Question

Block A, mass 5.00 kg, rests on a surface with μk = 0.600. A massless rope is attached to its right side, and runs over a pulley, treated as a thin ring, mass 1.00 kg and radius 5.00 cm, to Block B, mass 7.00 kg, which hangs from the rope and is held at rest. The rope does not slip over the pulley, and the pulley spins on a frictionless axle. Block B is released from rest, and after an unspecified period of time, it has descended 2.00 m. Find the velocity of Block B at this point.

Answer 1

oment of inertia of ring, I = m*r^2

let v is the speed of two blocks and w is the angular speed of pulley.

Apply,

workdone by friction = change in mechanical energy

fk*d(cos(180) = (1/2)*(mA+mB)*v^2 + (1/2)*I*w^2 - mB*g*d

mue_k*mA*g*d*(-1) = (1/2)*(mA+mB)*v^2 + (1/2)*m*r^2*w^2 - mB*g*d

-mue_k*mA*g*d = (1/2)*(mA+mB)*v^2 + (1/2)*m*v^2 - mB*g*d (since v = r*w)

-0.6*5*9.8*2 = (1/2)*(5+7)*v^2 + (1/2)*1*v^2 - 7*9.8*2

on solving the above equation we get

v = 3.47 m/s <<<<<<<<<<---------------Answer