Two snowy peaks are H = 790 m and h = 750 m above the valley between them. A ski run extends bewteen the peaks, with a total length of 3.1 km and an average slope of? = 30°.

(a) A skier starts from rest on the higher peak. At what speed will he arrive at the top of the lower peak if he just coasts without using the 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?

a)

Difference in height

= 790 - 750

= 40 m

. To know the speed at height of 750, the lower peak, we shall calculate loss in potential while coasting from height of 790 to 750 m

loss of p E

= mg (H - h)

1/2 m v² = mg (H - h)

v² = 2g (H - h)

= 2 x 9.8 x 40

= 784

v = 28 m / s

b)

Let coefficient of friction be μ.

friction force = μ mg cos 30

woke done by friction force = μ mg cos θ d, d is distance covered, μ is coefficient of friction

woke done

work done by friction force

3.1 x 10³ x μ mg cos 30

= loss of energy

= mg (H - h)

= work done by friction force = loss of energy

3.1 x 10³ x μ mg cos 30 = mg (H - h)

3.1 x 10³ x μ x cos 30 = (H - h) = 40

3100 x. 866 μ = 40

μ =.015

=