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March 27, 2015

March 27, 2015

Posted by **Zach** on Monday, February 25, 2008 at 11:12pm.

I posted this problem earlier and it was replied to with : The initial kinetic energy (1/2)M V^2 must equal or exceed the potential energy change at the top of the "hill", which is M g H.

Therefore V > sqrt (2 g H)

I was not able to find the correct answer.

- Physics -
**DanH**, Monday, February 25, 2008 at 11:36pmKE at base of hill gets converted into potential energy at the top, plus remaining kinetic energy of bicycle moving. Thus, KE(base) = PE(top) + KE(top).

(mv^2)/2 base = mgh + (mv^2)/2 top.

Note the masses cancel out.

v^2/2(base) = gh + v^2/2(top)

400/2 = (9.8)(4.4) + v^2/2(top).

So you solve for v(top). This is the speed at which she is traveling over the top of the hill.

- Physics -
**drwls**, Monday, February 25, 2008 at 11:43pmI apologize for misreading your question.

I forgot that there is an initial velocity of 20 m/s. I thought they were asking for the minimum velocity to coast to the top of the hill.

The velocity decreases from V1 to V2 while coasting uphill so that the change in kinetic energy equals the potential energy increase

( M/2)(V1^2 - V2^2) = M g H

V2^2 = V1^2 - 2 g H = 400 - 2*9.8*4.4

= 400 - 86.24 = 313.76

V2 = 17.7 m/s is the velocity at the top of the hill.

- Physics -
**pen**, Tuesday, May 11, 2010 at 6:09pmA cyclist approaches the bottom of a gradual hill at a speed of 15 m/s. The hill is 4.7 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.

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