Answer to Question #253955 in Physics for Kristine

Question #253955

A 62.0-π‘˜π‘” skier is moving at 6.50 π‘š/𝑠 on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.20 π‘š long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 π‘š 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?


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Expert's answer
2021-10-31T18:11:25-0400

(a) Work-energy theorem:


12mvi2=ΞΌmgx+12mvf2, vf=4.19 m/s\frac12mv_i^2=\mu mgx+\frac12mv_f^2,\\\space\\ v_f=4.19\text{ m/s}

at the end of the 4.2m-long rough gap. Then apply it one more time:


12mvf2+mgh=12mu2, u=8.16 m/s\frac12mv_f^2+mgh=\frac12 mu^2,\\\space\\ u=8.16\text{ m/s}

when she gets to the bottom of the hill.

(b) The internal energy is


U=ΞΌmgx=766 J.U=\mu mgx=766\text{ J}.


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