Question #253955

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 4.20 𝑚 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 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?


Expert's answer

(a) Work-energy theorem:


12mvi2=μmgx+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=μmgx=766 J.U=\mu mgx=766\text{ J}.


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