Question

A body of mass $3 \,kg$ is under a constant force which causes a displacement s in metre in it, given by the relation $s=\frac{1}{3}t^2$, where $t$ is in second. Work done by the force in $2 s$ is

• A. $\frac{5}{19} \,J$
• B. $\frac{3}{8} \,J$
• C. $\frac{8}{3} \,J$
• D. $\frac{19}{5} \,J$

Answer 1

Answer: (C)

$s=\frac{t^{2}}{3} ; \frac{d s}{d t}=\frac{2 t}{3} ; \frac{d^{2} s}{d t^{2}}=\frac{2}{3}$
Work done, $W=\int F d s=\int m \frac{d^{2} s}{d t^{2}} d s$
$=\int m \frac{d^{2} s}{d t^{2}} \frac{d s}{d t} d t=\int\limits_{0}^{2} 3 \times \frac{2}{3} \times \frac{2 t}{3} d t=\frac{4}{3} \int\limits_{0}^{2} t d t$
$=\frac{4}{3} \int\limits_{0}^{2} t d t=\frac{4}{3}\left|\frac{t^{2}}{2}\right|_{0}^{2}=\frac{4}{3} \times 2=\frac{8}{3} J$