Question

A particle that can move along the x-axis experiences an interaction force $F_x=\left(3x^2-5x\right)\mathrm{N},$where x is inm. Find an expression for the system’s potential energy.

Short answer

Expression for potential energy is$U\left(x\right)=\frac{5x^2}{2}-x^3$

Step-by-step solution

Given information

The interactive force is given as

$F_x=\left(3x^2-5x\right)\mathrm{N}$

Explanation

$$\begin{gathered} F_x=\left(3x^2-5x\right)\mathrm{N} \\ \mathrm{U}\left(\mathrm{x}\right)=-\int {\mathrm{F}}_{\mathrm{x}}\mathrm{dx} \\ =-\int \left(3{\mathrm{x}}^2-5\mathrm{x}\right)\mathrm{dx} \\ =-\int 3{\mathrm{x}}^2\mathrm{dx}+\int 5\mathrm{xdx} \\ =-3\left[\frac{{\mathrm{x}}^3}{3}\right]+5\left[\frac{{\mathrm{x}}^2}{2}\right] \\ \mathrm{U}\left(\mathrm{x}\right)=-{\mathrm{x}}^3+\frac{5{\mathrm{x}}^2}{2} \\ \mathrm{U}\left(\mathrm{x}\right)=\frac{5{\mathrm{x}}^2}{2}-{\mathrm{x}}^3 \end{gathered}$$