A skier moves down a $12 ^\circ$ slope at constant speed. What can you say about the coefficient of friction, $\mu_k$? Assume the speed is low enough that air resistance can be ignored.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The skier is going down this slope at constant speed which means the component of gravity directed down the slope, Fgx, is equal in magnitude to the friction force that's opposing the velocity. And the frictional force is going up the slope. And so, that's what we say here. And Fgx is mg sine theta because this Fgx is the opposite leg of this gravity triangle. And force of friction is muK times normal force. And the normal force equals the component of gravity perpendicular to the slope which is mg cos theta. And it's using cosine because this is the adjacent leg of this triangle and let me substitute into here from both of these lines. So, we have that the force of friction is mg cos theta times coefficient of kinetic friction. And then we have that Fgx is mg sine theta. So, that's we write here. And then we divide both sides by mg and they just disappear which is nice. And then also divide both sides by cos theta, and you get this line here that coefficient of kinetic friction is sine theta divided by cos theta. And you just need to memorize that sine theta over cosine is always tangent theta. And that means the coefficient of kinetic friction is tangent of the slope angle which is 12 degrees and that makes mu k equal to 0.21.