In the figure below, the hanging object has a mass of m 1 = 0.440 kg; the slidin
ID: 2017566 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.440 kg; the sliding block has a mass of m2 = 0.800 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius ofR2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
Explanation / Answer
Given thatm1 = 0.440 kg, m2 = 0.800 kg, M = 0.350 kg, R1 = 0.02 m, R2 = 0.03 m. µk = 0.250, vi = 0.820 m/s,
moment of inertia I = M(R1^2 + R2^2)/2 = 2.275*10^-4 kgm^2
(a) initial energy = (m1^vi^2)/2 + (m2^vi^2)/2 + I*wi^/2 = (m1 + m2)*vi^2/2 + I*(vi/R2)^2/2
= (m1 + m2 + I/R2^2)*vi^2/2 = 0.5 J
final energy = (m1 + m2 + I/R2^2)*vf^2/2 - m1*g*d = 1.49*vf^2/2 - 3.0184
work done by friction W = -k*m2*gd = 1.372 J
so
W = Final energy - initial energy 1.372 = (1.49*vf^2/2 - 3.0184) - 0.5 So the final velocitty, vf = 2.56 m/s
(b) The angular speed is wf = vf/R2 = 85.4 rad/s