Question

A force $F=-K(y\hat{i}+x\hat{j})$, 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. $-2K{a}^2$
• B. $2K{a}^2$
• C. $-K{a}^2$
• D. $K{a}^2$

Answer 1

The correct option is C

$-K{a}^2$

In going from (0,0) to (a, 0), the x-coordinate varies from 0 to a while the y-coordinate remains zero.
$\therefore$ Work done by force F along this path is $W_1={\int }_0^{a}F.dx={\int }_0^a-Kx\hat{j}.dx\hat{i}=0$ $(\because \hat{j}.\hat{i}=0)$
In going from (a, 0) to (a, a) the x-coordinate remains constant at x = a while the y-coordinate changes from 0 to a.
$\therefore$ Work done by force F along this path is $W_2={\int }_0^{a}F.dy={\int }_0^a-K(y\hat{i}+a\hat{j}).dy\hat{j}$ $=-Ka{\int }_0^{a}dy=-K{a}^2$
$(\because \hat{i}.\hat{j}=0,\hat{j}.\hat{j}=1)$
Since work is a scalar quantity, the total work done is $W=W_1+W_2=0-K{a}^2=-K{a}^2$
Hence, the correct choice is (c).