# Potential Energy of a single particle

1. Mar 1, 2009

### yb1013

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

A single conservative force acting on a particle varies as Fvec = (-Ax + Bx5)i-hat N, where A and B are constants and x is in meters. Accurately round coefficients to three significant figures.

(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use A, B, and x as appropriate.)

U = ____________

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.20 m to x = 3.80 m. (Use A, B, and x as appropriate.)

ΔU = ____________

ΔK = ____________

2. Relevant equations

Not sure if there are relevant equations but PE = mgh

3. The attempt at a solution

I really have never been this confused about a physics problem... I think the answer needs to be in an equation form? Im really not sure on this one guys, I tried to understand it, its just not coming to me. I did all the rest of the hw completely fine other than this one.

2. Mar 1, 2009

### Vuldoraq

To get you started: Potential energy is just the integral of the force.

3. Mar 1, 2009

### yb1013

So, Like -Ax^2+Bx^6?

4. Mar 1, 2009

### Vuldoraq

Just so. Don't forget to divide by the power though. Since the question gives you some initial conditions you can also find the constant of integration.

5. Mar 1, 2009

### yb1013

ok so for the first question the formula would be -Ax^2+B/5x^6 i think. But then they give you U and x are = 0 ... So i guess you cant just plug x in because that would be 0. How do you solve for potential energy?

6. Mar 1, 2009

### Vuldoraq

Try to use the LaTeX for equations, it's much clearer. For the potential you use,

$$\int F dx= U$$

For any indefinite integral you will have a constant of integration, for example,

$$\int x^{n} dx= \frac{x^{n+1}}{n+1} +C$$

When it gives you the value of U when x is zero it is basically saying plug these vaules into your P.E. equation to find your integration constant.

7. Mar 1, 2009

### yb1013

ok well i found part a.. but when i go to plug in 3.8 and 1.2 to get the answer, I come out with 7.94A-502.3B , and it comes up incorrect..

8. Mar 1, 2009

### Vuldoraq

Maybe you have a minus out of place? Is this what you are evaluating?

$$[\frac{-Ax^{2}}{2}+\frac{Bx^{6}}{6}]_{1.2}^{3.2}$$

9. Mar 1, 2009

### yb1013

Im evaluating A/2x^2-B/6x^6 , Im sorry, i tried using latex but ive never used it before and i just couldnt figure it out.. But yea thats what im evaluating, you had to multiply the whole thing by -1 because of the formula. So with A/2x^2-B/6x^6 i am taking (3.8) - (1.2), and i get what i told you before

10. Mar 1, 2009

### Vuldoraq

Check your calculation, I think you added the A's rather than subtracted the second from the first.

11. Mar 1, 2009

### yb1013

aahhhh, thank you sir, very wise!
haha alright well thanks again man and have a great night.

12. Mar 1, 2009