Question

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

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

Answer 1

Answer: (A)

$\begin{array}{c}\overrightarrow{F}=\left(x,y\right)=-\nabla U\left(x,y\right)\\ \nabla U\left(x,y\right)=\frac{\partial U}{\partial x}\hat{x}+\frac{\partial U}{\partial y}\hat{y}=\cos \left(x+y\right)\left(\hat{x}+\hat{y}\right)\\ F\left(x,y\right)=\left|\nabla U\left(x,y\right)\right|\\ F\left(0,\frac{\pi }{4}\right)=\left|\nabla U\left(x,y\right)\right|=\left|\cos \left(\frac{\pi }{4}\right)\right|\left|\hat{x}+\hat{y}\right|\\ F\left(0,\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\sqrt{1^2+1^1}=1unit\end{array}$

Hence,

option $\left(A\right)$ is correct answer.