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(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.Integrating this,we get $dU = -F \, dx$,so $U(x) = -\

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(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.Integrating this,we get $dU = -F \, dx$,so $U(x) = -\int F \, dx$.Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) \, dx = \int (kx - ax^3) \, dx$.Performing the integration,we get $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 + C$.Assuming $U(0) = 0$,we find $C = 0$,so $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 = \frac{1}{4}x^2(2k - ax^2)$.For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.For $0 For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative and decreases as $x^4$ dominates.This corresponds to a curve that starts at the origin,rises to a maximum,and then crosses the $x$-axis to become negative,which matches the graph in option $D$.

The force acting on a particle is given by the formula $F(x) = -kx + ax^3$. Find the graph of its potential energy $U(x)$ for $x \ge 0$.

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