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Phy 241
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I am studying for TEST 2 (PHY 241) and need help. The problem states: Coasting from rest down a certain hill, whose slope is variable, I reach a speed of 8.485m/s at the bottom. If I coast from rest down the first half of the hill, I reach a speed of 6m/s. Ignoring the effects of air resistance and friction: How fast would I therefore be going if I coasted from rest down the 2nd half of the hill? How high would I have to climb from the bottom to the halfway point?
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Ok...I know that I will be using energy conservation on this problem. 'dPE_AC='dPE_AB+'dPE_BC, and since it's frictionless, 'dKE_XY=-'dPE_XY and 'dPE_AC=-'dKE_AB+'dPE_BC. Wouldn't I also use W_nc=0, 'dKE+'dPE=W_nc=0? If so, that's going to be (0.5mvf^2-0=mg(hf-h0)=mgh0) and vf=sqrt(2gh0). However, I don't know what to do with this considering I'm not given a mass or any heights. All I know is: v01, v02, vf, and mid velocity. HELP!
Your reasoning does lead nicely to
0.5mvf^2-0=mg(hf-h0).
This could apply to either phase of motion, the first half of the hill or the entire hill, since both start with velocity v0 = 0. However m g (hf - h0) is the change in PE, so the equation should read (0.5mvf^2-0) + mg(hf-h0) = 0. Then, assuming hf to be zero (which is a good choice, though you should state it), you can rearrange to get h0 = .5 m vf^2 / (m g). Note that the mass divides out, so you don't need the mass.
You would actually be better off not to make any assuptions about h0 or hf.
(0.5mvf^2-0) + mg(hf-h0) = 0 is easily solved for h0 - hf, the change in vertical position. You get
h0 - hf = .5 m vf^2 / (m g),
and h0 - hf is clearly the height of the hill.
It would be even simpler to use just `dh for the change in altitude. The equation would be
(0.5mvf^2-0) + mg `dh = 0,
with solution
`dh = -.5 m vf^2 / (m g).
`dh is negative as you go down the hill.
If your interval is top of hill to bottom, then your h0 - hf, or - `dh, will give you the height of the hill.
If your interval is the first half of the hill, then `dh will be the change in vertical position for that portion of the hill.
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Good reasoning. You've pretty much got it, just need to finish and interpret. See my notes.