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

Vedclass · Accepted Answer

$-2k(x\hat{i} + y\hat{j} + z\hat{k})$

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(C) The force $\vec{F}$ is related to the potential energy $U$ by the negative gradient: $\vec{F} = -
abla U = -\left( \hat{i} \frac{\partial U}{\partial x} + \hat{j} \frac{\partial U}{\partial y} + \hat{k} \frac{\partial U}{\partial z} ight)$.Given $U = K(x^2 + y^2 + z^2)$.Calculating the partial derivatives:$\frac{\partial U}{\partial x} = 2Kx$,$\frac{\partial U}{\partial y} = 2Ky$,and $\frac{\partial U}{\partial z} = 2Kz$.Substituting these into the force equation:$\vec{F} = -(2Kx\hat{i} + 2Ky\hat{j} + 2Kz\hat{k})$.Therefore,$\vec{F} = -2K(x\hat{i} + y\hat{j} + z\hat{k})$.

$A$ particle moves in a potential region given by the formula $U = K(x^2 + y^2 + z^2)$. Find the force acting on the particle.

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