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Question

(A) The force $\vec{F}$ is related to the potential energy $U$ by the negative gradient: $\vec{F} = -
abla U = -\left( \frac{\partial U}{\partial x}\

Vedclass · Accepted Answer

$-x\hat{i} + z\hat{k}$

Vedclass · Answer

(A) The force $\vec{F}$ is related to the potential energy $U$ by the negative gradient: $\vec{F} = -
abla U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} ight)$.Given $U = \frac{1}{2}(x^2 - z^2)$,we calculate the partial derivatives:$F_x = -\frac{\partial U}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{1}{2}x^2 - \frac{1}{2}z^2 ight) = -\frac{1}{2}(2x) = -x$.$F_y = -\frac{\partial U}{\partial y} = 0$ (since there is no $y$ dependence).$F_z = -\frac{\partial U}{\partial z} = -\frac{\partial}{\partial z} \left( \frac{1}{2}x^2 - \frac{1}{2}z^2 ight) = -\frac{1}{2}(-2z) = z$.Thus,the force vector is $\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k} = -x\hat{i} + z\hat{k}$.

The potential energy of a certain particle is given by $U = \frac{1}{2}\,(x^2 - z^2)$. The force on it is:

Similar Questions

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