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Homework Statement
Suppose the roller coaster in the figure (h1 = 35 m, h2 = 13 m, h3 = 20) passes point A with a speed of 2.70 m/s. If the average force of friction is equal to one sixth of its weight, with what speed will it reach point B? The distance traveled is 35.0 m.
http://www.webassign.net/gianpse3/8-29alt.gif
Homework Equations
Work force of friction = [1/2mv2^2-1/2mv1^2] + [mgh2-mgh1]
The Attempt at a Solution
I took 1/6(9.8m/s^2)(35m) = [1/2v2^2-1/2(2.7m/s)^2] + [(9.8m/s^2)(13m)-(9.8m/s^2)(35m)] and got 23.5m/s which is not correct...what am I doing wrong?
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