JEE Main 2020 — Work Power Energy Question with Solution

It is given that a force $\overrightarrow{F}=-x\hat{\mathrm{i}}+y\hat{\mathrm{j}}$ acts on a particle. The particle is moved from point $A\left(1,0\right)$ to $B\left(0,1\right)$ along the line segment.

Work done by a variable force on the particle,
$W=\int \overrightarrow{F}·\overrightarrow{\mathrm{d}r}=\int \overrightarrow{F}·\left(\mathrm{d}x\hat{\mathrm{i}}+\mathrm{d}y\hat{\mathrm{j}}\right)$
(In two dimension, $\overrightarrow{\mathrm{d}r}=\mathrm{d}x\hat{\mathrm{i}}+\mathrm{d}y\hat{\mathrm{j}}$
and it is given $\overrightarrow{F}=-x\hat{\mathrm{i}}+y\hat{\mathrm{j}}$)

$W=\int \left(-x\hat{\mathrm{i}}+y\hat{\mathrm{j}}\right)·\left(\mathrm{d}x\hat{\mathrm{i}}+\mathrm{d}y\hat{\mathrm{j}}\right)$

$=\int -x\mathrm{d}x+y\mathrm{d}y=\int -x\mathrm{d}x+\int y\mathrm{d}y$.
As the particle is displaced from $A\left(1,0\right)$ to $B\left(0,1\right)$, $x$ varies from $1$ to $0$ and $y$ varies from $0$ to $1$.
So, work will be,

$W=-{\int }_1^{0}x\mathrm{d}x+{\int }_0^{1}y\mathrm{d}y=-{\left[\frac{x^2}{2}\right]}_1^0+{\left[\frac{y^2}{2}\right]}_0^1$

$=\frac{1}{2}\left(0+{\left(1\right)}^2+{\left(1\right)}^2-0\right)=1 \mathrm{J}$.