Question

A force $F=-K(x\hat{i}+y\hat{j})$ (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 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. $-K{a}^2$
• D. $K{a}^2$

Answer 1

Answer: (C)

$-K{a}^2$

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 $(a,a)$
$\overrightarrow{F}=-K(a\hat{i}+a\hat{j})$ and 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 $(a,a)$ $W=\overrightarrow{F}.\overrightarrow{r}$
$=-K(a\hat{i}+a\hat{j}).a\hat{j}=-K{a}^2$
So total work done $=-K{a}^2$