A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0

Question

A skier of mass 60.0kg starts from rest at the top of a ski slope of height 61.0m.

a) If frictional forces do -1.09 x 10^4 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g=9.80m/s^2.

The answer to this is 28.8 m/s, I need help on part B and part C.

b) Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is uk=0.150. If the patch is of width 67.0m and the average force of air resistance on the skier is 160N, 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 2.60minto it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Explanation / Answer

Her energy at start is m*g*h, all gravitational potential energy. At the bottom, her net energy is m*g*h - Ef, (Ef is given as 1.09*10^4 J) and this is all kinetic energy = 0.5*m*v^2. Therefore m*g*h - Ef = 0.5*m*v^2 v = v[2*(m*g*h - Ef)/m] On the horizontal path, the skier loses energy to two forces: friction Fr = µ*m*g and air resistance Fa. The energy lost is the total force times the distance traveled, so the (kinetic) energy remaining after the patch is 0.5*m*vr^2 = 0.5*m*v^2 - (µ*m*g + Fa)*W where w = patch width, µ = coeff of friction, vr is the final velocity. Solve for vr.

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