Answer to Question #271296 in Physics for Abby

Question

Someone goes skiing. He starts from rest at the top of a ski hill with an incline of 22 degrees up from the ground. If this person is going 26 m/s after 10 seconds, what is the coefficient of friction between his skis and the hill?

Answer 1

$F=ma\to mg\sin22°-\mu mg\cos22°=ma\to$

$a=g\sin22°-\mu g\cos22°$

$v=at\to v=g(\sin22°-\mu \cos22°)t\to\mu=\tan22°-\frac{v}{gt\cos22°}=$

$=\tan22°-\frac{26}{9.8\cdot 10\cdot\cos22°}=0.404-0.286=0.118$ . Answer