Question

A force $\overrightarrow{F}=-k(y\hat{i}+x\hat{j})$, where $k$ is a constant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken from the origin $(0,0)$ to the point $(a,0)$ and then from the point $(a,0)$ to the point $(a,a)$. The total work done by the force $\overrightarrow{F}$ on the particle is

• A. $k{a}^2$
• B. $-k{a}^2$
• C. $2k{a}^2$
• D. $-2k{a}^2$

Answer 1

The correct option is B $-k{a}^2$

Given $\overrightarrow{F}=-k(y\hat{i}+x\hat{j})$
and $\overrightarrow{r}=\hat{i}x+\hat{j}y$ so that $\overrightarrow{dr}=\hat{i}dx+\hat{j}dy$
Now work done is
$W=\int \overrightarrow{F}.d\overrightarrow{r}=-k(y\hat{i}+x\hat{j}).(\hat{i}dx+\hat{j}dy)$
$W=-k\int ydx+xdy=-k\int d\left(xy\right)$
$W=-k.xy$ .... (1)
Let the work done from $O$ to $A$ is $W_1$ and from $A$ to $B$ is $W_2$. For the path $O$ to $A$, $y=0$ , therefore putting $y=0$ in Eq. (1) we get
$W_1=0$
For the path $A$ to $B$, $x=a$ and $y$ varies from $0$ to $a$ , therefore
$W_2=-ka.a=-k{a}^2$
Total work done $W=W_1+W_2=-k{a}^2$
Hence, the correct answer is option (b).