Question

A particle is moved along a path $AB-BC-CD-DE-EF-FA$, as shown in figure in presence of a force $\overrightarrow{F}=(\alpha y\hat{i}+2\alpha x\hat{j})N$, where x and y are in meter and $\alpha =-1N{m}^{-1}$. The work done on the particle by this force $\overrightarrow{F}$ will bejoule. (write up to two decimal places)

Answer 1

Given, $\overrightarrow{F}=\alpha y\hat{i}+2\alpha x\hat{j}$
$\alpha =-1\therefore \overrightarrow{F}=-[y\hat{i}+2x\hat{j}]$
by the graph representation,


On taking line integration,
$W_{AB}=-[y\hat{i}+2x\hat{j}].\left[1\hat{i}\right]=-1$
$W_{BC}=-[y\hat{i}+2x\hat{j}].\left[-0.5\hat{j}\right]=1$
$W_{CD}=-[y\hat{i}+2x\hat{j}].\left[-0.5\hat{j}\right]=0.25$
$W_{DE}=-[y\hat{i}+2x\hat{j}].\left[-0.5\hat{j}\right]=0.5$
$W_{EF}=W_{FA}=0$
$\Delta W=W_{AB}+W_{BC}+W_{CD}+W_{DE}+W_{EF}+W_{FA}$
$\therefore \Delta W=+0.75J$