Question

A particle of mass $m$ moves with a variable velocity $v$, which changes with distance covered $x$ along a straight line as $v=k\sqrt{x}$, where $k$ is a positive constant. The work done by all the forces acting on the particle, during the first $t$ seconds is

• A. $\frac{m{k}^4}{t^2}$
• B. $\frac{m{k}^4{t}^2}{4}$
• C. $\frac{m{k}^4{t}^2}{8}$
• D. $\frac{m{k}^4{t}^2}{16}$

Answer 1

Answer: (C)

Given $v=k\sqrt{x}$
or $\frac{dx}{dt}=k\sqrt{x}$ or $x^{\frac{1}{2}}dx=kdt$
Integrating both sides, we get
$\frac{x^{\frac{1}{2}}}{\frac{1}{2}}=kt+C$; assuming $x\left(0\right)=0$
Therefore, $C=0$
$2\sqrt{x}=kt$
$\Rightarrow x=\frac{k^2{t}^2}{4}$ or $v=\frac{k^{2}t}{2}$
Therefore, work done,
$\Delta W=$ Increase in $KE$
$=\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(0\right)}^2=\frac{1}{2}m{\left[\frac{k^{2}t}{2}\right]}^2=\frac{1}{8}m{k}^4{t}^2$