A force `F=-K(yhati+xhatj)` (where K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point `(a, 0)`, and then parallel to the y-axis to the point `(a, a)`. The total work done by the force F on the particle is

Options

A. `-2Ka^(2)`
B. `2Ka^(2)`
C. `-Ka^(2)`
D. `Ka^(2)`

Answer

C

Explanation

(c ) total work done `W=fF.dr=f-K(yhati+xhatj).(dxhati+dyhati)`
`fK(ydx+xdy)=-Kf_((0","0))^((a","a)s(xy)=-K[(xy0]_((0","0))^((a","a))=-KA^(2)`