A 63.0-kg skier starts from rest at the top of a ski slope of height 68.0m . Par

Question

A 63.0-kg skier starts from rest at the top of a ski slope of height 68.0m .

Part B

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.21. If the patch is of width 63.0m and the average force of air resistance on the skier is 180N , how fast is she going after crossing the patch?

??

Part C

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.5m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

??

Explanation / Answer

From that height, she must have KE that is

KE = mgh = 41983 J

The work done by friction and air resistance is

Wncon = -uk m g d - Fair d

= -19508 J

Thus, the remaining KE is

KEf = 41983 - 19508 J

= 22475 J

As

v = sqrt(2KE/m),

v = 26.7 m/s   [ANSWER, PART B]

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As she has KE = 22475 and she stopped,

Wncon = Fd = -22475 J

As d = 2.5 m,

F = 8990 N   [ANSWER, PART C]

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