Question

- Problem Solving (40 points total, 10 points each) 12. A 75-kg skier stands at rest atop a hill. The coefficient of friction between the skis and snow is 0.06. The skier glides down the length of the hill to its bottom. What is the skier's velocity at the bottom of the hill? L = 32.5 m O = 40°

Answer 1

NA mgsin40 mgco 40 I=32.5m mg 40 deg

Consider the figure

$sin40=\frac{h}{l}$

$h=lsin40$

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As the skier slides down the hill, initial potential energy is converted into kinetic energy.

Due to friction, some portion of initial kinetic energy is lost.

$PE-W_{friction}=KE$

$PE-F_{friction}*l=KE$

$PE-\mu N*l=KE$

$mgh-\mu mgcos40*l=\frac{1}{2}mv^{2}$

$gh-\mu gcos40*l=\frac{1}{2}v^{2}$

$glsin40-\mu gcos40*l=\frac{1}{2}v^{2}$

$9.81m/s^{2}*32.5m*sin40-0.06*9.81m/s^{2}*cos40*32.5m=0.5*v^{2}$

$9.81*32.5*sin40-0.06*9.81*cos40*32.5=0.5*v^{2}$

$190.2827125=0.5*v^{2}$

$\sqrt{\frac{190.2827125}{0.5}}=v$

ANSWER: ${\color{Red} v=19.508m/s}$

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