Question

A body of mass $m={10}^{–2}kg$ is moving in a medium and experiences a frictional force $F=–k{v}^2$. Its initial speed is $v_0=10m{s}^{-1}$. If, after $10s$, its energy is $\frac{1}{8}$$m{v}_0^2$, the value of $k$ will be

• A. ${10}^{-3}kg{m}^{-1}$
  
• B. ${10}^{-3}kg{s}^{-1}$
  
• C. ${10}^{-4}kg{m}^{-1}$
  
• D. ${10}^{-1}kg{m}^{-1}{s}^{-1}$
  

Answer 1

The correct option is C ${10}^{-4}kg{m}^{-1}$


$m={10}^{-2}kg$

$F=-k{v}^2$

$v_0=10m{s}^{-1}$

Also
$u=v_0,v=\frac{v_0}{2}$

$a=\frac{f}{m}=-\frac{k{v}^2}{m}$

$a=\frac{dv}{dt}$

$\frac{-k{v}^2}{m}=\frac{dv}{dt}$

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

$\frac{-k}{m}\times 10=-[\frac{1}{v}{]}_{10}^5$

$\frac{k}{m}\times 10=\frac{1}{5}-\frac{1}{10}$

$k\times {10}^3=\frac{1}{10}$

$R={10}^{-4}kg{m}^{-1}$