Two snowy peaks are 842 m and 742 m above the valley between them. A ski run ext

Question

Two snowy peaks are 842 m and 742 m above the valley between them. A ski run extends down from the top of the higher peak and then back up to the top of the lower one, with a total length of 2.8 km and an average slope of 30.0°. A skier starts from rest on the higher peak. At what speed will she arrive at the top of the lower peak if she just coasts without using the poles? Ignore friction. Approximately what coefficient of kinetic friction between the snow and skis would make the skier stop just at the top of the lower peak?

Explanation / Answer

Apply conservation of energy

K1 + U1 = K2 + U2

0 + m*g*h1 = 0.5*m*v2^2 + m*g*h2

0.5*m*v2^2 = m*g*(h1-h2)

v2 = sqrt(2*g*(h1-h2))

= sqrt(2*9.8*(842 - 742))

= 44.27 m/s <<<<<<<<<<<<<<<<<<---------Answer

let mue_k is the coeffcient friction.

Normal force, N = m*g*cos(30)

frictional force = N*mue_k

now Apply workdone by friction = change in mechnical energy

Friction*d*cos(180) = -m*g*(h1-h2)

-N*mue_k*d = -m*g*(h1-h2)

m*g*cos(30)*mue_k*d = m*g*(h1-h2)

mue_k = (h1-h2)/(d*cos(30))

= (842-742)/(2800*cos(30))

= 0.04 <<<<<<<<<<<<<<<<<<---------Answer

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