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 $x-y$ 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 point $(a,a)$. Then, total work done by the force is

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

Answer 1

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

Given, $\overrightarrow{F}=-k(y\hat{i}+x\hat{j})=-ky\hat{i}-kx\hat{j}$
Applying the fundamental equation for work done as the force is variable,
$W=\int \overrightarrow{F}.\overrightarrow{dr}$
[$\overrightarrow{dr}=dx\hat{i}+dy\hat{j}$]
$\Rightarrow W=(-ky\hat{i}-kx\hat{j}).(dx\hat{i}+dy\hat{j})$
$\Rightarrow W=\int -kydx-kxdy...\left(1\right)$


For displacement along path $OA$, $y$ is constant and $x$ is variable.
$\Rightarrow y=0,dy=0$
$x$ gives from $0$ to $a$
From Eq.$\left(1\right)$,
$W_{OA}=\int 0-0=0...\left(2\right)$
For displacement along path $AB$, $x$ is constant and $y$ is variable.
$\Rightarrow x=a,dx=0$
Limit of $y$ goes from $0$ to $a$.
$\therefore$From Eq.$\left(1\right)$,
$W_{AB}=\int 0-kady$
$W_{AB}=-ka\underset{0}{\overset{a}{\int }}dy$
$W_{AB}=-ka[y{]}_0^a=-k{a}^2....(3)$

Therefore, total work done by the force is given as,
$W=W_{OA}+W_{AB}$
$\therefore W=-k{a}^2$