# Homework Help: Graph of potential energy

1. Oct 30, 2013

### darksyesider

[solved] graph of potential energy

1. The problem statement, all variables and given/known data

A mass of 5 kg follows the potential energy shown by the equation:

$$U(x) = .333(x-4)^3 - \dfrac{x^2}{4} + 5$$

1. Find the force on the mass at x=2
2. Describe the motion qualitatively if the mass is placed at x=4 m and released
3. With part 1, determine the max speed of the mass
4. With #1, determine where the mass will again come to rest if at all.
2. Relevant equations

mgh
1/2mv^2

3. The attempt at a solution

1. I got this pretty easily, as -3N
2. I really had no idea for this. I thought that it would eventually settle at x=5.6 m.
3. U(4)-U(5.68) = .5mv^2 so v=1.576
4. u(4)-U(5.868) = U(x_0) => x_0 = 7.162 m

Is this correct?

Last edited by a moderator: Oct 30, 2013
2. Oct 30, 2013

### Staff: Mentor

I assume you equation for the potential is in Joules, correct? From the equation for the potential, what is the equation for the force? To get a better understanding of what happens in part 2-4, plot a graph of the force as a function of distance. Or at least tell us the direction of the force when the mass is placed at x = 4. Will it continue moving in this direction? How far does it have to go before the force reverses direction (if at all)?

3. Oct 30, 2013

### darksyesider

I came up with $$F(x) = -dU/dx$$
I don't get intuitively what you mean.

Last edited: Oct 30, 2013
4. Oct 30, 2013

### Staff: Mentor

At x = 4, F = 2 N, so that, once the mass is released, it is going to start moving in the + x direction. You didn't show the details, so I didn't know whether you were doing it right, but, in part 3, it looks like you correctly calculated the location at which the force goes to zero (and reverses sign), and also calculated the velocity correctly. In part 4, you should have set U(xfinal)=U(4) to find location when the mass comes to rest again. It doesn't look like 7.162 satisfies this.