Two blocks of mass m A = 5kg and mB = 7Akg are connected by a cable of negligibl

Question

Two blocks of mass m A = 5kg and mB = 7Akg are connected by a cable of negligible mass on a pulley. The pulley rotates on a frictionless axle, the cable doesn't slip on the pulley, it has a radius r = 4cm and a mass mp = 10g. Block A is on an inclined ramp at an angle 9 = 30 degree above the ground and there is a coefficient of friction mu k = 0.12 between block A and the ramp. Block B sits a distance d = 3m from the ground. The system is released from rest and block B descends. Calculate the speed of block B

Explanation / Answer

Let v be the final velocity of block B. It can easily be seen that v is also the velocity of block A and tangential speed of pulley.

Change in energy of different components of the system:

Epulley = [(1/2)I2 = (1/2) * (mpr2/2) * (v/r)2] - 0 = (1/2) * (0.01*0.042/2) * (v/0.04)2 = 0.0025v2

EB = (mBv2/2) - (mBgh) = (7 * v2/2) - (7 * 9.8 * 3) = 3.5v2 - 205.8

EA = PEA + KEA = mAgh(sin30o) + mAv2/2 = [5 * 9.8 * 3 * (1/2)] + (5 * v2/2) = 73.5 + 2.5v2

Also,

Work done by friction on block A, WA = -k[mAg(cos30o)] * (h*sin30o)

=> WA = -0.12 * 9.8 * cos30o * 3 * sin30o = -1.53 J

Friction is the only external force that does work in this problem. Tension does not do any work since the cable does not slip on the pulley and work done by tension on block A and block B cancels each other out.

So, we have

Epulley + EB + EA = WA

=> 0.0025v2 + (3.5v2 - 205.8) + (73.5 + 2.5v2) = -1.53 J

=> 6.0025v2 = 205.8 - 73.5 - 1.53

=> v = 4.67 m/s

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