Posted by Anonymous on .
A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snow covered plateau when she encounters a rough patch 3.50 m long. The coefficient of kinetic friction between this patch and returning to frictionfree snow, she skis down an icy, frictionless hill 2.50 m high.
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?

COLLEGE PHYSICS 
drwls,
You seem to have omitted some words of the question, along with the coefficient of kinetic friction.
Subtract the work done crossing the rough patch from the initial kinetic energy, and then add the additional kinetic energy gained when going down the frictionless hill.
From the final kinetic energy, compute the final velocity. 
COLLEGE PHYSICS 
Anonymous,
Whoops! Here is the complete question.
A 62.0 kg skier is moving at 6.50 m/s on a frictionless, horizontal snowcovered plateau when she encounters a rough patch 3.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to frictionfree snow, she skis down an icy, frictionless hill 2.50 m high.
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch? 
COLLEGE PHYSICS 
drwls,
The work done (creating heat "internal energy") is
(Friction Force) *( distance)
= M g *0.300* 3.50 .
M is the mass and g is the acceleration of gravity.
The answer will be in Joules