Question

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

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

Answer 1

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

Given,

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

$v_0=10m/s$

$F=k{v}^{-2}$

The kinetic energy after $10sec$ is $\frac{1}{8}m{v}_0^2$

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

$v_f^2=\frac{v_0^2}{4}$

$v_f=\frac{v_0}{2}=\frac{10}{2}=5m/s$. . . . . . . .(1)

Frictional force is given by

$F=k{v}^{-2}$

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

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

$m{v}^{2}dv=kdt$

$m{\int }_0^5{v}^{2}dv=k{\int }_0^{10}dt$

$m\frac{v^3}{3}{|}_0^5=k.t{|}_0^{10}=10k$

$\frac{(5^3-0^0)}{3}=100k$

$k={10}^{-4}$