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The pulley in the attachment has radius 0.160 [tex]m[/tex] and a moment of inertia 0.480 [tex]{\rm kg} \cdot {\rm m}^{2}[/tex]. The rope does not slip on the pulley rim.

1.Use energy methods to calculate the speed of the 4.00-kg block just before it strikes the floor.

I'm thinking: [tex]K_1 + U_1 = K_2 + U_2[/tex].

I'm also thinking that [tex]K_1=U_2=0[/tex], which implies [tex]U_1=K_2[/tex].

Then, [tex]mgh=(1/2)(m)(v^2)[/tex]. Solving for v, I get [tex]v=\sqrt(2*9.80*5)=9.90[/tex] m/s.

But, this isn't correct.

What did I do wrong?

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# Homework Help: Using energy methods in a moment of inertia problem?

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