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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 $dU = -F dx$.$U(x) = -\int_0^

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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 $dU = -F dx$.$U(x) = -\int_0^x (-kx + ax^3) dx = \int_0^x (kx - ax^3) dx$.$U(x) = \left[ \frac{kx^2}{2} - \frac{ax^4}{4} \right]_0^x = \frac{kx^2}{2} - \frac{ax^4}{4}$.At $x = 0$,$U(0) = 0$.The slope of the $U-x$ graph is $\frac{dU}{dx} = -F = kx - ax^3$. At $x = 0$,the slope is $0$.For small $x$,$U(x) \approx \frac{kx^2}{2}$,which is a parabola opening upwards.Setting $U(x) = 0$,we get $\frac{kx^2}{2} = \frac{ax^4}{4} \Rightarrow x^2 = \frac{2k}{a} \Rightarrow x = \sqrt{\frac{2k}{a}}$.Beyond this point,$U(x)$ becomes negative. Thus,the graph starts at the origin,increases to a maximum,and then decreases,crossing the $x$-axis at $x = \sqrt{\frac{2k}{a}}$. This matches the shape shown in Graph $D$.

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