Question

The potential energy for a force field $\overrightarrow{F}$ is given by $U(x,y)=\sin (x+y)$. The force acting on the particle of mass $m$ at $(0,\frac{\pi }{4})$ is (in $N$)

Answer 1

Relation between potential energy and force is

$\overrightarrow{F}=-\frac{\partial U}{\partial x}\hat{i}-\frac{\partial U}{\partial y}\hat{j}-\frac{\partial U}{\partial z}\hat{k}$
Given $U(x,y)=\sin (x+y)$,

$\Rightarrow \overrightarrow{F}=-cos(x+y)\hat{i}-cos(x+y)\hat{j}$
$\Rightarrow {\overrightarrow{F}}_{(0,\frac{\pi }{4})}=-cos\frac{\pi }{4}\hat{i}-cos\frac{\pi }{4}\hat{j}$
$\Rightarrow {\overrightarrow{F}}_{(0,\frac{\pi }{4})}=-\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$

$\Rightarrow |\overrightarrow{F}|=\sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}=1N$