The potential energy for a force field vec{F} | vedclass

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(B) The force $\vec{F}$ is related to the potential energy $U$ by the relation $\vec{F} = -
abla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$

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(B) The force $\vec{F}$ is related to the potential energy $U$ by the relation $\vec{F} = -
abla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})$.Given $U(x, y) = \cos(x + y)$.Calculating the partial derivatives:$F_x = -\frac{\partial U}{\partial x} = -(-\sin(x + y)) = \sin(x + y)$.$F_y = -\frac{\partial U}{\partial y} = -(-\sin(x + y)) = \sin(x + y)$.At the point $(0, \pi/4)$:$F_x = \sin(0 + \pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$.$F_y = \sin(0 + \pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}$.Therefore,the force vector is $\vec{F} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.

The potential energy for a force field $\vec{F}$ is given by $U(x, y) = \cos(x + y)$. The force acting on a particle at the position given by coordinates $(0, \pi/4)$ is:

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