91Ó°ÊÓ

The angle of incline is $\mathrm{θ}=12°$.

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

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 $N=mg\cos \mathrm{θ}$.

The frictional force on the skier is:

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

The net force on the skier is:

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

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

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

Now, substituting the value of $\mathrm{θ}$in the above equation, you will get:

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

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