Question

Two blocks, A and B (with mass 60 kg and 115 kg, respectively), are connected by a string, as shown in the figure below. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between block A and the incline is ur = 0.27. Determine the change in the kinetic energy of block A as it moves from © to D, a distance of 24 m up the incline (and block B drops downward a distance of 24 m) if the system starts from rest. D) В

Answer 1

solution Given mass of block (A) m = bokg mass of block B m = 115 kg Coefficient of kinetic friction between block A cincline F. BiDi of block A het 'a' be acceleration of block & T is tension in string. Using Newton's Ind law aleurs blu mag. 60137 At Equations of motion NA = mag as 370 No = 60x9.8 (0.799) [:co37€ 0.799) No = 469.812 neton newton & T-(mg Sin 37° + 4) = maa. - ) = 60 a (Goog. 8x 0.602 + MkNA T (353.976 + 0.27 x 469.812) = 6o a T- 480,82 524 = 60 a F. B.D. og block B mos mog - T. = my a 115x9.8 T = 115a Scannedish T = lisa

Adding & 4 , 1127.00 - 480.825 24 = 60a +11sa 646.17476 = 175 a la = 3.69 m/s2/ from eqh 1127 - T = 115*3.69 - T - 1128 424.35 T 11 27.00 - 424.35 = 702.65 = Te T= 702.65 N c to D, net Now, when block Í mones from point force acting on it is given by E = T - (Ag Sin 37° + 1 x) 'net = 702.65 - (60x9.8 x0.60 +0.27X 469.8) For = 70265 - 480.83 Fhot = 221.82 N Using work - energy theorem - Change in the kinetic energy of block A = work done in moving from c to D sk - Fret (distance between c & D) Ak = ( 22 1.8 2) (24) lok = 5323.68 JL Ary. sk Scanner with = 5.32368 x 103 Cams ar ner k= 5.32 X103 J
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