Question

Potential energy for a conservative force $\overrightarrow{F}$ is given by $U(x,y)=\cos (x+y)$. Force acting on a particle at position given by coordinates $(0,\frac{\pi }{4})$ is

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

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

We know, $F_x=-\frac{dU}{dx}$
and $F_y=-\frac{dU}{dy}$
$\therefore$ $F_x=-\frac{dU}{dx}=-\frac{d}{dx}\cos (x+y)=\sin (x+y)\therefore F_x\text{at}(0,\frac{\pi }{4})\text{is}=\sin (0+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$
Similarly,
$F_y=-\frac{dU}{dy}=-\frac{d}{dy}\cos (x+y)=\sin (x+y)\therefore F_y\text{at}(0,\frac{\pi }{4})\text{is}=\sin (0+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$
$\Rightarrow \overrightarrow{F}=F_x\hat{i}+F_y\hat{j}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$