The potential energy for a force filed `vecF` is given by `U(x,y)=cos(x+y)`. The force acting on a particle at position given by coordinates `(0, pi//4)` is

Options

A. (a) `-1/sqrt2(hati+hatj)`
B. (b) `1/sqrt2(hati+hatj)`
C. (c) `(1/2hati+sqrt3/2hatj)`
D. (d) `(1/2hati-sqrt3/2hatj)`

Answer

B

Explanation

`F_x=-(delU)/(delX)=sin(x+y)=1/sqrt2`
`F_y=-(delU)/(delY)=sin(x+y)=1/sqrt2`
`vecF=1/sqrt2[hati+hatj]`