Q. A force of $5\ N$ acts on a $15\ kg$ body initially at rest. The work done by the force during the first second of motion of the body is
A) $5\ J$
B) $\frac{5}{6}\ J$
C) $6\ J$
D) $75\ J$
Explanation
(b) $W = Fs = F \times \frac{1}{2}at^{2}\left\lbrack from\ s = ut + \frac{1}{2}at^{2} \right\rbrack$
$\Rightarrow W = F\left\lbrack \frac{1}{2}\left( \frac{F}{m} \right)t^{2} \right\rbrack = \frac{F^{2}t^{2}}{2m} = \frac{25 \times (1)^{2}}{2 \times 15} = \frac{25}{30} = \frac{5}{6}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.