A 60.5 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered

Question

A 60.5 kg skier is moving at 6.00 m/s on a frictionless, horizontal snow-covered plateau when she encounters a rough patch 3.05 m long. The coefficient of kinetic friction between this patch and her skis is 0.340. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 3.45 m high.

Part A

How fast is the skier moving when she gets to the bottom of the hill?

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Part B

How much internal energy was generated in crossing the rough patch?

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Explanation / Answer

initial kinetic energy = 0.5*m*v^2 = 0.5*60.5*(6)^2 = 1089 J

energy lost in crossing frictional patch = f*d
= miu*m*g*d
= 0.34*60.5*9.8*3.05
=614.84 J

Gain in potential energy= m*g*h = 60.5*9.8*3.45 = 2045.5 J

Kibetic energy at bottom = 1089 - 614.84 + 2045.5 =2519.66 J
0.5*m*vf^2 = 2519.66
0.5*60.5*vf^2 = 2519.66
vf = 9.13 m/s
Answer: 9.13 m/s

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There is no way to calculate the internal energy

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