ Correct option is (d). 17. mgh + work done by the force of friction = 1 2 2 mv \ Work done by the force of friction =  1 2 2 mv mgh = ´ ´ æ è ç ö ø ÷  ´ ´ 1 2 1 2 1 10 1 2 ( ) =  8 J \ Correct option is (c). 18. U a x b x =  12 6 U ax bx =    12 6 For stable equilibrium, dU dx = 0 i.e., a x b x ( ) ( )    =   12 6 0 13 7 i.e., 6 12 7 12 b x a x   = i.e., x a b = æ è ç ö ø ÷ 2 1 6 \ Correct option is (a). 19. For the rotational equilibrium of the rod m g l k x l × = 2 i.e., x mg k = 2 \ PE stored in the spring = 1 2 2 kx = æ è ç ö ø ÷ 1 2 2 2 k mg k = ( ) mg k 2 8 \ Correct option is (c). 20. a m m m m g =  + 1 2 1 2 Speed (v) with which mass m 1 strikes the floor = + 0 2 2 gh =  + æ è ç ö ø ÷ 2 1 2 1 2 m m m m gh \ Correct option is (a). 21. F ax bx =  + 2 \  =  + dU dx ax bx 2 or dU ax bx dx =  ( ) 2 or U ax bx dx =  ò ( ) 2 or U ax bx c =  + 2 3 2 3 At x = 0, F = 0 \ U = 0 and so, c = 0 Thus, U ax bx =  2 3 2 3 U = 0, when ax bx 2 3 2 3 = i.e., at x a b = 3 2 Work, Energy and Power  131 C B A kx 1 kx 2 F = (k + k )x 12 q x mmg cos q mg sin q l kx l/2 mg dU dx = 0, when F = 0 i.e.,  + = ax bx 2 0 i.e., at x a b = Graph between U and x will be \ Correct option is (c). 22. W W Fs Fs A B = = 1 1 \ Correct option is (c). W W mv mv A B A B = = 1 2 1 2 4 1 2 2 Þ v v A B 2 2 4 1 = Þ v v A B = 2 1 W W A B = 1 1 \ K K A B = 1 1 23. U i at (1, 1) = + = k k ( ) 1 1 2 U f at (2, 3) = + = k k ( ) 2 3 5 W U U f i =  =  = 5 2 3 k k k \ Correct option is (b). 24. Gain in PE of spring = Loss of PE of block \ 1 2 2 k x mg h x max max ( ) = + …(i) \ From above Eq. (i), x max depends upon h and also x max depends upon k. KE of the block will be maximum when it is just at the point of touching the plank and at this moment there would no compression in the spring. Maximum KE of block = mgh \ Correct option is (c). 25. Gain in KE of chain = Decrease in PE of chain When the whole chain has justcome out of the tube. or 1 2 2 2 2 mv mg r = + æ è ç ö ø ÷ p p p \ v gr + + æ è ç ö ø ÷ 2 2 2 p p \ Correct option is (b). 26. Acceleration of the block will decrease as the block moves to the right and spring expands the velocity (v) of block will be maximum, when 1 2 1 2 2 2 mv kx = 132  Mechanics1 3a 2b X a b O x max h PE = 0 level for block 2r p CM (i) CM (f) pr pr/2Read More
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