JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Work, Power & Energy

Given $\vec{F}=\alpha y\hat{i}+2\alpha x\hat{j}$ with $\alpha =-1N/m$, so $\vec{F}=-y\hat{i}-2x\hat{j}$.

$W=\int \vec{F}\cdot d\vec{r}=\int \left(-ydx-2xdy\right)$

From the figure: $A(0,1)$, $B(1,1)$, $C(1,0.5)$, $D(0.5,0.5)$, $E(0.5,0)$, $F(0,0)$.

$A\to B$: $y=1$ constant, $dy=0$, $x:0\to 1$
$W_{AB}=-y\Delta x=-(1)(1-0)=-1$ J

$B\to C$: $x=1$ constant, $dx=0$, $y:1\to 0.5$ (downward, so $\Delta y=-0.5$)
$W_{BC}=-2x\Delta y=-2(1)(-0.5)=+1$ J

$C\to D$: $y=0.5$ constant, $dy=0$, $x:1\to 0.5$, so $\Delta x=-0.5$
$W_{CD}=-y\Delta x=-(0.5)(-0.5)=+0.25$ J

$D\to E$: $x=0.5$ constant, $dx=0$, $y:0.5\to 0$, so $\Delta y=-0.5$
$W_{DE}=-2x\Delta y=-2(0.5)(-0.5)=+0.5$ J

$E\to F$: $y=0$ throughout, so $-ydx=0$
$W_{EF}=0$

$F\to A$: $x=0$ throughout, so $-2xdy=0$
$W_{FA}=0$

$W_{\text{total}}=-1+1+0.25+0.5+0+0=0.75$ J

Note: the path is a closed loop, yet $W\ne 0$. This is because the force is non-conservative: $\frac{\partial F_x}{\partial y}=-1\ne \frac{\partial F_y}{\partial x}=-2$.