A 58.0-kg skier starts from rest at the top of a ski slope of height 62.0 m. Par

Question

A 58.0-kg skier starts from rest at the top of a ski slope of height 62.0 m.

Part B

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.22. If the patch is of width 67.0 m and the average force of air resistance on the skier is 160 N , 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 2.2 minto it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

B. Initially using energy conservation

KEi + UEi = KEf + PEf

KEf = mgh = 58*9.81*62 = 35276.76 J

velocity at bottom

KEf = 0.5*m*vf^2

vf = sqrt(2*KEf/m)

vf = sqrt(35276.76*2/58) = 34.87 m/sec

Now in horizontal diretion

KEh = KEf - (Wf + Wair) = KEf - (uk*m*g*d + f(air)*d)

KEh = 35276.76 - (0.22*58*9.81*67 + 160*67)

KEh = 16170 J

vh = sqrt(2*KEh/m)

vh = sqrt(2*16170/58) = 23.61 m/sec

C. using the work energy theoram

W = dKE

F*d = dKE

F = (KEh - KEs)/d

F = (16170 - 0)/2.2

F = 8085 N

Let me know if you have any doubt.

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