A 60.0 kg skier starts from rest at the top of a ski slope 60.0 m high.
The Attempt at a Solution
(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?
This one I got correct, it's 28.8 m/s.
(b)Now moving horizontally, the skier crosses a patch of soft snow, where µk = 0.2. If the patch is 85.0 m wide and the average force of air resistance on the skier is 160 N, how fast is she going after crossing the patch?
This one I also have correct, 6.35 m/s
(c) The skier hits a snowdrift and penetrates 2.9 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
I derived the equation F = m*(Velocity from answer B)^2/(2*2.9m). This gets me 417N. However, it is not the right answer. I also know that the sign is not the problem, as neither 417N or -417N give the right answer. I used the work energy theorem to get F*d = 1/2m(VB)^2, and then just divided the kinetic energy by the distance to get the average Force. I don't know what I am doing worng.