Answer to Question #253955 in Physics for Kristine

Question

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?

Answer 1

(a) Work-energy theorem:

$$\frac12mv_i^2=\mu mgx+\frac12mv_f^2,\\\space\\ v_f=4.19\text{ m/s}$$

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

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

when she gets to the bottom of the hill.

(b) The internal energy is

$$U=\mu mgx=766\text{ J}.$$