In the figure shown,the potential energy (U) | vedclass

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(D) For a particle to be in equilibrium,the force $F = -dU/dx$ must be zero,which corresponds to the points where the slope of the $U-x$ graph is zero

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) For a particle to be in equilibrium,the force $F = -dU/dx$ must be zero,which corresponds to the points where the slope of the $U-x$ graph is zero (i.e.,at the maxima or minima of the potential energy curve).$1$. At $x_2$,the potential energy $U$ is at a local maximum. $A$ point of maximum potential energy corresponds to unstable equilibrium because any small displacement results in a force that pushes the particle further away from the equilibrium position.$2$. At $x_1$ and $x_3$,the slope $dU/dx$ is not zero (the curve is crossing the x-axis with a non-zero slope). Therefore,the force $F = -dU/dx$ is not zero at these points,meaning the particle is not in equilibrium at $x_1$ or $x_3$.Since none of the points $x_1, x_2, x_3$ represent a stable equilibrium (which requires a local minimum of potential energy),the correct option is 'None of these'.

In the figure shown,the potential energy $(U)$ of a particle is plotted against its position $x$ from the origin. The particle at:

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