Question

A potential energy function is given by U(x)=(3.00J)x+(1.00J/m^2)x^3. What is the force function F(x) (in newtons) that is associated with this potential energy function? Express your answer in terms of x, where x in meters.

Answer 1

Answer:

$F(x)=-3 N -(3N/m^2) x^2$

Explanation:

The force is equal to the negative of the derivative of the potential energy function:

$F=-\frac{dU}{dx}$

In this problem, the potential energy function is

$U(x)=3 x + 1 x^3$

Therefore, the force function is:

$F(x)=-\frac{dU}{dx}=-\frac{d}{dx}(3x+1x^3)=-3-3x^2$

So, adding the units of measurement,

$F(x)=-3 N -(3N/m^2) x^2$