Question

A body of mass, $m={10}^{-2}\text{kg}$ is moving in a medium experiencing a frictional force, $F=-k{v}^2$. Its intial speed is, $v_0=20\text{m/s}$. If, after $10\text{seconds}$, its energy is $\frac{1}{8}m{v}_0^2$, then the value of $k$ will be

• A. $5\times {10}^{-3}\text{kg/s}$
• B. $5\times {10}^{-5}\text{kg/s}$
• C. ${10}^{-4}\text{kg/s}$
• D. $0.5\times {10}^{-3}\text{kg/s}$

Answer 1

The correct option is B $5\times {10}^{-5}\text{kg/s}$

Given that,

Mass, $m={10}^{-2}\text{kg}$

Initial velocity , $v_0=20\text{m/s}$

Time, $t=10\text{s}$

Frictional force , $F=-k{v}^2$

If ${\upsilon }_t$ is velocity of body after $10\text{s}$,

then

$\frac{1}{2}m{v}_t^2=\frac{1}{8}m{v}_o^2$

$\Rightarrow v_t=\frac{v_o}{2}$

From newton's second law of motion,
$F=ma$

Substituting the value of $F$, $-k{v}^2=m\frac{dv}{dt}$

$\Rightarrow -kdt=m\frac{dv}{v^2}$

Integrating the above equation between the time $(t_1=0\text{sec})$ to $(t_2=10\text{sec})$,

$\Rightarrow -k{\int }_{t_1}^{t_2}dt=m{\int }_{v_0}^{v_t}\left(\frac{dv}{v^2}\right)$

$\Rightarrow -k{\int }_0^{10}dt=m{\int }_{v_0}^{v_o/2}{v}^{-2}dv$

$\Rightarrow -kt{|}_0^{10}=m\times \frac{v^{-2+1}}{-2+1}{|}_{v_0}^{v_{0/2}}$

$\Rightarrow -k\left(10\right)=-{10}^{-2}[\frac{2}{v_0}-\frac{1}{v_0}]$

$\Rightarrow k={10}^{-3}\left[\frac{1}{v_0}\right]={10}^{-3}\left[\frac{1}{20}\right]$

$\therefore k=5\times {10}^{-5}\text{kg/m}$

Hence, option (b) is correct answer.