Francisco P. answered 10/08/14
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Meg,
Could you give a clearer explanation for the problem? Thanks.
Francisco P.
Then the net work would equal the change in kinetic energy: Wnet = ΔK
Wg + Wfriction = mgh - Ffrictiond = mgh - μmgdcos(θ) = Kf - Ki
mgh - μmgdcos(θ) = ½mvf2 - ½mvi2
vf2 = vi2 + 2gh - 2μgdcos(θ)
vf = √[vi2 + 2gh - 2μgdcos(θ)]
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10/09/14
Francisco P.
Looks like you were just missing a factor of 2 in the frictional part of your solution.
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10/09/14
Meg N.
10/09/14