Inclination of a hill

A skier starts down a hill of length l and an angle of inclination of 10˚. It then moves to a horizontal section of the track, which travels the same length l until it stops.
Determine the coefficient of sliding friction between the skis and the snow.

Correct answer:

k =  0.1736

Step-by-step explanation:

$$\begin{gathered} \alpha =10°\to \text{rad}=10°\cdot \frac{\pi }{180}=10°\cdot \frac{3.1415926}{180}=0.17453=\pi /18 \\ g=9.81{\text{m/s}}^2 \\ {s}_1=s_2=l \\ {s}_1=\frac{1}{2}\cdot a_1\cdot t_1^2 \\ {s}_2=\frac{1}{2}\cdot a_2\cdot t_2^2 \\ {a}_1=a_2{\text{m/s}}^2 \\ g\cdot \sin \alpha =g\cdot k \\ k=\sin \alpha =\sin 0.1745=0.1736 \end{gathered}$$

Try another example