Question

A body of mass $m$ starting from rest is acted on by a force producing a velocity $v=\sqrt{k\times s}$ where $k$ is a constant and $s$ is displacement. The work done by the force in the first $t$ seconds is:

• A. $\frac{m^2{k}^2{t}^2}{8}$
• B. $\frac{m{k}^2{t}^2}{4}$
• C. $\frac{m{k}^2{t}^2}{8}$
• D. $\frac{m^2{k}^{2}t}{4}$

Answer 1

The correct option is C $\frac{m{k}^2{t}^2}{8}$

$v=\frac{ds}{dt}=(ks{)}^{\frac{1}{2}}$ .....(1)
$\therefore s^{-\frac{1}{2}}ds=k^{\frac{1}{2}}dt$

Integrating both sides
$s^{\frac{1}{2}}=\frac{1}{2}{k}^{\frac{1}{2}}t+constant$

At $t=0$, $s=0$
$\therefore constant=0$
$\therefore s^{\frac{1}{2}}=\frac{1}{2}{k}^{\frac{1}{2}}t$
$\Rightarrow s=\frac{1}{4}k{t}^2$
Work done is equal to change in KE.
$\Delta \left(KE\right)=\frac{1}{2}m{v}^2=\frac{1}{2}mks=\frac{1}{2}mk\left(\frac{1}{4}k{t}^2\right)=\frac{m{k}^2{t}^2}{8}$