# Potential Energy function problem

1. Mar 18, 2004

### cristina

A force is given by Fx = Ax^(-3) (this is A times x to the power -3), where A= 8N.m^3.
a) For positive values of x, does the potential energy associated with the force increase or decrease with increasing x?
(I did imagine what would happen to a particle that is placed at rest at some point a and is then released. The professor concidered my answer wrong to be decreasing!)
b) Find the potential energy function U associated with the force such that U approches zero as x approaches infinity.
I didnt know how to answer b)

2. Mar 18, 2004

### HallsofIvy

Staff Emeritus
Cristina, you have posted a number of similar problems without any sign that you have tried them yourself. One difficulty with that is that we don't know what techniques you have available.

Generally speaking "potential energy" due to the position of an object is the same as the work necessary to move the object into that postion (From some other position, of course. That's why potential energy is always relative to some position.)

In this case, the force is 8x-3. For positive x, the force is positive, "pushing" an object to the right (increasing x). Since it is necessary to do work against the force to move to the right (decreasing x), potential energy increases as x decreases and decreases as x increases.

Since "work= force times distance" for constant force,
work= the integral of f(x)dx when f is a variable.

Here, potential energy= $$\int 8x^{-3}dx$$. Find that indefinite integral and choose the constant of integration so that it goes to 0 as x goes to infinity.

Last edited: Mar 18, 2004
3. Mar 18, 2004

### cristina

I am working on it, honestly I am, since hours ago!
We dont cover integrals in this course nor in the calculus one which is a pre for this course. and beleive or not its all about integrals here. would you give me a break, if I ask you what is an indefinite itegral?

4. Mar 19, 2004

### cristina

The indefinite integral is -4/(x^2)but I dont see a need for a constant since it goes to zero as x increase.?

5. Mar 19, 2004

### ShawnD

As far as I know, you are correct. To check that theory, divide each term by the highest power of x.

$$\lim_{x \to \infty} \frac{-4}{x^2}$$

$$\lim_{x \to \infty} \frac{(\frac{-4}{x^2})}{(\frac{x^2}{x^2})}$$

$$\frac{(\frac{-4}{\infty})}{1}}$$

$$\frac{0}{1}$$

6. Mar 19, 2004

### cristina

I am done with this one. thanks to you and hallsofIvy. would you mind helping me with the others please?!

7. Mar 19, 2004

### HallsofIvy

Staff Emeritus
There is always a "need for a constant"- the constant just happens to be 0 here!

8. Mar 19, 2004

### cristina

I did put -4/(x^2) + c and c = 0.