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The potential energy for a force filed v...

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

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))`

Text Solution

Verified by Experts

The correct Answer is:
B
`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)]`
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