Two blocks with masses m 1 = 2.50 kg and m 2 = 9.80 kg are attached over a pulle

Question

Two blocks with masses m1 = 2.50 kg and m2 = 9.80 kg are attached over a pulley with mass M = 3.20 kg hanging straight down as in Atwood's machine (see figure b). The pulley is a solid cylinder with radius 0.0530 m, and there is some friction on the axle. The system is released from rest, and the string moves without slipping over the pulley. If the larger mass is traveling at a speed of 2.42 m/s when it has dropped 1.00 m, how much mechanical energy was lost due to friction in the pulley's axle?
Wnc = J

Explanation / Answer

Find the total change in GPE of the Atwood's machine during the given interval. Then, find the total kinetic energy at the end state. It began from rest, so in theory, the total change in GPE equals the total kinetic energy at the end state, but since the axle has some friction, this will not be true. Define GPE to equal zero for the mass which is distance h below its highest point mentioned. State 1: large mass is high, small mass is low. System is entirely at rest. GPE1 = m2*g*h KE1 = 0 Since the rope is assumed inextensible, and no complex pulley arrangements exist, both masses travel the same distance State 2: large mass low, small mass high. Pulley is spinning, large mass is descending, small mass is rising. GPE2 = m1*g*h KE2 = 1/2*m2*v^2 + 1/2*m1*v^2 + 1/2*I*omega^2 As per no-slip condition, v = omega*r, thus omega = v/r Because it is treated as a uniform disk, I = 1/2*M*r^2 Thus: KE2 = 1/2*m2*v^2 + 1/2*m1*v^2 + 1/4*M*r^2*(v/r)^2 Summary: ?GPE = GPE1 - GPE2 ?KE = KE2 - KE1 ?GPE = g*h*(m2 - m1) ?KE = 1/4*(2*m2 + 2*m1 + M)*v^2 Heat lost to friction: Q = ?GPE - ?KE Thus: Q = g*h*(m2 - m1) - 1/4*(2*m2 + 2*m1 + M)*v^2 Data: g:=9.8 N/kg; m2:=9 kg; m1:=2 kg; h:= 1m; v:=2.5 m/s; M:=3 kg; Result: Q = 29.54 Joules

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