Two blocks with masses m1 = 1.80 kg and m2 = 9.60 kg are attached over a pulley

Question

Two blocks with masses m1 = 1.80 kg and m2 = 9.60 kg are attached over a pulley with mass M = 3.10 kg hanging straight down as in Atwood's machine (see Figure (b)). The pulley is a solid cylinder with radius 0.0510 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 2.05 m/s when it has dropped 1.00 m, how much mechanical energy was lost due to friction in the pulley's axle?

Wnc = ??

*What is the initial energy in the system? What is the final energy? What causes the change in energy of the system? Can you write the kinetic energy of the pulley in terms of the given speed of the 9.60 kg block? J

*Please be very detailed when working this problem... Thanks u*

Explanation / Answer

I = moment of inertia of pulley = (0.5) Mr2

initial energy = potential energy of masses = m1gh + m2gh = (1.8 + 9.6) 9.8 x 1 = 111.72 J

final energy = Kinetic energy of two blocks + potential energy of m1 + Rotational KE of pulley

Ef = (0.5) m1 V2 +(0.5) m2 V2 + m1 g (2h) + (0.5) Iw2

Ef = (0.5) (m1 + m2 ) V2 + 2 m1 g h + (0.5) ((0.5) Mr2) (V/r)2

Ef = (0.5) (m1 + m2 ) V2 + 2 m1 g h + (0.25) MV2

Ef = (0.5) (1.80 + 9.60 ) (2.05)2 + 2 (1.80) (9.8) (2) + (0.25) (3.10) (2.05)2 = 97.8 J

energy lost = 111.72 - 97.8= 13.92

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