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 first 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 ($\because y=0$)
$W_1={\int }_0^{a}F.dx={\int }_0^a-\left(Kx\hat{j}\right).dx\hat{i}{W}_1=0$
$(\because \hat{j}.\hat{i}=0)$

In going from $(a,0)$ to $(a,a)$, the x-coordinate remains constant at $x=a$ while y-coordinate changes from $0$ to $a$.
$\therefore$ Work done by force $F$ along this path is $(\because x=a)$


$W_2={\int }_0^{a}F.dy={\int }_0^a-K(y\hat{i}+a\hat{j}).dy\hat{j}$
$=-Ka{\int }_0^{a}dy{W}_2=-K{a}^2$
$(\because \hat{i}.\hat{j}=0,\hat{j}.\hat{j}=1)$

The total work done is:
$W=W_1+W_2=0-K{a}^2=-K{a}^2$.