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(D) The total mechanical energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $V(x)$,given by $E = K + V(x)$.Since $K =

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Particle has zero acceleration where $\frac{dV(x)}{dx} = 0$

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(D) The total mechanical energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $V(x)$,given by $E = K + V(x)$.Since $K = \frac{1}{2}mv^2$,we have $E = \frac{1}{2}mv^2 + V(x)$.The force acting on the particle is given by $F = -\frac{dV(x)}{dx}$.According to Newton's second law,$F = ma$,so $ma = -\frac{dV(x)}{dx}$.Acceleration $a = -\frac{1}{m} \frac{dV(x)}{dx}$.For the acceleration to be zero,we must have $\frac{dV(x)}{dx} = 0$.Therefore,the particle has zero acceleration where the derivative of the potential energy with respect to position is zero.

$A$ particle moves in a one-dimensional field with total mechanical energy $E$. If the potential energy of the particle is $V(x)$,then:

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