Potential energy function U(x) in one dimension

AP MECHANICS -C LEARNING OBJECTIVE CON 1.C -Describe the force within a system and the potential energy of a system.

So far we have seen that a potential energy function U(x) can be found in the cases where conservative forces are acting in the system. We proved that in such cases the work done by the forces in a closed loop is always zero. For a one-dimensional case where the U varies only as a function of either x, y or z. The corresponding F(x) produces zero work on a closed path in agreement with the law of conservation of ME.

$\int_{x1}^{x2} F(x) dx = [ F^* ( x2 ) – F^* ( x1 ) ] = 0,\mbox{ if }x_1 = x_2$

But if the potential energy function varies with some other variable such as velocity U(v) or both x and v, U(x,v) it will not satisfy the requirement.

Example of potential energy function
$ U(x) = Ax $

$U(x) = A/x$

$U(x) = A/x^2$

$U(x) = 2x^4 + x^2$

or more complicated looking form

$U(x) = ( b/x^2 ) – ( 2c/x )$

$U(x) = U_0 [e^{-2(x-x_0/b} – 2e^{-(x-x_o)/b} ]
$

Where, A, $U_0$, a, b, c, $x_o$ etc are constants. As long as U(x) is a unique function of x the requirements of mechanical energy conservation will be satisfied.

Potential energy function