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

• A. $1N$
• B. $\sqrt{2}N$
• C. $\frac{1}{\sqrt{2}}N$
• D. $0$

Answer 1

The correct option is A $1N$

Relation between potential energy and force is

$\overrightarrow{F}=-\frac{dU}{dx}\hat{i}-\frac{dU}{dy}\hat{j}-\frac{dU}{dz}\hat{k}$
Given $U(x,y)=\sin (x+y)$,

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

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