Question

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

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

Answer 1

The correct option is

D

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

The force is always tangential to the curve at every point and the magnitude is constant. The direction of force is opposite to the direction of the motion of the particle.

The magnitude of the force = $F=\sqrt{x^2+y^2}$

Total distance covered = $s=\frac{\pi a}{2}$

Hence $W=\overrightarrow{F}.\overrightarrow{s}=-Fs=-\sqrt{x^2+y^2}\times \frac{\pi a}{2}=\frac{\pi a^2}{2}$ as x = a and y = 0.