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The potential energy of a force field $\hat F$ is given by $v(x,y)=\cos (x+y)$ . The force acting on the particle at position given by coordinate $(0, \large\frac{\pi}{4})$ is

\[(a)\;\frac{-1}{\sqrt 2}(\hat i+\hat j) \quad (b)\;\frac{1}{\sqrt 2}(\hat i+\hat j) \quad (c)\;\bigg(\frac{1}{2}\hat i+\frac{\sqrt 3}{2}\hat j\bigg) \quad (d)\;\bigg(\frac{1}{2}\hat i-\frac{\sqrt 3}{2}\hat j\bigg)\]

1 Answer

Let U be the potential energy
$U(x,y)=\cos (x +y)$
$\therefore F_x=\large\frac{-\partial u}{\partial x}$
$\quad =\large\frac{-\partial }{\partial x}$$ \cos (x+y)$
$\quad=\sin (x+y)$
at $(0,\large\frac{\pi}{4})\quad$$ F_x=\large\frac{1}{\sqrt 2}$
$F_y=\large\frac{-\partial u}{\partial y}$
$\quad=\sin (x+y)$
at $(0, \large\frac{\pi}{4}) \quad $$F_y=\large\frac{1}{\sqrt 2}$
$\therefore F= F_x \hat i+F_y \hat j$
$\quad= \large\frac{1}{\sqrt 2}$$(\hat i+\hat j)$
Hence b is the correct answer. 
answered Aug 6, 2013 by meena.p
edited Jun 11, 2014 by lmohan717

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