A 55.0-kg skier starts from rest at the top of a ski slope 90.0 m high. (a) If f

Question

A 55.0-kg skier starts from rest at the top of a ski slope 90.0 m high.

(a) If frictional forces do 10.5 kJ of work on her as she descends, how fast is she going at the bottom of the slope?

(b) Now moving horizontally, the skier crosses a patch of soft snow, where µk = 0.2. If the patch is 84.0 m wide and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch?

(c) The skier hits a snowdrift and penetrates 2.1 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Will rate.

Explanation / Answer

Solution-

(a) Let v be the velocity of skier at the bottom . We have
0.5*55*v^2 = 55*9.8*90 -10500 = 27720 or
v = sq rt[38010/27.5] = 37.18 m/s

(b)Now Let the skier move with velocity V after the snow patch, we have
0.5*55*[v^2 - V^2] = 160*84 + 0.2*55*9.8*84 = 22495.2
=>27.5*[1382.35 - V^2] = 22763.2 or
V^2 = 1382.35 - 827.75 = 554.6 or V = 23.55 m/s

(c)Let us assume the required force be F. We have
2.1*F = 0.5*55*V^2 = 27.5*554.60 = 15251.5 or F = 7262.62 N

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