Question

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 position given by coordinates $(0, \pi / 4)$ is -

• A. $-\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$
• B. $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$
• C. $\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right)$
• D. $\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right)$

Answer 1

Answer: (B)

$F_{x}=-\frac{\partial U}{\partial x}=\sin (x+y)$
$F_{y}=-\frac{\partial U}{\partial x}=\sin (x+y)$
$\left.F_{x}=\sin (x+y)\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}}$,
$\left.F_{y}=\sin (x+y)\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}}$
$\therefore F=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$