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 the point $(a,a)$. Find the total work done by the forces $\overrightarrow{F}$ on the particle.

• A. Ka²
• B. -2 Ka²
• C. 2 ka²
• D.

Answer 1

The correct option is D

$\overrightarrow{F}=-k(y\hat{i}+x\hat{j})$

$\overrightarrow{dx}=(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$w=\int \overrightarrow{F}.\overrightarrow{dS}$

= $\int -k(y\hat{i}+x\hat{j})(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$w=\int [kydx-kxdy]$

Now particle is starting from (0, 0) origin and going to (a, 0)

So x varies from 0 - a

Y varies from 0 - 0
So $w=\int -kydx+\int -kxdy$

= $-ky\underset{0}{\overset{a}{\int }}dx-kx\underset{0}{\overset{0}{\int }}dy$

(here y is not changing (0 - 0))

$\Rightarrow -K0.[x{]}_0^2-kx[y{]}_0^0$

$\Rightarrow 0-0=0$

Now particle goes from (a, 0) to (a, a)

So x varies from a - a

Y varies from 0 - a

$w=\int -kydx+\int -kxdy$

$-ky\underset{a}{\overset{a}{\int }}-kx\underset{0}{\overset{a}{\int }}dy$

= $-ky[x{]}_a^a-kx[y{]}_0^a$

=$0-kaa$

$W=-k{a}^2$