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. `-(1)/(sqrt2)(hat(i)+hat(j))`
B. `(1)/(sqrt2)(hat(i)+hat(j))`
C. `((1)/(2)hat(i)+(sqrt3)/(2)hat(j))`
D. `((1)/(2)hat(i)-(sqrt3)/(2)hat(j))`

Answer

B

Explanation

`F_(x)=-(delU)/(delx)=sin(x+y)` and
`F_(y)=-(delU)/(dely)=sin(x+y)`
so `F_(x)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)`
`F_(y)=[sin(x+y)]_(((0,pi/4)))=(1)/(sqrt2)`
`vec(F)=(1)/(sqrt2)[hat(i)+hat(j)]`