# Potential Energy Problem

1. Sep 20, 2003

### Moxin

A particle moves along the x axis under the influence of a variable force F(x) = 6.8x^2 + 3.0x where the force is measured in Newtons and the distance in meters. What is the potential energy associated with this force at x = 2.0 m? Assume that U(x) = 0 J at x = 0 m.

The formula I was trying to work with (from the book) is

dU = -F*ds where U is the potential energy and s is the displacement..

Anyways,.. I went on ahead a plugged 2 into the equation, and I got 33.2, the Force.. then I multiplied the negative of that by 2.. the displacement, but apparently the answer is wrong. So I'm guessin the derivative of displacement is indeed different than the regular displacement..so I guess my question here is, how do i find the derivative of displacement? Sorry I'm jus now starting calculus this year and am not completely familiar with the concept yet..

2. Sep 21, 2003

### HallsofIvy

Staff Emeritus
Exactly. One of the basic rules of physics is: whatever mathematics you need for this semester's physics course, you will study NEXT semester!

dU= f(s)ds is the correct generalization of "Work= force times distance" which is only true when force is a constant.

To go from "dU= f(s)ds" to "U= ...", you need to find the anti-derivative or integral. In this case you have:F(x)= 6.8 x so you need to find the integral of 6.8 x dx from x=0 to x= 2.

Have you yet learned the derivative rule (xn)'= n xn-1? or it's anti-derivative twin int xndx= (1/(n+1))xn+1?

If not, you may just need to content yourself with looking at a table of integrals in your calculus text book (they are often on the inside of the covers).

3. Sep 21, 2003

### HallsofIvy

Staff Emeritus
Another point just occured to me: as long as the "rate of change" is itself constant and the force function is linear (as here), you can use the "average force". What is the force when x= 0? What is the force when x= 2? What is the average of those two numbers? Use that as the force and multiply by 2.

4. Sep 21, 2003

### Moxin

Well, I tried this method since I still didn't really understand how to get the derivative of x, and ended up getting it wrong again, then I realized it's because the force function isn't linear.. you left off half the eq'n lol its parabolic (F(x) = 6.8x^2 + 3.0x) ..So I guess now I have to try to understand your first suggestion.

And Yeh, I just learned the derivative rule the other day, I know how to apply it when given an eq'n (i.e. F(x)).. but not just with a lone variable (x).. wouldn't it just be one everytime? So that dU = F(s) ? So the eq'n would be simply dU = F(s) ?

5. Sep 21, 2003

### HallsofIvy

Staff Emeritus
You're right. I completely misread the force function. You have
F(x) = 6.8x^2 + 3.0x so the potential energy is the anti-derivative of 6.8x^2 + 3.0x- the function whose derivative is 6.8x^2+ 3.0x.

Okay, you say you have learned the power rule: the derivative of
x^n is n x^(n-1). What we need to do here is reverse that.

Notice that the derivative reduces the power by 1. If the derivative is x^2, then we must have had x^3 to start with. But x^3 alone won't work: it's derivative is 3x^2. We need to get rid of that "3". Okay, multiply by (1/3). What is the derivative of
(1/3)x^3? To get the anti-derivative of 6.8x^2, just multiply that (1/3)x^3 by 6.8.

Similarly, if the derivative is x, then we must have had x^2 to start with. But the derivative of x^2 is 2x. We need (1/2)x^2 to wind up with (1/2)(2x)= x. To get the anti-derivative of 3.0x, just multiply by 3: 1.5x^2 has derivative what?

Of course, since the derivative of f + g is the derivative of f plus the derivative of g, to find the function whose derivative is
6.8 x^2+ 3.0x, just add the two functions above.

By the way, the derivative of any constant is 0 (a constant function doesn't increase or decrease) so we could add any constant to what we got before and not change its derivative. That's the reason potential energy is always relative!