Question

A Force $\overrightarrow{F}=-2k[y\hat{i}+x\hat{j}]N$ ,where $k$ is 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 $y-$ axis to the point $(a,a)$. The total work done by the force $\overrightarrow{F}$ on the particle is

• A. $-4k{a}^2$
• B. $2k{a}^2$
• C. $-2k{a}^2$
• D. $4k{a}^2$

Answer 1

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

For an infinitesimally small displacement $\overrightarrow{dr}$ ,
Work done by $\overrightarrow{F}$ will be, $dW=\overrightarrow{F}.\overrightarrow{dr}\Rightarrow dW=-2k[y\hat{i}+x\hat{j}].[dx\hat{i}+dy\hat{j}]\Rightarrow dW=-2k[ydx+xdy]=-2kd\left[xy\right]$
Now, integrating within limits,

${\int }_0^{W}dW=-2k{\int }_{(0,0)}^{(a,a)}d\left(xy\right)\Rightarrow W=-2k[xy{]}_{0,0}^{a,a}=-2k{a}^2$