Question

A force $F=-K(y\hat{i}+x\hat{j})$ (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 $F$ on the particles is

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

Answer 1

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

Using partial differentiation:
$F(x,y)=-Ky\hat{i}-Kx\hat{j}$
$\frac{\partial F}{\partial y}=-K$
$\frac{\partial F}{\partial x}=-K$
$\frac{\partial F}{\partial y}=\frac{\partial F}{\partial x}=-K$
This means force is conservative. Hence, work done is independent of the path.
Let's take straight line path between two pointt

Equation of this line will be $y=x$
Lets replace $x$ by $y$ and $y$ by $x$ in force equation
we get, $F(x,y)=-Kx\hat{i}-Ky\hat{j}$

we Know, Work $\left(W\right)=\int F.dr$
From above graph we can express, $d\overrightarrow{r}=dx\hat{i}+dy\hat{j}$
$W={\int }_{(0,0)}^{(a,a)}(-kx\hat{i}-ky\hat{j}).(dx\hat{i}+dy\hat{j})$
$W={\int }_0^a-Kxdx+{\int }_0^a-Kydy$
$W={\left[\frac{-K{x}^2}{2}\right]}_0^a+{\left[\frac{-K{y}^2}{2}\right]}_0^a$
$W=\frac{-K{a}^2}{2}+\left(\frac{-K{a}^2}{2}\right)$
$W=-K{a}^2$