# Homework Help: Potential Energy Question

1. Mar 11, 2005

### Azytzeen

A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy U(x) given by U(x)= (1.49J/m^4)x^4.

What is the force (magnitude and direction) when the particle is at position x=-0.835m?

I can't seem to get the right answer for this question. Plugging -0.835m in for x and then solving for F does not work. Maybe my logic is flawed. Please tell me how to reason this out. Thanks in advance.
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Last edited: Mar 12, 2005
2. Mar 11, 2005

### James R

F = -dU/dx.

Do the derivative first, then plug in the x value.

3. Mar 12, 2005

### Azytzeen

Ok, got that. Thanks very much. Here's another one that stumped me.

A skier of mass 58.0kg starts from rest at the top of a ski slope of height 70.0m.
If frictional forces do -1.08*10^4 of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g=9.80m/s^2.

I used the formula 1/2*m*v_1^2+m*g*y_1+W_other=1/2*m*v_2^2+m*g*y_2

So I cancelled out the kinetic energy in the first part as she starts from not moving, and I solved for v_2, which is the correct answer of 31.62 m/s. But part 2 of the problem got me.

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is mu=0.170. If the patch is of width 63.0m and the average force of air resistance on the skier is 180N, how fast is she going after crossing the patch?

For this part, I used the same equation, and plugged in 31.62m/s for v_1, and for the W_other, I put in -(180N+0.17*63m*58kg*9.8m/s^2). Then I solved for v_2, but the answer is wrong. This is a 10 attempt problem, and I have one try left. So please help on this again. Thanks!

After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.10m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

This is part c, which can be solved only after I got part B. Please advise me on how to approach this problem. Thanks again.

4. Mar 12, 2005

### Staff: Mentor

You are mixing force and work in that expression. (Look at your units.) The first term (180N) is the force of air resistance, but you need the work done. (You left out the distance.)