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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 pull

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Answer #1

given

m1 = 0.48 kg

m2 = 0.82 kg

M = 0.35 kg

R1 = 0.02 m

R2 = 0.3 m

\muk = 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 = \muk 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

wf = 58.5 rad/sec

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