A particle is moved along a path AB-BC-CD-DE-EF-FA, as shown in figure, in presence of a force  $\overrightarrow{F}=\left(\alpha y\widehat{i}+2\alpha x\widehat{j}\right)N,$ where x and y are in meter and  $\alpha =-1N{m}^{-1}$. The work done on the particle by this force $\overrightarrow{F}$ will be $n\times 0.25$ Joule.Find the value of n

$\begin{array}{l}\text{ For }AB:y=\text{ const }\\ \overrightarrow{F}=-y\widehat{i}-2x\widehat{j}\\ \Rightarrow \overrightarrow{F}=-\widehat{i}-2x\widehat{j}\\ \overrightarrow{dr}=dx\widehat{i}\\ \text{ So, }{W}_{AB}={\int }_{x=0}^1 \overrightarrow{F}\cdot \overrightarrow{dr}={\int }_0^1 -dx=-1J\end{array}$

$\begin{array}{l}\text{ For }\mathrm{BC}:\mathrm{x}=\mathrm{const}\\ \overrightarrow{F}=-y\widehat{i}-2\widehat{j}\\ \overrightarrow{dr}=dyj\\ W_{BC}={\int }_{y=1}^{0.5} -2dy\\ \Rightarrow W_{BC}=+1\mathrm{J}\end{array}$

$\begin{array}{l}\text{ For CD:y=const }\\ \overrightarrow{F}=-0.5\widehat{i}-2xj\text{ and }d\overrightarrow{r}=dx\widehat{i}\\ W_{CD}={\int }_1^{0.5} -0.5dx=0.25\mathrm{J}\end{array}$

$\begin{array}{l}\text{ For }\mathrm{DE}:\mathrm{x}=\text{ const }\\ \overrightarrow{F}=-y\widehat{i}-\widehat{j}\\ \overrightarrow{dr}=dy\widehat{j}\\ W_{DE}={\int }_{y=0.5}^0 -dy\\ =+0.5\mathrm{J}\end{array}$

$\begin{array}{l}\text{ For }\mathrm{EF}:\mathrm{y}=0\\ \overrightarrow{F}=-2\times \widehat{j}\\ d\overrightarrow{r}=dx\widehat{i}\\ \Rightarrow W_{EF}=0\\ \text{ For }FA:\mathrm{x}=0\\ \overrightarrow{F}=-y\widehat{i}\\ d\overrightarrow{r}=dy\widehat{j}\\ \Rightarrow W_{FA}=0\\ W_{\mathrm{Net}}=-1+1+0.25+0.5=0.75\mathrm{J}\end{array}$