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# M2 Work, Power, Energy watch

1. I have tried working this out by working out the power in each case and then trying to equate them somehow but so far have had no luck in proving the expression.

"A car of mass m has a maximum power H. Its maximum speeds on the flat and up a hill of inclinatin alpha to the horizonal are v1 and v2. Assuming that the only resistance is one proportional to the speed, show that the maximum speed of the car down the same hill is v1²/v2."
2. anyone?
3. I've tried it but just get v3 = 2v1 - v2.

You have v3 = v1²/v2, which gives you the ratio (v3/v1) = (v1/v2), but I can't see how to get that !
4. yea i get to the exact same point as you, it's really frustrating
5. OK, first define resistance as R = kv.

So along slope, H = Fv where F= R = kv1 so H = kv1^2

This also gives v1^2 = H/k which we´ll need later.

Next, up slope
This time F = R + mgsina = Kv2 + mgsina

So H = kv2^2 + mgsina.v2 -------------------------(2)

Similarly for the downwards one

H = kv3^2 - mgsina.v3 ________________________(3)

equating 2 and 3 gives us

kv3^2 - mgsina.v3 = kv2^2 + mgsina.v2

So kv3^2 - kv2^2 = mgsina.v2 + mgsina.v3

So k (v3^2 - v2^2) = mgsina (v2 + v3)

Factorise LHD, diff of 2 sqs

k (v3 + v2) (v3 - v2) = mgsina (v2 + v3)

Cancel the bracket

k (v3 - v2) = mgsina

Ok, now in eqn (3), subs this value for mgsina and you get

v3 = H /( k v2) and remembering that H/k is v1^2 there's your answer.

Hurrah!

(There may well be a quicker way to get there ....)
6. thanks a lot rsk

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Updated: October 23, 2006
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