7. +-1 points SerPSET9 10.P.053 N In the figure below, the hanging object has a

Question

7. +-1 points SerPSET9 10.P.053 N In the figure below, the hanging object has a mass of m-0.430 kg: the sliding block has a mass of m2-0.885 kg; and the pulley is a hollow cylinder with a mass of M-0.350 kg, an inner radius of 10.020 0 m, and an outer radius of R2 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 0.820 m/s toward the pulley when it passes a reference point on the table (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. m/s (b) Find the angular speed of the pulley at the same moment rad/s

Explanation / Answer

h = x = 0.7 m

work done by gravitational force Wg = m1*g*h

work done by frictioanl force Wf = -uk*m2*g*x

moment of inertia of pulley I = (1/2)*M*(R1^2 + R2^2) = (1/2)*0.35*(0.03^2+0.02^2) = 22.75*10^-5 kg m^2

initial kinetic energy of system = Ki = (1/2)*m1*vi^2 + (1/2)*m2*vi^2 + (1/2)*I*wi^2

wi = vi/R2

initial kinetic energy of system = Ki = (1/2)*(m1+m2)*vi^2 + (1/2)*I*(vi/R2)^2

final kinetic energy of system = Kf = (1/2)*(m1+m2)*vf^2 + (1/2)*I*(vf/R2)^2

work energy relation

work done = change in KE

Wg + Wf = Kf - Ki

m1*g*h - uk*m2*g*x = (1/2)*(m1+m2)*(vf^2-vi^2) + (1/2)*I*(vf^2-vi^2)/R2^2

0.43*9.8*0.7 - 0.25*0.885*9.8*0.7 = (1/2)*(0.43+0.885)*(vf^2-0.82^2) + (1/2)*22.75*10^-5*(vf^2-0.82^2)/0.03^2

vf = 1.52m/s

(b)

angular speed wf = vf/R2 = 1.52/0.03 = 50.7 rad/s

DONE please check the answer. any doubts post in comment box

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