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(D) The force $F$ is given by $F = -\frac{dU}{dx}$. Equilibrium occurs where $F = 0$,which implies $\frac{dU}{dx} = 0$. This corresponds to the points

Vedclass · Accepted Answer

Position $D$ is stable equilibrium.

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(D) The force $F$ is given by $F = -\frac{dU}{dx}$. Equilibrium occurs where $F = 0$,which implies $\frac{dU}{dx} = 0$. This corresponds to the points where the slope of the $U-x$ graph is zero,which are points $B$ and $D$.At point $B$,the potential energy is at a local minimum,so $\frac{d^2U}{dx^2} > 0$,which indicates stable equilibrium.At point $D$,the potential energy is at a local maximum,so $\frac{d^2U}{dx^2} At point $C$,the slope $\frac{dU}{dx}$ is at its maximum positive value.Therefore,the statement 'Position $D$ is stable equilibrium' is incorrect.

The potential energy of a particle varies with distance $x$ as shown in the graph. Select the wrong statement.

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