# Potential Energy Problem

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..

Homework Helper
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 textbook (they are often on the inside of the covers).

Homework Helper
Another point just occurred 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.

Moxin
Originally posted by HallsofIvy
Another point just occurred 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.

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) ?

Homework Helper
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!