Question

A particle of mass $m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $a$. A force $\overrightarrow{F}=y\hat{i}-x\hat{j}N$ acts on the particle, where $x,y$ denote the coordinates of position of the particle. Calculate the work done by this force, if the particle displaces from point $A(a,0)$ to point $B(0,a)$ along the circular path.

• A. $\frac{\pi a^2}{4}J$
• B. $\frac{\pi a^2}{2}J$
• C. $-\frac{\pi a^2}{2}J$
• D. $-\frac{\pi a^2}{4}J$

Answer 1

The correct option is C $-\frac{\pi a^2}{2}J$

Work done by variable force is $W=\overrightarrow{F}.\overrightarrow{dr}$
$W=\overrightarrow{F}.(dx\hat{i}+dy\hat{j}+dz\hat{k})$
$W=\int F_{x}dx+\int F_{y}dy=\int ydx-\int xdy$

Equation of circle passing through point $A$, point $B$ with centre origin is $x^2+y^2=a^2$
As $x,y$ points lie on the circle, we can substitute
$x=a\cos \theta ,y=a\sin \theta ,$
$dx=-a\sin \theta d\theta ,dy=a\cos \theta d\theta$



$W=-\int a^{2}si{n}^2\theta d\theta -\int a^{2}co{s}^2\theta d\theta$
$W=-a^2\int d\theta$
As particle moves from $A$ to $B$, $\theta$ varies from $0$ to ${90}^{\circ }$
$W=-a^2{\int }_0^{\pi /2}d\theta$

$W=-\frac{\pi a^2}{2}J$