A force `F=-k(y hati + x hatj)` (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
(a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)`

Explanation

For motion of the particle from (0,0)to(a,0),
`vec(F)=-K(0hat(i)+ahat(j))rArrvec(F)=-Kaj`
Displacement, `vec(r )=(ahat(i)+0hat(j))-(0hat(i)+0hat(j))=ahat(i)`
so work done from (0,0)to(a,0) is given by
`W=vec(F).vec(r )=-Kahat(j).ahat(i)=0`
For motion (a,0) to (a,a)
`vec(F)= -K(ahat(i)+hat(j))`
and displacement, `vec(r )=(ahat(i)+ahat(j))-(ahat(i)+0hat(j))=ahat(j)`
So work done from `(a,0)` to`(a,a)`
`W=vec(F).vec(r)= -K(ahat(i)+ahat(j)).ahat(j)= -Ka^(2)`
So total work done `= -Ka^(2)`