- #1

PacFan01

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## Homework Statement

[/B]

A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disc of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground?

## Homework Equations

ΔE = 0

E

_{i}= E

_{f}(closed system)

E = K + U

^{G}+ E

_{th}+ E

_{s}

U

_{i}

^{G}= K

_{transf}+ K

_{rotf}

K

_{trans}= ½mv

^{2}

K

_{rot}= ½Iω

^{2}

I

_{Solid Cylinder}= ½mr

^{2}

ω = v/r

## The Attempt at a Solution

[/B]

First, I step up the energy equation and substitute in my energy expressions. I've used the subscript "1" to represent the 0.20-kg mass, the subscript "2" to represent the 0.25-kg mass, and the subscript "p" to represent the pulley.

U

_{1i}

^{G}+ U

_{2i}

^{G}= U

_{1f}

^{G}+ U

_{2f}

^{G}+ K

_{1f}+ K

_{2f}+ K

_{pf}

m

_{1}gh

_{1i}+ m

_{2}gh

_{2i}= m

_{1}gh

_{1f}+ m

_{2}gh

_{2f}+ ½m

_{1}v

_{1}

^{2}+ ½m

_{2}v

_{2}

^{2}+ ½Iω

^{2}

I know what the next step is, but I'm not sure how to get there algebraically.

I should end up with:

(m

_{1}- m

_{2})gΔh = ½(m

_{1}+ m

_{2})v

_{f}

^{2}+ ¼m

_{p}v

_{f}

^{2}

I understand that v

_{1}, v

_{2}, and v

_{p}are equal, so no problem there. I see that how we substitute ½mr

^{2}for I, and v/r for ω.

__I don't understand how we end up with (m__I understand that we pull the common variable g out, and I can kind of understand how the h-values become Δh. However, it looks like this when I set it up:

_{1}- m_{2})gΔh.m

_{1}gh

_{1i}+ m

_{2}gh

_{2i}- m

_{1}gh

_{1f}- m

_{2}gh

_{2f}= ½m

_{1}v

_{1}

^{2}+ ½m

_{2}v

_{2}

^{2}+ ½Iω

^{2}

Which becomes:

g(m

_{1}h

_{1i}+ m

_{2}h

_{2i}- m

_{1}h

_{1f}- m

_{2}h

_{2f}) = ½m

_{1}v

_{1}

^{2}+ ½m

_{2}v

_{2}

^{2}+ ½Iω

^{2}

Any help is greatly appreciated.