JEE > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Work, energy and Power- 2

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Page 1 21. NR mgR mv s cos ( ) 180 1 2 2 ° - + = a ( cos ) ( cos ) mg R mgR mv s a a - + = 1 2 2 or 1 2 1 2 2 mv mgR s = - ( cos ) a or v gR s 2 2 2 = sin a or v gR s = 2 sin a (v s is the speed with which sphere hits ground) 1 2 1 2 2 2 mv mgR mv w s = - = - mgR m gR 1 2 2 2 sin a = - mgR ( sin ) 1 2 a = mgR cos 2 a \ v gR w = 2 cos a (v w is the speed of wedge when the sphere hits ground) 22. For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 45 12 10 1 2 45 1 2 1050 3 2 ´ ´ ´ = ´ ´ + ´ - 9.8 v ´ + - ´ - [( ) ] 75 24 75 10 2 2 6 i.e., 5.292 22.5 2.192 = + v 2 or 22.5 3.0996 v 2 = or v 2 = 0.13776 or v = - 0.371 ms 1 (b) With friction (when mechanical energy does not remain conserved) 23. De crease in PE of 1 kg mass = Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass 1 10 1 4 10 2 1 2 1 2 ´ ´ = ´ ´ ´ + ´ ´ m k ( ) 0.3 or 10 80 = + m k 0.045 or m k = - 10 80 0.045 =0.124 24. f = force of friction while disc in slipping over inclined surface = ° m mg cos 30 f ¢ = force of friction while disc is slipping over plane surface = m mg Now, decreases in PE of disc = Work done against frictional force mg s fs f 2 = + ¢ ( ) 0.5 or mg s mg s mg 2 30 = ° + ( cos ) ( ) m m 0.5 or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30 Þ s = ´ - ° m m 0.5 0.5 cos 30 = ´ - ´ 0.15 0.5 0.5 0.15 3 2 =0.2027 m Work performed by frictional forces over the whole distance = - mg s 2 = - ´ ´ 50 1000 10 2 0.2027 = -0.051 J 25. m g m g T m a A A A sin cos q m q - - = 2 or 300 3 5 300 4 5 2 30 ´ - ´ ´ - = 0.2 T a 118 | Mechanics-1 0.50 m Stops f' 30° s (say) f N s/2 a 2R N a R mg a 3 mg B T T 2T T T mmg cos q A T T a A S 4 q m g cos q A 2T a B Page 2 21. NR mgR mv s cos ( ) 180 1 2 2 ° - + = a ( cos ) ( cos ) mg R mgR mv s a a - + = 1 2 2 or 1 2 1 2 2 mv mgR s = - ( cos ) a or v gR s 2 2 2 = sin a or v gR s = 2 sin a (v s is the speed with which sphere hits ground) 1 2 1 2 2 2 mv mgR mv w s = - = - mgR m gR 1 2 2 2 sin a = - mgR ( sin ) 1 2 a = mgR cos 2 a \ v gR w = 2 cos a (v w is the speed of wedge when the sphere hits ground) 22. For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 45 12 10 1 2 45 1 2 1050 3 2 ´ ´ ´ = ´ ´ + ´ - 9.8 v ´ + - ´ - [( ) ] 75 24 75 10 2 2 6 i.e., 5.292 22.5 2.192 = + v 2 or 22.5 3.0996 v 2 = or v 2 = 0.13776 or v = - 0.371 ms 1 (b) With friction (when mechanical energy does not remain conserved) 23. De crease in PE of 1 kg mass = Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass 1 10 1 4 10 2 1 2 1 2 ´ ´ = ´ ´ ´ + ´ ´ m k ( ) 0.3 or 10 80 = + m k 0.045 or m k = - 10 80 0.045 =0.124 24. f = force of friction while disc in slipping over inclined surface = ° m mg cos 30 f ¢ = force of friction while disc is slipping over plane surface = m mg Now, decreases in PE of disc = Work done against frictional force mg s fs f 2 = + ¢ ( ) 0.5 or mg s mg s mg 2 30 = ° + ( cos ) ( ) m m 0.5 or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30 Þ s = ´ - ° m m 0.5 0.5 cos 30 = ´ - ´ 0.15 0.5 0.5 0.15 3 2 =0.2027 m Work performed by frictional forces over the whole distance = - mg s 2 = - ´ ´ 50 1000 10 2 0.2027 = -0.051 J 25. m g m g T m a A A A sin cos q m q - - = 2 or 300 3 5 300 4 5 2 30 ´ - ´ ´ - = 0.2 T a 118 | Mechanics-1 0.50 m Stops f' 30° s (say) f N s/2 a 2R N a R mg a 3 mg B T T 2T T T mmg cos q A T T a A S 4 q m g cos q A 2T a B or 180 48 2 30 - - = T a or 132 2 30 - = T a …(i) Also, T m g m a B B - = or T a - = 50 5 or 2 100 10 T a - = …(ii) Adding Eq. (i) and Eq. (ii), 32 40 = a or a = - 0.8 ms 2 Speed ( ) v of block A after it moves 1 m down the plane v as 2 2 = or v 2 2 1 = ´ ´ 0.8 or v = - 1.12 ms 1 26. Work done by frictional force acting on block = -m mgs = - ´ ´ ´ 0.25 9.8 7.8 3 5 . = - 66.88 J \ Increase in thermal energy of block-floor system = 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring = Maximum KE of block 1 2 2 kx max = 66.88 \ x max . = ´ 2 66 88 640 m Maximum compression in the spring = 0.457 m 27. Decrease in PE of mass m 2 = Work done against friction by mass m 1 + Increase KE of mass m 1 + Increase in KE of mass m 2 m g m g m v m v 2 1 1 2 2 2 2 4 4 1 2 1 2 ´ = ´ + + m 5 10 4 10 10 4 1 2 10 5 2 ´ ´ = ´ ´ ´ + + 0.2 ( )v Solving, v = - 4 1 ms Three types of Equilibrium 28. (a) F dU dr = - Point dU dr F A + ive - ive B + ive - ive C - ive + ive D - ive + ive E zero zero (b) x = 2 m point is of unstable equilibrium (U being + ive) x=6 m point is of stable equilibrium (U being lowest - ive) 29. U x x = - + 3 3 4 6 For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU dx =0 i.e., d dx x x 3 3 4 6 0 - + æ è ç ö ø ÷ = or x 2 4 0 - = or x=±2 dU dx d dx x x 2 2 4 4 2 = - = ( ) At x=+2 m, d U dx 2 2 2 2 4 = ´ + = + ( ) \ U is minium. At x = - 2 m, d U dx 2 2 2 2 4 = ´ - = - ( ) \ U is maximum. \ x = + 2 m point of stable equilibrium. x = - 2 m point of unstable equilibrium. Work, Energy and Power | 119 m = 5 kg 2 m 1 10 kg Page 3 21. NR mgR mv s cos ( ) 180 1 2 2 ° - + = a ( cos ) ( cos ) mg R mgR mv s a a - + = 1 2 2 or 1 2 1 2 2 mv mgR s = - ( cos ) a or v gR s 2 2 2 = sin a or v gR s = 2 sin a (v s is the speed with which sphere hits ground) 1 2 1 2 2 2 mv mgR mv w s = - = - mgR m gR 1 2 2 2 sin a = - mgR ( sin ) 1 2 a = mgR cos 2 a \ v gR w = 2 cos a (v w is the speed of wedge when the sphere hits ground) 22. For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 45 12 10 1 2 45 1 2 1050 3 2 ´ ´ ´ = ´ ´ + ´ - 9.8 v ´ + - ´ - [( ) ] 75 24 75 10 2 2 6 i.e., 5.292 22.5 2.192 = + v 2 or 22.5 3.0996 v 2 = or v 2 = 0.13776 or v = - 0.371 ms 1 (b) With friction (when mechanical energy does not remain conserved) 23. De crease in PE of 1 kg mass = Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass 1 10 1 4 10 2 1 2 1 2 ´ ´ = ´ ´ ´ + ´ ´ m k ( ) 0.3 or 10 80 = + m k 0.045 or m k = - 10 80 0.045 =0.124 24. f = force of friction while disc in slipping over inclined surface = ° m mg cos 30 f ¢ = force of friction while disc is slipping over plane surface = m mg Now, decreases in PE of disc = Work done against frictional force mg s fs f 2 = + ¢ ( ) 0.5 or mg s mg s mg 2 30 = ° + ( cos ) ( ) m m 0.5 or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30 Þ s = ´ - ° m m 0.5 0.5 cos 30 = ´ - ´ 0.15 0.5 0.5 0.15 3 2 =0.2027 m Work performed by frictional forces over the whole distance = - mg s 2 = - ´ ´ 50 1000 10 2 0.2027 = -0.051 J 25. m g m g T m a A A A sin cos q m q - - = 2 or 300 3 5 300 4 5 2 30 ´ - ´ ´ - = 0.2 T a 118 | Mechanics-1 0.50 m Stops f' 30° s (say) f N s/2 a 2R N a R mg a 3 mg B T T 2T T T mmg cos q A T T a A S 4 q m g cos q A 2T a B or 180 48 2 30 - - = T a or 132 2 30 - = T a …(i) Also, T m g m a B B - = or T a - = 50 5 or 2 100 10 T a - = …(ii) Adding Eq. (i) and Eq. (ii), 32 40 = a or a = - 0.8 ms 2 Speed ( ) v of block A after it moves 1 m down the plane v as 2 2 = or v 2 2 1 = ´ ´ 0.8 or v = - 1.12 ms 1 26. Work done by frictional force acting on block = -m mgs = - ´ ´ ´ 0.25 9.8 7.8 3 5 . = - 66.88 J \ Increase in thermal energy of block-floor system = 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring = Maximum KE of block 1 2 2 kx max = 66.88 \ x max . = ´ 2 66 88 640 m Maximum compression in the spring = 0.457 m 27. Decrease in PE of mass m 2 = Work done against friction by mass m 1 + Increase KE of mass m 1 + Increase in KE of mass m 2 m g m g m v m v 2 1 1 2 2 2 2 4 4 1 2 1 2 ´ = ´ + + m 5 10 4 10 10 4 1 2 10 5 2 ´ ´ = ´ ´ ´ + + 0.2 ( )v Solving, v = - 4 1 ms Three types of Equilibrium 28. (a) F dU dr = - Point dU dr F A + ive - ive B + ive - ive C - ive + ive D - ive + ive E zero zero (b) x = 2 m point is of unstable equilibrium (U being + ive) x=6 m point is of stable equilibrium (U being lowest - ive) 29. U x x = - + 3 3 4 6 For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU dx =0 i.e., d dx x x 3 3 4 6 0 - + æ è ç ö ø ÷ = or x 2 4 0 - = or x=±2 dU dx d dx x x 2 2 4 4 2 = - = ( ) At x=+2 m, d U dx 2 2 2 2 4 = ´ + = + ( ) \ U is minium. At x = - 2 m, d U dx 2 2 2 2 4 = ´ - = - ( ) \ U is maximum. \ x = + 2 m point of stable equilibrium. x = - 2 m point of unstable equilibrium. Work, Energy and Power | 119 m = 5 kg 2 m 1 10 kg 30. F dU dx = - i.e., U F dx = - ò = - (Area under F x - graph) The corresponding U Vs x graph will be as shown in figure Thus, point C corresponds to stable equilibrium and points A and E correspond to unstable equilibrium. 31. - q charge placed at origin is in equilibrium as is two equal and opposite forces act on it. (a)But, if we displace it slightly say towards +ive x side, the force on it due to charge to B will decreases while that due to A will increase. Due to net force on - q towards right the change - q will never come back to original O, its origin position. Thus, the equilibrium of the charge -q is unstable if it is slightly displaced along x-axis. (b) If charge - q is displaced slightly along Y axis, the net force on it will be along origin O and the particle will return to its original position. And as such the equilibrium of the - q is stable. 32. (a) Velocity at t = 0 s is 0 ms -1 Velocity at t=2 s is 8 ms -1 (using v at = + 0 ) \ v av ms = - 4 1 (as acceleration is constant) P F v av av = ´ =mav av = ´ ´ = 1 4 4 16 W (b) Velocity at t = 4 s is 16 ms -1 (using v u at = + ) \ Instantaneous power of the net force at t = 4 s will be P mav = = ´ ´ 1 4 16 =64 W 33. Power = F v min max P rv = max \ v P r max = Objective Questions (Level 1) Single Correct Option 1. In KE = 1 2 2 mv m is always +ive and v 2 is +ive. (even if v is - ive). \ KE is always + ive. Below reference level ( ) PE = 0 PE is - ive. \ Mechanical energy which is sum of KE and PE may be - ive. \ Correct option is (a). 120 | Mechanics-1 F x x E C D A B U + – + q – q F A (+ a, 0, 0) F x + q B (– a, 0, 0) y + q – q F net A F' + q B (– a, 0, 0) F' O + q – q F 2 F > F 12 O r F max Initially r F = (r) min Finally v max Page 4 21. NR mgR mv s cos ( ) 180 1 2 2 ° - + = a ( cos ) ( cos ) mg R mgR mv s a a - + = 1 2 2 or 1 2 1 2 2 mv mgR s = - ( cos ) a or v gR s 2 2 2 = sin a or v gR s = 2 sin a (v s is the speed with which sphere hits ground) 1 2 1 2 2 2 mv mgR mv w s = - = - mgR m gR 1 2 2 2 sin a = - mgR ( sin ) 1 2 a = mgR cos 2 a \ v gR w = 2 cos a (v w is the speed of wedge when the sphere hits ground) 22. For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 45 12 10 1 2 45 1 2 1050 3 2 ´ ´ ´ = ´ ´ + ´ - 9.8 v ´ + - ´ - [( ) ] 75 24 75 10 2 2 6 i.e., 5.292 22.5 2.192 = + v 2 or 22.5 3.0996 v 2 = or v 2 = 0.13776 or v = - 0.371 ms 1 (b) With friction (when mechanical energy does not remain conserved) 23. De crease in PE of 1 kg mass = Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass 1 10 1 4 10 2 1 2 1 2 ´ ´ = ´ ´ ´ + ´ ´ m k ( ) 0.3 or 10 80 = + m k 0.045 or m k = - 10 80 0.045 =0.124 24. f = force of friction while disc in slipping over inclined surface = ° m mg cos 30 f ¢ = force of friction while disc is slipping over plane surface = m mg Now, decreases in PE of disc = Work done against frictional force mg s fs f 2 = + ¢ ( ) 0.5 or mg s mg s mg 2 30 = ° + ( cos ) ( ) m m 0.5 or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30 Þ s = ´ - ° m m 0.5 0.5 cos 30 = ´ - ´ 0.15 0.5 0.5 0.15 3 2 =0.2027 m Work performed by frictional forces over the whole distance = - mg s 2 = - ´ ´ 50 1000 10 2 0.2027 = -0.051 J 25. m g m g T m a A A A sin cos q m q - - = 2 or 300 3 5 300 4 5 2 30 ´ - ´ ´ - = 0.2 T a 118 | Mechanics-1 0.50 m Stops f' 30° s (say) f N s/2 a 2R N a R mg a 3 mg B T T 2T T T mmg cos q A T T a A S 4 q m g cos q A 2T a B or 180 48 2 30 - - = T a or 132 2 30 - = T a …(i) Also, T m g m a B B - = or T a - = 50 5 or 2 100 10 T a - = …(ii) Adding Eq. (i) and Eq. (ii), 32 40 = a or a = - 0.8 ms 2 Speed ( ) v of block A after it moves 1 m down the plane v as 2 2 = or v 2 2 1 = ´ ´ 0.8 or v = - 1.12 ms 1 26. Work done by frictional force acting on block = -m mgs = - ´ ´ ´ 0.25 9.8 7.8 3 5 . = - 66.88 J \ Increase in thermal energy of block-floor system = 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring = Maximum KE of block 1 2 2 kx max = 66.88 \ x max . = ´ 2 66 88 640 m Maximum compression in the spring = 0.457 m 27. Decrease in PE of mass m 2 = Work done against friction by mass m 1 + Increase KE of mass m 1 + Increase in KE of mass m 2 m g m g m v m v 2 1 1 2 2 2 2 4 4 1 2 1 2 ´ = ´ + + m 5 10 4 10 10 4 1 2 10 5 2 ´ ´ = ´ ´ ´ + + 0.2 ( )v Solving, v = - 4 1 ms Three types of Equilibrium 28. (a) F dU dr = - Point dU dr F A + ive - ive B + ive - ive C - ive + ive D - ive + ive E zero zero (b) x = 2 m point is of unstable equilibrium (U being + ive) x=6 m point is of stable equilibrium (U being lowest - ive) 29. U x x = - + 3 3 4 6 For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU dx =0 i.e., d dx x x 3 3 4 6 0 - + æ è ç ö ø ÷ = or x 2 4 0 - = or x=±2 dU dx d dx x x 2 2 4 4 2 = - = ( ) At x=+2 m, d U dx 2 2 2 2 4 = ´ + = + ( ) \ U is minium. At x = - 2 m, d U dx 2 2 2 2 4 = ´ - = - ( ) \ U is maximum. \ x = + 2 m point of stable equilibrium. x = - 2 m point of unstable equilibrium. Work, Energy and Power | 119 m = 5 kg 2 m 1 10 kg 30. F dU dx = - i.e., U F dx = - ò = - (Area under F x - graph) The corresponding U Vs x graph will be as shown in figure Thus, point C corresponds to stable equilibrium and points A and E correspond to unstable equilibrium. 31. - q charge placed at origin is in equilibrium as is two equal and opposite forces act on it. (a)But, if we displace it slightly say towards +ive x side, the force on it due to charge to B will decreases while that due to A will increase. Due to net force on - q towards right the change - q will never come back to original O, its origin position. Thus, the equilibrium of the charge -q is unstable if it is slightly displaced along x-axis. (b) If charge - q is displaced slightly along Y axis, the net force on it will be along origin O and the particle will return to its original position. And as such the equilibrium of the - q is stable. 32. (a) Velocity at t = 0 s is 0 ms -1 Velocity at t=2 s is 8 ms -1 (using v at = + 0 ) \ v av ms = - 4 1 (as acceleration is constant) P F v av av = ´ =mav av = ´ ´ = 1 4 4 16 W (b) Velocity at t = 4 s is 16 ms -1 (using v u at = + ) \ Instantaneous power of the net force at t = 4 s will be P mav = = ´ ´ 1 4 16 =64 W 33. Power = F v min max P rv = max \ v P r max = Objective Questions (Level 1) Single Correct Option 1. In KE = 1 2 2 mv m is always +ive and v 2 is +ive. (even if v is - ive). \ KE is always + ive. Below reference level ( ) PE = 0 PE is - ive. \ Mechanical energy which is sum of KE and PE may be - ive. \ Correct option is (a). 120 | Mechanics-1 F x x E C D A B U + – + q – q F A (+ a, 0, 0) F x + q B (– a, 0, 0) y + q – q F net A F' + q B (– a, 0, 0) F' O + q – q F 2 F > F 12 O r F max Initially r F = (r) min Finally v max 2. Yes, this is work-en ergy the o rem. \ Correct option is (a). 3. On a body placed on a rough surface if an external force is applied the static body does not move then the work done by frictional force will be zero. 4. W = × ® ® F r 21 = × - ® ® ® F r r ( ) 2 1 = + + × - + - + + ( ) [( ) ( )] ^ ^ ^ ^ ^ ^ ^ ^ ^ i j k i j k i j k 2 3 2 = + + × - + ( ) ( ) ^ ^ ^ ^ ^ i j k j k 2 3 2 = - + 4 3 = - 1 J \ Correct option is (b). 5. \ W F dx = ò = - + ò 7 2 3 0 5 2 x x dx = - + = [ ] 7 135 2 3 0 5 x x x J \ Correct option is (d). 6. P = × ® ® F v = + + × - + ( ) ( ) ^ ^ ^ ^ ^ ^ 10 10 20 5 3 6 i j k i j k = - + 50 30 120 =140 W \ Correct option is (c). 7. Work done in displacing the body = Area under the curve = ´ + ´ + ´ - + ´ æ è ç ö ø ÷ ( ) ( ) ( ) 1 10 1 5 1 5 1 10 2 = 15 J \ Correct option is (b). 8. P W t mgh mv t = = + 1 2 2 = ´ ´ + ´ ´ - - 800 10 10 1 2 800 20 1 2 1 2 kg ms m kg ms min ( ) = ´ ´ + ´ ( ) [ ( ) ] 800 10 10 400 20 60 2 = + 80000 160000 60 = 4000 W Option (c) is correct. 9. x t = 3 3 \ v dx dt t = = 2 and a dv dt t = = 2 F ma t t = = ´ = 2 2 4 W F v = = ´ = 4 4 2 3 t t t \ Work done by force in first two seconds = × = = ò 4 3 0 2 t dt t t = × é ë ê ù û ú 4 4 4 0 2 t =16 J \ Correct option is (c). 10. Range = 4 ´ height or u g u g 2 2 2 2 4 2 sin sin q q = × or sin sin 2 2 2 q q = or 2 2 0 2 sin cos sin q q q - = or sin (cos sin ) q q q - = 0 or cos sin q q - = 0 (as sin q ¹ 0) or tan q = 1 i.e., q = ° 45 Now, K = KE at highest point = 1 2 2 m u ( cos ) q = 1 2 2 2 mu cos q = ° 1 2 45 2 2 mu cos = × 1 2 1 2 2 mu \ 1 2 2 2 mu K = i.e., Initial KE = 2K \ Correct option is (b). Work, Energy and Power | 121 Page 5 21. NR mgR mv s cos ( ) 180 1 2 2 ° - + = a ( cos ) ( cos ) mg R mgR mv s a a - + = 1 2 2 or 1 2 1 2 2 mv mgR s = - ( cos ) a or v gR s 2 2 2 = sin a or v gR s = 2 sin a (v s is the speed with which sphere hits ground) 1 2 1 2 2 2 mv mgR mv w s = - = - mgR m gR 1 2 2 2 sin a = - mgR ( sin ) 1 2 a = mgR cos 2 a \ v gR w = 2 cos a (v w is the speed of wedge when the sphere hits ground) 22. For 45 kg mass to drop 12 mm, the increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 45 12 10 1 2 45 1 2 1050 3 2 ´ ´ ´ = ´ ´ + ´ - 9.8 v ´ + - ´ - [( ) ] 75 24 75 10 2 2 6 i.e., 5.292 22.5 2.192 = + v 2 or 22.5 3.0996 v 2 = or v 2 = 0.13776 or v = - 0.371 ms 1 (b) With friction (when mechanical energy does not remain conserved) 23. De crease in PE of 1 kg mass = Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass 1 10 1 4 10 2 1 2 1 2 ´ ´ = ´ ´ ´ + ´ ´ m k ( ) 0.3 or 10 80 = + m k 0.045 or m k = - 10 80 0.045 =0.124 24. f = force of friction while disc in slipping over inclined surface = ° m mg cos 30 f ¢ = force of friction while disc is slipping over plane surface = m mg Now, decreases in PE of disc = Work done against frictional force mg s fs f 2 = + ¢ ( ) 0.5 or mg s mg s mg 2 30 = ° + ( cos ) ( ) m m 0.5 or mgs mg ( cos ) 0.5 (0.5) - ° = m m 30 Þ s = ´ - ° m m 0.5 0.5 cos 30 = ´ - ´ 0.15 0.5 0.5 0.15 3 2 =0.2027 m Work performed by frictional forces over the whole distance = - mg s 2 = - ´ ´ 50 1000 10 2 0.2027 = -0.051 J 25. m g m g T m a A A A sin cos q m q - - = 2 or 300 3 5 300 4 5 2 30 ´ - ´ ´ - = 0.2 T a 118 | Mechanics-1 0.50 m Stops f' 30° s (say) f N s/2 a 2R N a R mg a 3 mg B T T 2T T T mmg cos q A T T a A S 4 q m g cos q A 2T a B or 180 48 2 30 - - = T a or 132 2 30 - = T a …(i) Also, T m g m a B B - = or T a - = 50 5 or 2 100 10 T a - = …(ii) Adding Eq. (i) and Eq. (ii), 32 40 = a or a = - 0.8 ms 2 Speed ( ) v of block A after it moves 1 m down the plane v as 2 2 = or v 2 2 1 = ´ ´ 0.8 or v = - 1.12 ms 1 26. Work done by frictional force acting on block = -m mgs = - ´ ´ ´ 0.25 9.8 7.8 3 5 . = - 66.88 J \ Increase in thermal energy of block-floor system = 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring = Maximum KE of block 1 2 2 kx max = 66.88 \ x max . = ´ 2 66 88 640 m Maximum compression in the spring = 0.457 m 27. Decrease in PE of mass m 2 = Work done against friction by mass m 1 + Increase KE of mass m 1 + Increase in KE of mass m 2 m g m g m v m v 2 1 1 2 2 2 2 4 4 1 2 1 2 ´ = ´ + + m 5 10 4 10 10 4 1 2 10 5 2 ´ ´ = ´ ´ ´ + + 0.2 ( )v Solving, v = - 4 1 ms Three types of Equilibrium 28. (a) F dU dr = - Point dU dr F A + ive - ive B + ive - ive C - ive + ive D - ive + ive E zero zero (b) x = 2 m point is of unstable equilibrium (U being + ive) x=6 m point is of stable equilibrium (U being lowest - ive) 29. U x x = - + 3 3 4 6 For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU dx =0 i.e., d dx x x 3 3 4 6 0 - + æ è ç ö ø ÷ = or x 2 4 0 - = or x=±2 dU dx d dx x x 2 2 4 4 2 = - = ( ) At x=+2 m, d U dx 2 2 2 2 4 = ´ + = + ( ) \ U is minium. At x = - 2 m, d U dx 2 2 2 2 4 = ´ - = - ( ) \ U is maximum. \ x = + 2 m point of stable equilibrium. x = - 2 m point of unstable equilibrium. Work, Energy and Power | 119 m = 5 kg 2 m 1 10 kg 30. F dU dx = - i.e., U F dx = - ò = - (Area under F x - graph) The corresponding U Vs x graph will be as shown in figure Thus, point C corresponds to stable equilibrium and points A and E correspond to unstable equilibrium. 31. - q charge placed at origin is in equilibrium as is two equal and opposite forces act on it. (a)But, if we displace it slightly say towards +ive x side, the force on it due to charge to B will decreases while that due to A will increase. Due to net force on - q towards right the change - q will never come back to original O, its origin position. Thus, the equilibrium of the charge -q is unstable if it is slightly displaced along x-axis. (b) If charge - q is displaced slightly along Y axis, the net force on it will be along origin O and the particle will return to its original position. And as such the equilibrium of the - q is stable. 32. (a) Velocity at t = 0 s is 0 ms -1 Velocity at t=2 s is 8 ms -1 (using v at = + 0 ) \ v av ms = - 4 1 (as acceleration is constant) P F v av av = ´ =mav av = ´ ´ = 1 4 4 16 W (b) Velocity at t = 4 s is 16 ms -1 (using v u at = + ) \ Instantaneous power of the net force at t = 4 s will be P mav = = ´ ´ 1 4 16 =64 W 33. Power = F v min max P rv = max \ v P r max = Objective Questions (Level 1) Single Correct Option 1. In KE = 1 2 2 mv m is always +ive and v 2 is +ive. (even if v is - ive). \ KE is always + ive. Below reference level ( ) PE = 0 PE is - ive. \ Mechanical energy which is sum of KE and PE may be - ive. \ Correct option is (a). 120 | Mechanics-1 F x x E C D A B U + – + q – q F A (+ a, 0, 0) F x + q B (– a, 0, 0) y + q – q F net A F' + q B (– a, 0, 0) F' O + q – q F 2 F > F 12 O r F max Initially r F = (r) min Finally v max 2. Yes, this is work-en ergy the o rem. \ Correct option is (a). 3. On a body placed on a rough surface if an external force is applied the static body does not move then the work done by frictional force will be zero. 4. W = × ® ® F r 21 = × - ® ® ® F r r ( ) 2 1 = + + × - + - + + ( ) [( ) ( )] ^ ^ ^ ^ ^ ^ ^ ^ ^ i j k i j k i j k 2 3 2 = + + × - + ( ) ( ) ^ ^ ^ ^ ^ i j k j k 2 3 2 = - + 4 3 = - 1 J \ Correct option is (b). 5. \ W F dx = ò = - + ò 7 2 3 0 5 2 x x dx = - + = [ ] 7 135 2 3 0 5 x x x J \ Correct option is (d). 6. P = × ® ® F v = + + × - + ( ) ( ) ^ ^ ^ ^ ^ ^ 10 10 20 5 3 6 i j k i j k = - + 50 30 120 =140 W \ Correct option is (c). 7. Work done in displacing the body = Area under the curve = ´ + ´ + ´ - + ´ æ è ç ö ø ÷ ( ) ( ) ( ) 1 10 1 5 1 5 1 10 2 = 15 J \ Correct option is (b). 8. P W t mgh mv t = = + 1 2 2 = ´ ´ + ´ ´ - - 800 10 10 1 2 800 20 1 2 1 2 kg ms m kg ms min ( ) = ´ ´ + ´ ( ) [ ( ) ] 800 10 10 400 20 60 2 = + 80000 160000 60 = 4000 W Option (c) is correct. 9. x t = 3 3 \ v dx dt t = = 2 and a dv dt t = = 2 F ma t t = = ´ = 2 2 4 W F v = = ´ = 4 4 2 3 t t t \ Work done by force in first two seconds = × = = ò 4 3 0 2 t dt t t = × é ë ê ù û ú 4 4 4 0 2 t =16 J \ Correct option is (c). 10. Range = 4 ´ height or u g u g 2 2 2 2 4 2 sin sin q q = × or sin sin 2 2 2 q q = or 2 2 0 2 sin cos sin q q q - = or sin (cos sin ) q q q - = 0 or cos sin q q - = 0 (as sin q ¹ 0) or tan q = 1 i.e., q = ° 45 Now, K = KE at highest point = 1 2 2 m u ( cos ) q = 1 2 2 2 mu cos q = ° 1 2 45 2 2 mu cos = × 1 2 1 2 2 mu \ 1 2 2 2 mu K = i.e., Initial KE = 2K \ Correct option is (b). Work, Energy and Power | 121 11. P t t = - + 3 2 1 2 i.e., dW dt t t = - + 3 2 1 2 dW t t dt = - + ( ) 3 2 1 2 W t t dt = - + ò ( ) 3 2 1 2 2 4 = × - + é ë ê ù û ú 3 3 2 2 3 2 2 4 t t t or = - + [ ] t t t 3 2 2 4 = - + - - + ( ) ( ) 4 4 4 2 2 2 3 2 3 2 = - 52 6 = 46 DK = 46 J Correct option is (b). 12. K i = ´ ´ = 1 2 10 10 500 2 J Work done by retarding force, W xdx = - ò 0.1 20 30 = - é ë ê ù û ú 0.1 x 2 20 30 2 = - ´ - 0.05 [( ) ( ) ] 30 20 2 2 = -25 J Final kinetic energy = + K W i = + - 500 25 ( ) = 475 J \ Correct option is (a). 13. KE of 12 kg mass : KE of 6 kg mass = 1 2 1 2 12 2 6 2 m v m v : [Acceleration being same (equal to g) both will have same velocities] = m m 12 6 : =12 6 : =2 1 : 14. W k x x = - 1 2 2 2 1 2 [ ] = ´ ´ æ è ç ö ø ÷ - æ è ç ö ø ÷ é ë ê ù û ú 1 2 5 10 10 100 5 100 3 2 2 = ´ ´ - = 5 10 2 10 100 25 3 4 ( ) 18.75 Nm \ Correct option is (c). 15. x = 2.0 m to 3.5 m dU dx = - = - 6 4 1.5 \ F = + 4 N x = 3.5 m to 4.5 m dU dx = = 2 1 2 \ F = -2 N x = 4.5 m to 5.0 m dU dx =0 \ F =0 N Work done = ´ + - ´ + ( ) ( ) 4 2 1 0 1.5 = 4 J \ 1 2 4 2 mv = or 1 2 1 4 2 ´ ´ = v \ v = - 2 2 1 ms 16. KE at highest point = ° 1 2 45 2 m u ( cos ) = æ è ç ö ø ÷ 1 2 1 2 2 2 mu = æ è ç ö ø ÷ 1 2 1 2 2 mu = E 2 \ Correct option is (a). 17. Work done by person = - [Work done by gravitational pull on rope + gravitational pull on bucket] = - æ è ç ö ø ÷ + - é ë ê ù û ú mg h Mgh 2 ( ) = + æ è ç ö ø ÷ M m gh 2 \ Correct option is (a). 18. 1 2 2 mv Fx = …(i) 1 2 2 mv Fx ¢ = ¢ …(ii) \ x x v v ¢ = ¢ 2 2 = ( ) 2 2 2 v v = 4 Þ x x ¢ = 4 \ Correct option is (b). 122 | Mechanics-1Read More

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