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 (2, 0) and then parallel to the y-axis
to the point (2, 2). The total work done by the forces $\overrightarrow{F}$ on the particle is

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

Answer 1

The correct option is C $-4K$

For motion of the particle from (0, 0) to (a, 0)
$\overrightarrow{F}=-K(0\hat{i}+a\hat{j})\Rightarrow \overrightarrow{F}=-Ka\hat{j}$
Displacement $\overrightarrow{r}=(a\hat{i}+0\hat{j})-(0\hat{i}+0\hat{j})=a\hat{i}$
So work done from (0, 0) to (a, 0) is given by
$W=\overrightarrow{F}.\overrightarrow{r}=-Ka\hat{j}.a\hat{i}=0$
For motion (a, 0) to (a, a) displacement
$\overrightarrow{r}=(a\hat{i}+a\hat{j})-(a\hat{i}+0\hat{j})=a\hat{j}$
So work done from (a, 0) to (a, a) $W=\overrightarrow{F}.\overrightarrow{r}$
So total work done $=-K{a}^2$