Question

A force $F=-k(y\hat{i}+x\hat{j})$(where A: 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

Answer: (C)

$dW=F\cdot ds$, where $ds=dx\hat{i}+dy\hat{j}+dz\hat{k}$
and $F=-k(y\hat{i}+x\hat{j})$
$\therefore dW=-k(ydx+xdy)=-kd(xy)$
$\therefore W=\underset{0,0}{\overset{a,a}{\int }}dW=-k\underset{0,0}{\overset{a,a}{\int }}d(xy)$
$=-k[xy{]}_{0,0}^{a,a}$
$W=-k{a}^2$
Alternate Answer
While moving from $(0,0)$ to $(a,0)$ along positive $x$-axis,

$y=0\therefore F=-kx\hat{j}$ i.e. force is in negative $y$-direction while the displacement is in positive $x$-direction. Therefore, $W_1=0$ (Force $\perp$ displacement).
Then, it moves from $(a,0)$ to $(a,a)$ along a line parallel to $y$-axis $(x=+a)$. During this $F=-k(y\hat{i}+a\hat{j})$
The first component of force, $-ky\hat{i}$ will not contribute any work, because this component is along negative $x$-direction $(-\hat{i})$ while displacement is in positive $y$-direction $(a,0)$ to $(a,a)$
The second component of force i.e. $-ka\hat{j}$ will perform
negative work $\left[\begin{array}{c}{F}_y=-ka\hat{j}\\ s=a\hat{j}\end{array}\right],W_2=(-ka)(a)=-k{a}^2$
$\therefore W=W_1+W_2=-k{a}^2$