Question

A force $F=-k(yi+xj)$ (where $k$ is a positive constant) acts on a particle moving in the $xy$ - 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 $-k{a}^2$

While moving from $(0,0)$ to $(a,0)$
Along positive $x$ - axis,$y=0\Rightarrow \overrightarrow{F}=-kx\hat{j}$
i.e. force is in negative $y$ - direction while displacement is in positive $x$ - direction.
$\therefore W_1=0$
Because force is perpendicular to displacment.
Then particle moves from $(a,0)$ to $(a,a)$ along a line parallel to $y$ - axis $(x=+a)$ during this $\overrightarrow{F}=-k(y\hat{i}+a\hat{j})$
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
$\therefore W_2=(-ka\hat{j})\left(a\hat{j}\right)=(-ka)\left(a\right)=-k{a}^2$
So, net work done on the particle
$W=W_1+W_2=0+(-k{a}^2)=-k{a}^2$

Alternate solution

Given $F=-k(y\hat{i}+x\hat{j})$
We know that
$W=\int \overrightarrow{F}.\overrightarrow{ds}={\int }_{(0,0)}^{(a,a)}-k(y\hat{i}+x\hat{j}).(dx\hat{i}+dy\hat{j})={\int }_{(0,0)}^{(a,a)}-k(ydx+xdy)$W=$-{\int }_{(0,0)}^{(a,a)}kd\left(xy\right)=-k{a}^2$