In the figure, the hanging object has a mass of m_1 = 0.440 kg; the sliding bloc

Question

In the figure, the hanging object has a mass of m_1 = 0.440 kg; the sliding block has a mass of m_2 = 0.845 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R_1 = 0.0200 m, and an outer radius of R_2 = 0.0300 m (I = 1/2 (R_262 + R_2^2). Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is mu_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 v_i = 0.820 m/s toward the pulley when it passes a reference point on the table. Use energy methods to predict its speed after it has moved to a second point, 0.700 m away

Explanation / Answer

friction force = uk m2 g

Applying work - energy theorem,

Work done by all forces on system = change in KE

Work done by friction + work done by gravity = Kf - Ki

- uk m2 g d + m1 g d = [ (m1 + m2) vf^2 /2 + I wf^2 /2 ] - [ (m1 + m2) vi^2 /2 + I wi^2 /2 ]

-0.250 x 0.845 x 9.8 x 0.7 + (0.440 x 9.8 x 0.7) = [ (0.440 + 0.845) vf^2 /2 + (0.350 (0.03^2 + 0.02^2))(vf / 0.03)^2 / 2] - [ (0.0440 + 0.845)(0.820^2)/2 + (0.350 (0.03^2 + 0.02^2))(0.820 / 0.03)^2 / 2 ]

- 1.45 + 3.02 = (0.6425 vf^2 + 0.253 vf^2) -(0.4321 + 0.17)

1.57 = 0.8955 vf^2 - 0.602

vf = 1.56 m/s .........Ans

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