A skier moves down a 12° slope at constant speed. What can you say about the coefficient of friction,${\mu }_{\mathrm{k}}$? Assume the speed is low enough that air resistance can be ignored.

The angle of incline is $\theta =12°$.

The coefficient of static friction is ${\mu }_{\mathrm{k}}$.

Let the mass of the skier be $m$.

The diagram below shows that the motion of the skier is downward directed and the frictional force is upward.

The skier moves down at a constant speed, which means that there is no acceleration of the skier; in other words, the net force on the skier is zero.

The normal force acting on the skier is $N=mg\cos \theta$.

The frictional force on the skier is:

$\begin{array}{c}{f}_{\mathrm{k}}={\mu }_{\mathrm{k}}N\\ ={\mu }_{\mathrm{k}}mg\cos \theta \end{array}$

The net force on the skier is:

$\begin{array}{c}F=mg\sin \theta -f_{\mathrm{k}}\\ =mg\sin \theta -{\mu }_{\mathrm{k}}mg\cos \theta \\ =mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)\end{array}$

The skier is moving at a constant speed; therefore, the net force on him is zero, i.e.,$F=0$. Then,

$\begin{array}{c}mg\left(\sin \theta -{\mu }_{\mathrm{k}}\cos \theta \right)=0\\ \sin \theta -{\mu }_{\mathrm{k}}\cos \theta =0\\ \sin \theta ={\mu }_{\mathrm{k}}\cos \theta \\ {\mu }_{\mathrm{k}}=\tan \theta \end{array}$

Now, substituting the value of $\theta$in the above equation, you will get:

$\begin{array}{c}{\mu }_{\mathrm{k}}=\tan 12°\\ =0.21\end{array}$

Hence, the coefficient of kinetic friction is${\mu }_{\mathrm{k}}=0.21$