In the figure below, the hanging object has a mass of m1 = 0.400 kg; the sliding

Question

In the figure below, the hanging object has a mass of m1 = 0.400 kg; the sliding block has a mass of m2 = 0.845 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 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 vi = 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.

i couldnt upload image cuz i keep getting an error but the hollow cylinder has 2 radius being r1 and r2 which its moment of inertia will be 1/2m(r^2+r^2)

fors of friction on the block during that interval is ukmgh

i tried to get the rigth answer and so far i keep getting wrong on the webassign

here are the my previes answers

1.49

1.46

1.783

1.734

1.823

the advanced equaton editer sucks

5 stars to whoever explain it to me and get the right answer thanks so much in advanced!!

Explanation / Answer

The total energy in the system is given by T = 1/2 m2 Vo^2 + 1/2 m1 Vo^2 + 1/2 I wo^2 = 1/2 m2 Vf^2 + 1/2 m1 Vf^2 + 1/2 I wf^2 - m1gd + uk m2g d where Vo,Vf are the initial and final velocity of the blocks, wo/wf are the initial and final angular velocity of the pulley, I is the moment of inertia of the pulley. w is related to V as Vo = R2 wo and Vf = R2 wf. Also, the moment of inertia is given by I = 1/2 m (R2^2 + R1^2) You have all the data now, you can go ahead and solve for Vf in the energy equation above. After that, you can calculate wf.

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