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.

Homework Equations

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(V_B)^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.

 

[mfb]

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.[/size]

 

[gneill]

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.

 

[haruspex]

Part c is a wrong question. This comes up regularly. You cannot compute the "average force" from the energy and the distance. Yes, you can compute an average over distance, but consider this: average acceleration is defined as an average over time, Δv/Δt. So the reasonable definition of average force is also over time; for constant mass it's mΔv/Δt. That is not in general the same as an average ov distance.