A force `F=-k(^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)`
(a) `-2ka^(2)` , (b) `2ka^(2) , (c)`-ka^(2)` , (d) `ka^(2)`
Correct Answer – A::B::C
`dW=F.dr`, where `dr=dxhat(i) + dyhat(j) + dzhat(k)`
and `F=k(yhati + xhatj)`
`:. dW=-k(ydx + xdy)=-kd(xy)`
`:. W=-int_(0,0)^(a,a)dW =-kint_(0,0)^(a,a)d(xy)`
`=-k[xy]_(0,0)^(a,a)`,
`W=-ka^(a)`
`:.` The correct option is `(c)`.