Question

A force $F=-K(y\hat{i}+x\hat{j})$ (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 $x$-axis to the point $(a,0)$ and then parallel to $y$-axis to the point $(0,a)$. 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

Answer: (C)

$-K{a}^2$

Given
$\overrightarrow{F}=-K(y\overrightarrow{i}+x\overrightarrow{j})$
Consider the small work done by the force $F$ in displacing the particle through area $dS$ as

$dW=\overrightarrow{F}.\overrightarrow{dS}$

where, $\overrightarrow{dS}=dx\hat{i}+dy\hat{j}$

$\therefore dW=(-K(y\hat{i}+x\hat{j}\left)\right).(dx\hat{i}+dy\hat{j})$

$\therefore dW=-K(ydx+xdy)=-Kd\left(xy\right)$ .....................(1)

Since, first the particle is taken along the position x-axis to the point (a, 0) and then parallel to y-axis to the point (0, a) hence, integrating for whole area we get

$W={\int }_{x=0}^{x=a}{\int }_{y=0}^{y=a}dW$

$W={\int }_{x=0}^{x=a}{\int }_{y=0}^{y=a}(-Kd(xy\left)\right)$.............from(1)

$W=-K\left({\left|x\right|}_0^a\right)\left({\left|y\right|}_0^a\right)$

$W=-K{a}^2$