A 61.0 kg skier starts from rest at the top of a ski slope of height 70.0 m. A.)

Question

A 61.0 kg skier starts from rest at the top of a ski slope of height 70.0 m.

A.) If frictional forces do -1.04 x 10^4 J 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 the coefficient of friction is 0.18. If the patch is of width 65.0 m and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch?

C.) After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.2 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

A) Work done by gravity + work done by friction = Change in K.E.

61 x 9.81 x 70 - 1.04 x 10^4 = 61v^2 /2 - 0

v = 32.13 m/s

B) Friction Force = u.mg = 0.18 x 61 x 9.81 = 107.71 N

Work done by firction + work done by air resistance = Change inK.E.

- 107.71 x 65 - 160 x 65 = 61v^2/2 - 61 x 32.13^2 /2

v = 21.49 m/s

C) using vf^2 - vi^2 = 2ad

0 - 21.49^2 = 2 x a x 2.2

a = 104.96 m/s2

F = ma = 6402.33 N

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