# Physics Problem Conservation of Energy

John_Smithson

## Homework Statement

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.

Mentor
I get 417 N, too.

If the deceleration is not uniform and the force gets averaged over time, the value can be different (and you cannot calculate it with additional information), but that cannot be the issue here.

Mentor
If I do the calculations using g = 9.807 m/s the result is 414 N. I wonder how sensitive the marking bot is to sig figs here?

John_Smithson
Even though the highest sig figs in the problem is 3, it does expect you to use g = 9.807 instead, and 414N was correct. Thanks.