A skier of mass 58.0 starts from rest at the top of a ski slope of height 66.0 .?

Question

If frictional forces do −1.01×104 of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be = 9.80 .
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is = 0.250. If the patch is of width 70.0 and the average force of air resistance on the skier is 180 , how fast is she going after crossing the patch?

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.30 into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

Answer 1

Please label with proper units.
and
With a frictional force of −1.01E+4 she must as be nailed to the frozen ground.

Here are some theoretical aspects

Potential energy at the top
Pe=mgh
Kinetic energy at the bottom
Ke =0.5 mV^2
Work done against friction
Wf=fs=uN s=u mg cos(A)

If there was no friction
Pe=Ke
If there is a friction
Pe=Ke+Wf

What is unknown is the angle of the incline or the path s.