Question

A body of mass $3kg$ 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 $2s$ 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)

$s=\frac{t^2}{3};\frac{ds}{dt}=\frac{2t}{3};\frac{d^{2}s}{d{t}^2}=\frac{2}{3}$
Work done, $W=\int Fds=\int m\frac{d^{2}s}{d{t}^2}ds$
$=\int m\frac{d^{2}s}{d{t}^2}\frac{ds}{dt}dt=\underset{0}{\overset{2}{\int }}3\times \frac{2}{3}\times \frac{2t}{3}dt=\frac{4}{3}\underset{0}{\overset{2}{\int }}tdt$
$=\frac{4}{3}\underset{0}{\overset{2}{\int }}tdt=\frac{4}{3}{\left|\frac{t^2}{2}\right|}_0^2=\frac{4}{3}\times 2=\frac{8}{3}J$