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 Oct 29th 2008, 09:11 PM #1 Junior Member   Join Date: Oct 2008 Posts: 8 Work and Energy Question Rene, whose mass is 85 kg, skis down a hill, passing a point Z with a kinetic energy of 9700J. If friction is ignored, to what maximum height, h, can Rene ski? Relevant Equations ΔEp = -ΔEk -ΔEp = ΔEk for relevant equations I am not quite sure because one teacher showed us to use the one above while another teacher showed us the one below. Attempt For this question I am assuming that you use the above one because you are trying to find the final height "hf" for this question. So what I did was: -ΔEp = ΔEk -(mghf - mghi) = (1/2)mvf(squared) - (1/2)mvi(squared) I continued to plug in the numbers and isolate the final height "hf", but then I encountered a problem because I don't know what the final velocity is "vf" so I don't know how to proceed.
Oct 31st 2008, 07:51 PM   #2
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 Originally Posted by Inertialforce Question Rene, whose mass is 85 kg, skis down a hill, passing a point Z with a kinetic energy of 9700J. If friction is ignored, to what maximum height, h, can Rene ski? Relevant Equations ΔEp = -ΔEk -ΔEp = ΔEk for relevant equations I am not quite sure because one teacher showed us to use the one above while another teacher showed us the one below. Attempt For this question I am assuming that you use the above one because you are trying to find the final height "hf" for this question. So what I did was: -ΔEp = ΔEk -(mghf - mghi) = (1/2)mvf(squared) - (1/2)mvi(squared) I continued to plug in the numbers and isolate the final height "hf", but then I encountered a problem because I don't know what the final velocity is "vf" so I don't know how to proceed.
Using
$\displaystyle 9700= mv^2/2$
then you can get the velocity at that time

-ΔEp = ΔEk
$\displaystyle -mgh=mv^2/2$
$\displaystyle h=-v^2/2g$

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