Q. A body of mass $3\ kg$ is under a force, which causes a displacement in it given by $S = \frac{t^{3}}{3}$ (in $m$). Find the work done by the force in first 2 seconds
A) $2\ J$
B) $3.8\ J$
C) $5.2\ J$
D) $24\ J$
Explanation
(d) $S = \frac{t^{3}}{3}\ \therefore dS = t^{2}dt \Rightarrow a = \frac{d^{2}S}{dt^{2}} = \frac{d^{2}}{dt^{2}}\left\lbrack \frac{t^{3}}{3} \right\rbrack = 2t\ m/s^{2}$
Now work done by the force $W = \int_{0}^{2}{F.dS} = \int_{0}^{2}{ma.dS}$
$\int_{0}^{2}{3 \times 2t \times t^{2}dt} = \int_{0}^{2}{6t^{3}}dt = \frac{3}{2}\left\lbrack t^{4} \right\rbrack_{0}^{2} = 24\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.