Relation between potential energy U and conservative force F

AP MECHANICS -C LEARNING OBJECTIVE CON 1.B -Derive an expression that represents the relationship between a conservative force acting in a system on an object to the potential energy of the system using the methods of calculus.
1.F -Derive an expression for the gravitational potential energy of a system consisting of a satellite or large mass (e.g., an asteroid) and the Earth at a great distance from the Earth.

Consider a conservative force F(x) and its associated potential energy function U(x). Suppose that the force F(x) varies with position x in any arbitrary way but has a unique value at any given value of x. The work done by this force as an object moves from $x_1$ to $x_2$ a small distance dx such that $dx = ( x_2 – x_1 ) $ then the work done by F(x) during the displacement will be:

$dW = F(x) dx$

This work done by the conservative force will happen on expense of potential energy that is $U(x_1) – U(x_2)$, hence we can write:

$U(x_1) – U(x_2) = dW$

Or change in potential energy $dU = U(x_2) – U(x_1) $, hence

$dW = -dU = F (x) dx$

or

$F (x) = – \frac{dU}{dx}$

The above equation enables us to calculate the F(x) as a function of x if the potential energy function U(x) is known. From Equation 12, we see that the force F(x) is positive wherever the potential is a decreasing function of x, that is, wherever the derivative dU/dx is negative. Conversely, the force F (x) is negative wherever the potential is an increasing function of x, that is, wherever the derivative dU/dx is positive.

Let’s check the validity of the above expression for the known cases.

Potential energy of a body near Earth’s surface

U = mgh and F = – dU/dh = – mg

Potential energy of elastic forces

$U = \frac{kx^2}{2} \mbox{ and } F = -dU/dx = – Kx $

We can also use the equation to calculate the U(x) if the force is known and varies only with position. Two fundamental forces gravitation and electrical force vary with distance and hence are conservative forces in nature. In order to calculate the force we will have to tweak the equation a bit :

$U(x) = – \int F(x) dx$

Potential due to gravitational force

$ F = G \frac{m_1m_2}{r^2}, \mbox{ then } U = – \int Fdr \mbox{ and } U = G \frac{m_1m_2}{r} $

Calculating Force from Potential energy and vice versa
Force is a negative derivative of the potential energy function. In order to find the force we need to differentiate U (x), with respect to x and then apply a negative sign.

$ F(x) = – \frac{dU}{dx} $

Otherway around if we want to know the U(x) we need to integrate F(x) dx to obtain the potential energy function:

$ U(x) = – \int F(x) dx$

Relationship between force and potential energy