Page 1 Introductory Exercise 6.1 1. Work done by F ® = × ® ® F x | | ,| | F N ® ® = = F N, | | W ® = W | | T ® = T and | | x ® = x = ® ® | | | |cos F x p = - F x Work done by | | N ® = × ® ® N x = ® ® | | | |cos N x p 2 =0 Work done by W W x ® ® ® = × = × ® ® | || |cos W x p 2 =0 Work done by T ® = × ® ® T x = ® ® | | | |cos T x 0 = Tx 2. Work done by F ® = × ® ® F l = × ® ® | || |cos F l p = - F l Work done by N N l ® ® ® = × = × = ® ® | || |cos N l p 2 0 Work done by | | W W l ® ® ® = × = + æ è ç ö ø ÷ ® ® | || |cos W l p a 2 = -W l sina Work done by | | T T l ® ® ® = × = ® ® | || |cos T l b = T l cos b 6 Work, Energy and Power T N F W â€“ x T ® ® ® ® ® ® a F N T b l W |F| = F, |N| = N |W| = W, |T| = T |l| = l ® ® ® ® ® ® ® ® ® ® ® Page 2 Introductory Exercise 6.1 1. Work done by F ® = × ® ® F x | | ,| | F N ® ® = = F N, | | W ® = W | | T ® = T and | | x ® = x = ® ® | | | |cos F x p = - F x Work done by | | N ® = × ® ® N x = ® ® | | | |cos N x p 2 =0 Work done by W W x ® ® ® = × = × ® ® | || |cos W x p 2 =0 Work done by T ® = × ® ® T x = ® ® | | | |cos T x 0 = Tx 2. Work done by F ® = × ® ® F l = × ® ® | || |cos F l p = - F l Work done by N N l ® ® ® = × = × = ® ® | || |cos N l p 2 0 Work done by | | W W l ® ® ® = × = + æ è ç ö ø ÷ ® ® | || |cos W l p a 2 = -W l sina Work done by | | T T l ® ® ® = × = ® ® | || |cos T l b = T l cos b 6 Work, Energy and Power T N F W â€“ x T ® ® ® ® ® ® a F N T b l W |F| = F, |N| = N |W| = W, |T| = T |l| = l ® ® ® ® ® ® ® ® ® ® ® 3. W T g ® ® ® - = m 4 | | g ® = g and | | l ® = l \ | | T g g ® ® ® = - m m 4 = ® 3 4 m g Work done by string = × ® ® T l = ® ® | || |cos T l p = ½ ½ ½ ½ ½ ½ ® ® 3 4 mg l cosp = - 3 4 mgl 4. m N F = ° cos45 â€¦(i) N W = = ® ® | |, | | N W s F = = ® ® | |, | | s F , g = ® | | g N F W + ° = sin45 â€¦(ii) Substituting value of N from Eq. (ii) in Eq. (i). m ( sin ) cos W F F - ° = ° 45 45 or 1 4 1 2 1 2 W F F - æ è ç ö ø ÷= or W F F - = 2 4 2 or 5 2 F W = or F W mg = = 2 5 2 5 Work done by force F = × ® ® F s = ° ® ® | | | |cos F s 45 = F s 1 2 = æ è ç ö ø ÷ 2 5 1 2 mg s = mgs 5 = ´ ´ 1.8 10 2 5 =7.2 J Work done by friction = × ® ® m N s = ® ® m p | | | |cos N s = -m Ns = - ° F s cos45 = -7.2 J Work done by gravity = × ® ® g s = × ® ® | || |cos g s p 2 =0 5. F mg = ° sin45 = ´ ´ 1 10 1 2 =5 2 N Displacement of lift in 1s = 2 m Work, Energy and Power | 109 W N mN s F cos 45° F F F sin 45° 45° 45° mg mg cos 45° F â€“1 2ms sin 45° W = mg T a = g 4 l ® ® ® Page 3 Introductory Exercise 6.1 1. Work done by F ® = × ® ® F x | | ,| | F N ® ® = = F N, | | W ® = W | | T ® = T and | | x ® = x = ® ® | | | |cos F x p = - F x Work done by | | N ® = × ® ® N x = ® ® | | | |cos N x p 2 =0 Work done by W W x ® ® ® = × = × ® ® | || |cos W x p 2 =0 Work done by T ® = × ® ® T x = ® ® | | | |cos T x 0 = Tx 2. Work done by F ® = × ® ® F l = × ® ® | || |cos F l p = - F l Work done by N N l ® ® ® = × = × = ® ® | || |cos N l p 2 0 Work done by | | W W l ® ® ® = × = + æ è ç ö ø ÷ ® ® | || |cos W l p a 2 = -W l sina Work done by | | T T l ® ® ® = × = ® ® | || |cos T l b = T l cos b 6 Work, Energy and Power T N F W â€“ x T ® ® ® ® ® ® a F N T b l W |F| = F, |N| = N |W| = W, |T| = T |l| = l ® ® ® ® ® ® ® ® ® ® ® 3. W T g ® ® ® - = m 4 | | g ® = g and | | l ® = l \ | | T g g ® ® ® = - m m 4 = ® 3 4 m g Work done by string = × ® ® T l = ® ® | || |cos T l p = ½ ½ ½ ½ ½ ½ ® ® 3 4 mg l cosp = - 3 4 mgl 4. m N F = ° cos45 â€¦(i) N W = = ® ® | |, | | N W s F = = ® ® | |, | | s F , g = ® | | g N F W + ° = sin45 â€¦(ii) Substituting value of N from Eq. (ii) in Eq. (i). m ( sin ) cos W F F - ° = ° 45 45 or 1 4 1 2 1 2 W F F - æ è ç ö ø ÷= or W F F - = 2 4 2 or 5 2 F W = or F W mg = = 2 5 2 5 Work done by force F = × ® ® F s = ° ® ® | | | |cos F s 45 = F s 1 2 = æ è ç ö ø ÷ 2 5 1 2 mg s = mgs 5 = ´ ´ 1.8 10 2 5 =7.2 J Work done by friction = × ® ® m N s = ® ® m p | | | |cos N s = -m Ns = - ° F s cos45 = -7.2 J Work done by gravity = × ® ® g s = × ® ® | || |cos g s p 2 =0 5. F mg = ° sin45 = ´ ´ 1 10 1 2 =5 2 N Displacement of lift in 1s = 2 m Work, Energy and Power | 109 W N mN s F cos 45° F F F sin 45° 45° 45° mg mg cos 45° F â€“1 2ms sin 45° W = mg T a = g 4 l ® ® ® Work done by force of friction ( ) F = × ® ® F s = ° ® ® | | | |cos F s 45 = ° = × × = F scos 45 5 2 2 1 2 10 Nm 6. Total work-done by spring on both masses = PE of the spring when stretched by 2 0 x = 1 2 2 0 2 k x ( ) = 2 0 2 k x \ Work done by spring on each mass = 2 2 0 2 kx =kx 0 2 7. Work done = Area under the curve = + + + A A A A 1 2 3 4 = - ´- é ë ê ù û ú + - ´ é ë ê ù û ú + ´ 2 10 2 2 10 2 10 2 [ ] + ´ é ë ê ù û ú 2 10 2 =30 Nm Introductory Exercise 6.2 1. a s = - = - - - 20 2 10 1 2 ms ms \ F ma = = ´ - - 2 10 2 kg ms = - 20 N s = Area under curve = ´ ´ - 1 2 25 20 1 ms =20 m \ Work done = = - F s ( ) ( ) 20 20 N m = - 400 Nm 2. According to A (inertial frame) Acceleration of P a = \ Force on P ma = Work done over s displacement = mas Now, v u as 2 2 2 = + =2as (Q u =0) \ Gain in KE = = = 1 2 1 2 2 2 mv m as mas \ DK W = (Work -Energy theorem) According to B (non-inertial frame) work done = Work done by F + Work done by f p (pseudo force) = + - ( ) ( ) mas mas = 0 As P is at rest , DK = 0 \ DK W = (Work -Energy theorem) Note In in er tial frames one has to also to con sider work-done due to pseudo forces, while ap ply ing Work-en ergy the o rem. 3. v x =a \ a dv dt x dx dt = =a 1 2 =a a 1 2 x x = a 2 2 F ma m = = a 2 2 \ W m b = a 2 2 110 | Mechanics-1 m m m m x 0 2 4 â€“ 4 â€“ 2 A 1 A 2 A 3 A 4 10 â€“ 10 x (m) F(N) Q P a A m Q P F a B f P Page 4 Introductory Exercise 6.1 1. Work done by F ® = × ® ® F x | | ,| | F N ® ® = = F N, | | W ® = W | | T ® = T and | | x ® = x = ® ® | | | |cos F x p = - F x Work done by | | N ® = × ® ® N x = ® ® | | | |cos N x p 2 =0 Work done by W W x ® ® ® = × = × ® ® | || |cos W x p 2 =0 Work done by T ® = × ® ® T x = ® ® | | | |cos T x 0 = Tx 2. Work done by F ® = × ® ® F l = × ® ® | || |cos F l p = - F l Work done by N N l ® ® ® = × = × = ® ® | || |cos N l p 2 0 Work done by | | W W l ® ® ® = × = + æ è ç ö ø ÷ ® ® | || |cos W l p a 2 = -W l sina Work done by | | T T l ® ® ® = × = ® ® | || |cos T l b = T l cos b 6 Work, Energy and Power T N F W â€“ x T ® ® ® ® ® ® a F N T b l W |F| = F, |N| = N |W| = W, |T| = T |l| = l ® ® ® ® ® ® ® ® ® ® ® 3. W T g ® ® ® - = m 4 | | g ® = g and | | l ® = l \ | | T g g ® ® ® = - m m 4 = ® 3 4 m g Work done by string = × ® ® T l = ® ® | || |cos T l p = ½ ½ ½ ½ ½ ½ ® ® 3 4 mg l cosp = - 3 4 mgl 4. m N F = ° cos45 â€¦(i) N W = = ® ® | |, | | N W s F = = ® ® | |, | | s F , g = ® | | g N F W + ° = sin45 â€¦(ii) Substituting value of N from Eq. (ii) in Eq. (i). m ( sin ) cos W F F - ° = ° 45 45 or 1 4 1 2 1 2 W F F - æ è ç ö ø ÷= or W F F - = 2 4 2 or 5 2 F W = or F W mg = = 2 5 2 5 Work done by force F = × ® ® F s = ° ® ® | | | |cos F s 45 = F s 1 2 = æ è ç ö ø ÷ 2 5 1 2 mg s = mgs 5 = ´ ´ 1.8 10 2 5 =7.2 J Work done by friction = × ® ® m N s = ® ® m p | | | |cos N s = -m Ns = - ° F s cos45 = -7.2 J Work done by gravity = × ® ® g s = × ® ® | || |cos g s p 2 =0 5. F mg = ° sin45 = ´ ´ 1 10 1 2 =5 2 N Displacement of lift in 1s = 2 m Work, Energy and Power | 109 W N mN s F cos 45° F F F sin 45° 45° 45° mg mg cos 45° F â€“1 2ms sin 45° W = mg T a = g 4 l ® ® ® Work done by force of friction ( ) F = × ® ® F s = ° ® ® | | | |cos F s 45 = ° = × × = F scos 45 5 2 2 1 2 10 Nm 6. Total work-done by spring on both masses = PE of the spring when stretched by 2 0 x = 1 2 2 0 2 k x ( ) = 2 0 2 k x \ Work done by spring on each mass = 2 2 0 2 kx =kx 0 2 7. Work done = Area under the curve = + + + A A A A 1 2 3 4 = - ´- é ë ê ù û ú + - ´ é ë ê ù û ú + ´ 2 10 2 2 10 2 10 2 [ ] + ´ é ë ê ù û ú 2 10 2 =30 Nm Introductory Exercise 6.2 1. a s = - = - - - 20 2 10 1 2 ms ms \ F ma = = ´ - - 2 10 2 kg ms = - 20 N s = Area under curve = ´ ´ - 1 2 25 20 1 ms =20 m \ Work done = = - F s ( ) ( ) 20 20 N m = - 400 Nm 2. According to A (inertial frame) Acceleration of P a = \ Force on P ma = Work done over s displacement = mas Now, v u as 2 2 2 = + =2as (Q u =0) \ Gain in KE = = = 1 2 1 2 2 2 mv m as mas \ DK W = (Work -Energy theorem) According to B (non-inertial frame) work done = Work done by F + Work done by f p (pseudo force) = + - ( ) ( ) mas mas = 0 As P is at rest , DK = 0 \ DK W = (Work -Energy theorem) Note In in er tial frames one has to also to con sider work-done due to pseudo forces, while ap ply ing Work-en ergy the o rem. 3. v x =a \ a dv dt x dx dt = =a 1 2 =a a 1 2 x x = a 2 2 F ma m = = a 2 2 \ W m b = a 2 2 110 | Mechanics-1 m m m m x 0 2 4 â€“ 4 â€“ 2 A 1 A 2 A 3 A 4 10 â€“ 10 x (m) F(N) Q P a A m Q P F a B f P 4. DK = Work done by F + Work done by gravity = × × + × × 80 4 0 5 4 cos cos g p = + - 320 200 ( ) or K K f i - =120 J or K f =120 J (as K i = 0) 5. Change in KE = Work done 1 2 1 2 mv mgR mgR = - + ( cos ) sin q q Þ v gR = - + 2 1 ( cos sin ) q q 6. DK W = or 0 1 2 0 2 0 - = - ò mv Ax dx x or - = - 1 2 2 0 2 2 mv A x Þ x v m A = 0 7. (a) If T mg = , the block will not get ac cel er ated to gain KE. The value of T must be greater that Mg. \ Ans. False (b) As some negative work will be done by Mg, the work done by T will be more that 40 J. \ Ans. False (c) Pulling force F will always be equal to T, as T is there only because of pulling. \ Ans. True (d) Work done by gravity will be negative Ans. False Introductory Exercise 6.3 1. In Fig. 1 Spring is having its natural length. In Fig. 2 A is released. A goes down byx . Spring get extended by x. Decrease in PE of A is stored in spring as its PE. \ mAg x k x = 1 2 2 Now, for the block B to just leave contact with ground kx mg = i.e., 2m g mg A = Þ m m A = 2 Work, Energy and Power | 111 F = 80 N 5 g 4m g F mg ma = mg R(1 â€“ cosq) O R sin q R mg mg q T T T F Hand Mg Fig. 1 Fig. 2 B A m Ground x B A m Ground mg A T T T T T T mg Page 5 Introductory Exercise 6.1 1. Work done by F ® = × ® ® F x | | ,| | F N ® ® = = F N, | | W ® = W | | T ® = T and | | x ® = x = ® ® | | | |cos F x p = - F x Work done by | | N ® = × ® ® N x = ® ® | | | |cos N x p 2 =0 Work done by W W x ® ® ® = × = × ® ® | || |cos W x p 2 =0 Work done by T ® = × ® ® T x = ® ® | | | |cos T x 0 = Tx 2. Work done by F ® = × ® ® F l = × ® ® | || |cos F l p = - F l Work done by N N l ® ® ® = × = × = ® ® | || |cos N l p 2 0 Work done by | | W W l ® ® ® = × = + æ è ç ö ø ÷ ® ® | || |cos W l p a 2 = -W l sina Work done by | | T T l ® ® ® = × = ® ® | || |cos T l b = T l cos b 6 Work, Energy and Power T N F W â€“ x T ® ® ® ® ® ® a F N T b l W |F| = F, |N| = N |W| = W, |T| = T |l| = l ® ® ® ® ® ® ® ® ® ® ® 3. W T g ® ® ® - = m 4 | | g ® = g and | | l ® = l \ | | T g g ® ® ® = - m m 4 = ® 3 4 m g Work done by string = × ® ® T l = ® ® | || |cos T l p = ½ ½ ½ ½ ½ ½ ® ® 3 4 mg l cosp = - 3 4 mgl 4. m N F = ° cos45 â€¦(i) N W = = ® ® | |, | | N W s F = = ® ® | |, | | s F , g = ® | | g N F W + ° = sin45 â€¦(ii) Substituting value of N from Eq. (ii) in Eq. (i). m ( sin ) cos W F F - ° = ° 45 45 or 1 4 1 2 1 2 W F F - æ è ç ö ø ÷= or W F F - = 2 4 2 or 5 2 F W = or F W mg = = 2 5 2 5 Work done by force F = × ® ® F s = ° ® ® | | | |cos F s 45 = F s 1 2 = æ è ç ö ø ÷ 2 5 1 2 mg s = mgs 5 = ´ ´ 1.8 10 2 5 =7.2 J Work done by friction = × ® ® m N s = ® ® m p | | | |cos N s = -m Ns = - ° F s cos45 = -7.2 J Work done by gravity = × ® ® g s = × ® ® | || |cos g s p 2 =0 5. F mg = ° sin45 = ´ ´ 1 10 1 2 =5 2 N Displacement of lift in 1s = 2 m Work, Energy and Power | 109 W N mN s F cos 45° F F F sin 45° 45° 45° mg mg cos 45° F â€“1 2ms sin 45° W = mg T a = g 4 l ® ® ® Work done by force of friction ( ) F = × ® ® F s = ° ® ® | | | |cos F s 45 = ° = × × = F scos 45 5 2 2 1 2 10 Nm 6. Total work-done by spring on both masses = PE of the spring when stretched by 2 0 x = 1 2 2 0 2 k x ( ) = 2 0 2 k x \ Work done by spring on each mass = 2 2 0 2 kx =kx 0 2 7. Work done = Area under the curve = + + + A A A A 1 2 3 4 = - ´- é ë ê ù û ú + - ´ é ë ê ù û ú + ´ 2 10 2 2 10 2 10 2 [ ] + ´ é ë ê ù û ú 2 10 2 =30 Nm Introductory Exercise 6.2 1. a s = - = - - - 20 2 10 1 2 ms ms \ F ma = = ´ - - 2 10 2 kg ms = - 20 N s = Area under curve = ´ ´ - 1 2 25 20 1 ms =20 m \ Work done = = - F s ( ) ( ) 20 20 N m = - 400 Nm 2. According to A (inertial frame) Acceleration of P a = \ Force on P ma = Work done over s displacement = mas Now, v u as 2 2 2 = + =2as (Q u =0) \ Gain in KE = = = 1 2 1 2 2 2 mv m as mas \ DK W = (Work -Energy theorem) According to B (non-inertial frame) work done = Work done by F + Work done by f p (pseudo force) = + - ( ) ( ) mas mas = 0 As P is at rest , DK = 0 \ DK W = (Work -Energy theorem) Note In in er tial frames one has to also to con sider work-done due to pseudo forces, while ap ply ing Work-en ergy the o rem. 3. v x =a \ a dv dt x dx dt = =a 1 2 =a a 1 2 x x = a 2 2 F ma m = = a 2 2 \ W m b = a 2 2 110 | Mechanics-1 m m m m x 0 2 4 â€“ 4 â€“ 2 A 1 A 2 A 3 A 4 10 â€“ 10 x (m) F(N) Q P a A m Q P F a B f P 4. DK = Work done by F + Work done by gravity = × × + × × 80 4 0 5 4 cos cos g p = + - 320 200 ( ) or K K f i - =120 J or K f =120 J (as K i = 0) 5. Change in KE = Work done 1 2 1 2 mv mgR mgR = - + ( cos ) sin q q Þ v gR = - + 2 1 ( cos sin ) q q 6. DK W = or 0 1 2 0 2 0 - = - ò mv Ax dx x or - = - 1 2 2 0 2 2 mv A x Þ x v m A = 0 7. (a) If T mg = , the block will not get ac cel er ated to gain KE. The value of T must be greater that Mg. \ Ans. False (b) As some negative work will be done by Mg, the work done by T will be more that 40 J. \ Ans. False (c) Pulling force F will always be equal to T, as T is there only because of pulling. \ Ans. True (d) Work done by gravity will be negative Ans. False Introductory Exercise 6.3 1. In Fig. 1 Spring is having its natural length. In Fig. 2 A is released. A goes down byx . Spring get extended by x. Decrease in PE of A is stored in spring as its PE. \ mAg x k x = 1 2 2 Now, for the block B to just leave contact with ground kx mg = i.e., 2m g mg A = Þ m m A = 2 Work, Energy and Power | 111 F = 80 N 5 g 4m g F mg ma = mg R(1 â€“ cosq) O R sin q R mg mg q T T T F Hand Mg Fig. 1 Fig. 2 B A m Ground x B A m Ground mg A T T T T T T mg 2. Decrease in PE = mg l 2 \ 1 2 2 2 mv mg l = i.e., v g l = 3. OA = 50 cm \ Extension in spring (when collar is at A) = 50 cm - 10 cm = 0.4 m Extension in spring (collar is at B) = 30 cm - 10 cm = 20 cm = 0.2 m KE of collar at B = - PEofspring PEofspring (collarat ) (collarat ) A B = ´ ´ - 1 2 2 2 K [( ) ( ) ] 0.4 0.2 or 1 1 2 500 1 2 m mv B = ´ ´ 2 or v B = ´ 500 012 10 . =245 . s -1 Extension in spring (collar arrives at C) = + - [ ( ) ( ) ] 30 20 10 2 2 cm = 0.26 m KE of collar at C = PE of spring - PE of spring (Collar at A) (Collar at C) = ´ ´ - 1 2 500 0 4 026 2 2 [( . ) ( . ) ] or 1 2 1 2 500 00924 2 mv C = ´ ´ . or v c = ´ 500 100 0.0924 = - 2.15ms 1 4. Work done by man = + m gh Mgh 2 = + æ è ç ö ø ÷ m M gh 2 5. When block of man M goes down by x, the spring gets extended by x. Decrease in PE of man M is stored in spring as its PE. \ Mgx k x = 1 2 2 or kx Mg = 2 For the block of man m to just slide kx mg mg = ° + ° sin cos 37 37 m or 2 3 5 3 4 4 5 Mg mg mg = + or M m = 3 5 Introductory Exercise 6.4 1. Velocity at time t = 2 s v gt = = ´ = - 10 2 20 1 ms Power = Force ´ velocity = mgv = ´ ´ 1 10 20 =200 W 2. Velocity at time = a t = × F m t \ v Ft m av = 2 (acceleration being constant) P F v av av = ´ = F t m 2 2 112 | Mechanics-1 A C 20 cm B 40 cm 30 cm O m m 37° 37° mg T T T T T T x mg sin 37° mmg cos 37°Read More

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