A 60.0-kg skier starts from rest at the top of a ski slope of height 66.0 m.
A) If frictional forces do −1.02×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
Answer is 30.9 m/s
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 70.0 m and the average force of air resistance on the skier is 180 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?
work of non-conservative forces = KEf -KEi
The Attempt at a Solution
I did this.
worknc = Kef-KEi
work done by air = -180 * 70.0 = -12600 J
work done by friction is (0.21)(60.0)(9.8)(70.0)cos180 = -8643.6 J
-21243.6 = 30v^2 - .5(60.0)(30.9)^2
v = 15.6 (I actually received 15.7, but plugged into a previous part, it seems that 15.6 is more accurate).
C) I did:
Fsnow(2.2) = -.5(60.0)(15.6)^2
Are these right? non-conservative means only friction and air, and I think the only force in part C is the snow.