91Ó°ÊÓ

The work done by the friction force is equal to the loss in mechanical energy.

Given data:

The angle of incline is\(\theta = {28^ \circ }\).

The length of the incline is\(L = 85\;{\rm{m}}\).

The coefficient of friction is\(\mu = 0.090\).

The initial speed of the ski is\({v_1} = 0\)as it starts from rest.

Assumptions:

Let h be the initial height of the ski.

Let them\({v_2}\)be the speed of the ski at the base of the incline.

Let the ski travel x distance along with the level.

From the figure,

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

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

Then, the friction force on the ski is as follows:

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

The total mechanical energy of the ski at the top of the incline is calculated as follows:

\(\begin{aligned}{E_1} &= mgh + \frac{1}{2}mv_1^2\\ &= mgh + \left( {\frac{1}{2}m \times {0^2}} \right)\\ &= mgh\\ &= mgL\sin \theta \end{aligned}\)

The total mechanical energy of the ski at the bottom of the incline is calculated as follows:

\(\begin{aligned}{E_2} &= \left( {mg \times 0} \right) + \frac{1}{2}mv_2^2\\ &= \frac{1}{2}mv_2^2\end{aligned}\)

Work done against the friction force is calculated as follows:

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

Now from energy conservation,

\(\begin{aligned}{W_{\rm{f}}} &= {E_1} - {E_2}\\\mu mgL\cos \theta &= mgL\sin \theta - \frac{1}{2}mv_2^2\\\frac{{v_2^2}}{2} &= gL\sin \theta - \mu gL\cos \theta \\{v_2} &= \sqrt {2gL\left( {\sin \theta - \mu \cos \theta } \right)} \end{aligned}\)

Substituting all the values in the above equation,

\(\begin{aligned}{v_2} &= \sqrt {2 \times \left( {9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}} \right) \times \left( {85\;{\rm{m}}} \right) \times \left( {\sin {{28}^ \circ } - 0.090\cos {{28}^ \circ }} \right)} \\ &= 25.49\;\frac{{\rm{m}}}{{\rm{s}}}\end{aligned}\)

Hence, the speed of the ski is \(25.49\;\frac{{\rm{m}}}{{\rm{s}}}\)at the bottom of the incline.

When the ski is at the flat surface, the normal force is equal to the weight.

Then, the friction force on the ski is \(f' = \mu mg\) .

Work done by the friction force to travel a distance xis calculated as follows:

\(\begin{aligned}W &= f'x\\ &= \mu mgx\end{aligned}\)

From energy conservation,

\(\begin{aligned}W &= \frac{1}{2}mv_2^2\\\mu mgx &= \frac{1}{2}mv_2^2\\x &= \frac{{v_2^2}}{{2\mu g}}\end{aligned}\)

Substituting the value in the above equation,

\(\begin{aligned}x &= \frac{{{{\left( {25.49\;\frac{{\rm{m}}}{{\rm{s}}}} \right)}^2}}}{{2 \times 0.090 \times 9.80\;\frac{{\rm{m}}}{{{{\rm{s}}^{\rm{2}}}}}}}\\ &= 368.33\;{\rm{m}}\end{aligned}\)

Hence, the ski travels 368.33 m along the level.