Question

A potential energy function is given by u(x) = (3.00 j/m)x (1.00 j/m3)x3. what is the force function f(x) that is associated with this potential energy function?

Answer 1

The force function associated with the given potential energy function is found by taking the negative derivative of the potential energy function with respect to position, resulting in F(x) = -3.00 J/m - 3.00 J/m³x².

Step-by-step explanation:

The question asks for the force function associated with a given potential energy function, specifically u(x) = (3.00 J/m)x + (1.00 J/m3)x3. In physics, particularly in the context of mechanics, the force function is the negative derivative of the potential energy function with respect to position (x). Mathematically, to find the force function F(x), we differentiate u(x) with respect to x and then apply a negative sign:

F(x) = -µ(u(x))/µx = -µ[(3.00 J/m)x + (1.00 J/m3)x3]/µx

After taking the derivative, we find:

F(x) = -(3.00 J/m) - 3(1.00 J/m3)x2

F(x) = -3.00 J/m - 3.00 J/m3x2

Therefore, the force function that is associated with the potential energy function u(x) is F(x) = -3.00 J/m - 3.00 J/m3x2.