Q. A body of mass 3 kg acted upon by a constant force is displaced by $S$ metre, given by relation$\ S = \frac{1}{3}t^{2}$, where t is in second. Work done by the force in 2 seconds is
A) $\frac{8}{3}\ J$
B) $\frac{19}{5}J$
C) $\frac{5}{19}J$
D) $\frac{3}{8}\ J$
Explanation
(a) Given that, $S = \frac{1}{3}t^{2}$
$v = \frac{dS}{dt} = \frac{2}{3}t;\ a = \frac{d^{2}S}{dt^{2}} = \frac{2}{3}$
$F = ma = 3 \times \frac{2}{3} = 2N;\ \ \ \ Work = 2 \times \frac{1}{3}t^{2}$
At t =2
Work$\ = 2 \times \frac{1}{3} \times 2 \times 2 = \frac{8}{3}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.