A force F = -k(y i + x j ) (where k is 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

Q:A force $\large \vec{F} = -k(y \hat{i} + x \hat{j})$ (where k is 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) – 2Ka2

(b) 2ka2

(c) – ka2

(d) ka2

Ans: (c)

Sol: $\large dW = \vec{F}.\vec{ds} $

$\large dW = -k(y \hat{i} + x \hat{j}).(dx\hat{i}+ dy \hat{j} + dz \hat{k})$

$\large dW = -k(ydx + x dy) = – k d(x y)$

$\large W = -k\int_{0,0}^{a,a} d(x y)$

$\large = -k[x y]_{0,0}^{a,a}$

W = –ka2