Asked by lok
A 58.0-kg skier starts from rest at the top of a ski slope of height 70.0 .If frictional forces do −1.09×104 of work on her as she descends, how fast is she going at the bottom of the slope?
Answers
Answered by
drwls
I will have to assume the height is in meters and the friction work is in Joules. Don't be sloppy by omitting units.
Her kinetic energy will be
(M*V^2)/2 = M g H - 1.09*10^4 J
You know M, g and H.
Solve for V
Her kinetic energy will be
(M*V^2)/2 = M g H - 1.09*10^4 J
You know M, g and H.
Solve for V
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