# In the figure below, the hanging object has a mass of m1 -0.480 kg; the sliding...

In the figure below, the hanging object has a mass of m1 -0.480 kg; the sliding block has a mass of m2 0.820 kg; and the pulley is a hollow cylinder with a mass of M0.350 kg, an inner radius of R10.020 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 Hk0.250. 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 of0.820 m/s toward the pulley when it passes a reference point on the table. R2 RI mg my (a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all Intermedlate results to at least four-digit accuracy to minlmize roundoff error. m/s (b) Find the angular speed of the pulley at the same moment. rad/s Need Help?Read It

given

m1 = 0.48 kg

m2 = 0.82 kg

M = 0.35 kg

R1 = 0.02 m

R2 = 0.3 m

$\mu$k = 0.25

vi = 0.82 m/sec

a )

d = 0.7 m away

I = 0.5 M ( R12 + R22 )

= 0.5 x 0.35 x ( 0.022 + 0.032 )

= 0.0002275

K1i = 1/2 m1vi2

= 0.5 x 0.48 x 0.822

= 0.1613 J

U1i = m1 g d

= 0.48 x 9.8 x 0.7

U1i = 3.2928 J

K2i = 1/2 m2 vi2

= 0.5 x 0.82 x 0.822

= 0.275 J

Kroti = 1/2 I wi2

= 0.5 x 0.0002275 x ( vi/R2 )2

= 0.5 x 0.0002275 x ( 0.82/0.03 )2

Kroti = 0.0849 J

the frictional force fk = $\mu$k m2 g

= 0.25 x 0.82 x 9.8

fk = 2 N

applying law of conservation energy is

( K1i + K2i + U1i + U2i + Krot i ) - fk d = K1f + K2f + Krot f

( 0.1613 + 0.275 + 3.2928 + 0 ) - 2 x 0.7 = 0.5 x 0.48 x vf2 + 0.5 x 0.82 x vf2 + 0.5 x

0.0002275 x ( vf / R2 )2

2.3291 = 0.24 x vf2 + 0.41 x vf2 + 0.126 vf2

0.24 x vf2 + 0.41 x vf2 + 0.126 vf2 = 2.3921

vf2 = 2.3921/0.776

vf2 = 3.0826

vf = 1.755 m/sec

b )

wf = vf / R2

= 1.755 / 0.03

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