Question

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

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

Answer 1

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

Given $\overrightarrow{F}=-k(y\hat{i}+x\hat{j}),(k>0)$
It is given that, particle is taken along x-axis from origin to the point $(a,0)$ and then parallel to the y-axis upto $(a,a)$.


Work done along OA: $x$ changes from $0$ to $a$, $y=0$
Since $F$ is variable,
Work done $\left(W_{OA}\right)=\int \overrightarrow{F.}\overrightarrow{ds}$
$={\int }_{x_1}^{x_2}{F}_{x}dx+{\int }_{y_1}^{y_2}{F}_{y}dy\because dy=0W={\int }_{x_1}^{x_2}-kydx=0(\because y=0)\Rightarrow W_{OA}=0$

Work done along AB: $x=a$, $y$ changes from $0$ to $a$.
$\therefore W_{AB}={\int }_{x_1}^{x_2}{F}_{x}dx+{\int }_{y_1}^{y_2}{F}_{y}dy\because x\text{is constant},dx=0\Rightarrow W_{AB}={\int }_{y_1}^{y_2}{F}_{y}dy={\int }_0^a-kxdy$
$=-ka{\int }_0^{a}dy=-k{a}^2$
($\because x=a$)

Total work done= work done along OA + work done along AB
$=0+(-k{a}^2)$
$=-k{a}^2$