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

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

Vedclass · Answer

(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:$\frac{\partial U}{\partial x} = -\sin(x + y)$$\frac{\partial U}{\partial y} = -\sin(x + y)$Thus,the force components are:$F_x = -(-\sin(x + y)) = \sin(x + y)$$F_y = -(-\sin(x + y)) = \sin(x + y)$At the point $(0, \frac{\pi}{4})$:$F_x = \sin(0 + \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$$F_y = \sin(0 + \frac{\pi}{4}) = \sin(\frac{\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 $(0, \frac{\pi}{4})$ is

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