Question

A body of mass $m={10}^{-2}kg$ is moving in 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. ${10}^{-1}kg{m}^{-1}{s}^{-1}$
• B. ${10}^{-3}kg{m}^{-1}$
• C. ${10}^{-3}kg{s}^{-1}$
• D. ${10}^{-4}kg{m}^{-1}$

Answer 1

Answer: (D)

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

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

$F=-k{v}^2$

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

Initial energy $=\frac{1}{2}m{v}_0^2$

Final energy $=\frac{1}{8}m{v}_0^2$

Using energy conservation,

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

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

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

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

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

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

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

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

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