Potential energy of a particle in a system

[Claudia Sanchez]

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!

 

[HallsofIvy]

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

 

[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