# Potential energy of a particle in a system

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1. Oct 8, 2014

### Claudia Sanchez

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1. A single conservative force acting on a particle within a system varies as
= (− Ax + Bx6)ihat N, where A and B are constants, is in newtons, and x is in meters.
(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0.
(b) Find the change in potential energy and change in kinetic energy as the particle moves from
x = 1.30 m to x = 3.60 m.

2. I got Ax^2/2 - b^7/2 but it was wrong so I'm really confused on how to go about this problem

Thanks!!

Last edited: Oct 8, 2014
2. Oct 8, 2014

### HallsofIvy

Staff Emeritus
The potential energy corresponding to force $\vec{F}$ is a scalar function $\phi(x,y)$ such that $\nabla \phi= -\vec{F}$. Of course, $\nabla \phi$ is defined as $\frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}$. But it is not clear to me what "F" is- is "N" a vector? If so, what vector?

If you mean that $\vec{F}= (-Ax+ Bx^6)\vec{i}$ then we must have
$$\frac{\partial \phi}{\partial x}= Ax- Bx^6$$
$$\frac{\partial \phi}{\partial y}= 0$$.

From $\partial \phi/\partial x= Ax- Bx^6$, we have
$$\phi(x,y)= \frac{A}{2}x^2- \frac{B}{7}x^7+ p(y)$$
But then
$$\frac{\partial \phi}{\partial y}= p'(y)= 0$$
so that p(y) is actually a constant:
$$\phi(x, y)= \frac{A}{2}x^2- \frac{B}{7}x^7+ C$$

Last edited: Oct 8, 2014
3. Oct 8, 2014

### Claudia Sanchez

Hi, thanks for the help! that was the same answer I got but it was wrong and I don't really understand what I'm supposed to do