Question

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

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

Answer 1

Answer: (A)

We use
$\overrightarrow{ F } =\nabla U =-\cos (x+y) \hat{ i }-\cos (x+y) \hat{ j }$
$\left.\overrightarrow{ F }\right|_{\left(0, \frac{\pi}{4}\right)} =-\frac{1}{\sqrt{2}} \hat{ i }-\frac{1}{\sqrt{2}} \hat{ j } $
$|\overrightarrow{ F }|_{\left.0, \frac{\pi}{4}\right)} =1$