A force `F= -khati+xhatj` where k is a positive constant, acts on a praricle moving in the xy 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 formce on the particle is

Options

A. (a) `-2Ka^2`
B. (b) `2Ka^2`
C. (c) `-Ka^2`
D. (d) `Ka^2`

Answer

D

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`