Page 4 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 F = 80 N 5 g 4m g F mg ma = mg R(1 – 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 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 F = 80 N 5 g 4m g F mg ma = mg R(1 – 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 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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