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

`w=intf_xdx+int f_ydy=-k int y dx-k int xdy`

From 0 to A, `dy=0`, `y=0`
So `W_(OA)=0`
From A to B: `dx=0`, `x=a`
`W_(AB)=-kaint_0^ady=-ka^2`
Hence, total work done
`W_T=W_(OA)+W_(AB)=0-ka^2-ka^2`