91Ó°ÊÓ

The work done by the frictional force is equal to the loss in mechanical energy between the two points. The frictional force on the skier is directed downward as the skier moves upward.

Given data:

The angle of incline is\(\theta = 19^\circ \).

The length of the incline is\(L = 15\;{\rm{m}}\).

The initial speed of the skier is\({v_1} = 11.0\;\frac{{\rm{m}}}{{\rm{s}}}\)at the bottom of the incline.

The final velocity of the skier is\({v_2} = 0\)on the top of the incline.

Assumptions:

Let h be the height of the incline.

Let the coefficient of friction be\(\mu \).

From the figure, you get:

\(\begin{aligned}\frac{h}{L} &= \sin \theta \\h &= L\sin \theta \end{aligned}\)

The normal force on the skier is \(N = mg\cos \theta \).

Then, the friction force on the skier is:

\(\begin{aligned}{f_{\rm{k}}} &= \mu N\\ &= \mu mg\cos \theta \end{aligned}\)

The total mechanical energy of the skier at the bottom of the incline is:
\(\begin{aligned}{E_1} &= \left( {mg \times 0} \right) + \frac{1}{2}mv_1^2\\ &= \frac{1}{2}mv_1^2\end{aligned}\)

The total mechanical energy of the skier on the top of the incline is:

\(\begin{aligned}{E_2} &= mgh + \frac{1}{2}mv_2^2\\ &= mgh + \left( {\frac{1}{2}m \times {0^2}} \right)\\ &= mgh\\ &= mgL\sin \theta \end{aligned}\)

The work done against the frictional force is:

\(\begin{aligned}{W_{\rm{f}}} &= {f_{\rm{k}}}L\\ &= \mu mgL\cos \theta \end{aligned}\)

Now, from energy conservation, you can write:

\(\begin{aligned}{W_{\rm{f}}} &= {E_1} - {E_2}\\\mu mgL\cos \theta &= \frac{1}{2}mv_1^2 - mgL\sin \theta \\\mu gL\cos \theta &= \frac{1}{2}v_1^2 - gL\sin \theta \\\mu &= \frac{{v_1^2}}{{2gL\cos \theta }} - \tan \theta \end{aligned}\)

Now, substituting all the values in the above equation, you will get:

\(\begin{aligned}\mu &= \frac{{{{\left( {11.0\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}}{{2 \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^2}}}} \right) \times \left( {15\;{\rm{m}}} \right) \times \cos {{19}^ \circ }}} - \tan {19^ \circ }\\ &= 0.09\end{aligned}\)

Hence, the average coefficient of friction is 0.09.