Question

(II) A ski starts from rest and slides down a 28° incline 85 m long. (a) If the coefficient of friction is 0.090, what is the ski’s speed at the base of the incline? (b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? Use energy methods.

Short answer

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

The ski travels 368.33 m along the level.

Step-by-step solution

Given data

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.

Calculation for part (a)

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.

Calculation for part (b)

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.