Two snowy peaks are at heights$\mathit{H}\mathbf{=}\mathbf{850}\mathbf{}\mathit{m}$ and$\mathit{h}\mathbf{=}\mathbf{750}\mathbf{}\mathbf{m}$ above the valley between them. A ski run extends between the peaks, with a total length of 3.2 km and an average slope of$\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$ (Fig. 8-61). (a) A skier starts from rest at the top of the higher peak. At what speed will he arrive at the top of the lower peak if he coasts without using ski poles? Ignore friction. (b) Approximately what coefficient of kinetic friction between snow and skis would make him stop just at the top of the lower peak?

i) Height H =850 m

ii) Height h =750 m

iii) The length of the ski run is,$L=3.2\mathrm{km}=3200\mathrm{m}$ .

iv) The average slope $\theta =30°$.

Formula:

i) The potential energy is,$U=mgh$

ii) The kinetic energy is,

$K=\frac{1}{2}m{v}^2$

For the skier at the bottom of the peaks, the initial potential energy is given by

$U_i=mgH$

And the final potential energy is given by,

$U_f=mgh$

The initial kinetic energy of the skier is, and final kinetic energy is given by

$K_f=\frac{1}{2}m{v}^2$

Normal force does no work and friction is negligible, hence applying mechanical energy conservation write the equation as:

$U_i+K_i=U_f+K_f$

From this, we get,

$mgH=mgh+\frac{1}{2}m{v}^2$

Solve for the velocity as:

localid="1661232124504" $$\begin{gathered} \Rightarrow v^2=2g(H-h) \\ \Rightarrow v=\sqrt{2g(H-h)} \\ \Rightarrow v=\sqrt{2\times 9.8\times (850-750)} \\ \Rightarrow v=\sqrt{1960} \\ \Rightarrow v=44\hspace{0.33em}\frac{m}{s} \end{gathered}$$

The normal force of the slope on the skier is given by,

$F_N=mgcos\theta$

Then, the friction force is given by,

$$\begin{gathered} f={\mu }_k{F}_N \\ \Rightarrow f={\mu }_{k}mgcos\theta \end{gathered}$$

Thermal energy generated by the force of friction is given by,

$fd={\mu }_{k}mgdcos\theta$

d is the total distance along the path.

Since, the skier gets to the top of the lower peak with no kinetic energy, increase in the kinetic energy is equal to decrease in kinetic energy. Hence,

${\mu }_{k}mgdcos\theta =mg(H-h)$

$$\begin{gathered} \Rightarrow {\mu }_k=\frac{(H-h)}{dcos\theta } \\ \Rightarrow {\mu }_k=\frac{(850-750)}{3.2\times {10}^3\times cos30} \\ ={\mu }_k=\frac{100}{2.77\times {10}^3} \\ ={\mu }_k=0.036 \end{gathered}$$