Question

A body of mass ${10}^{-2}kg$ is moving in a medium and experiences a frictional force $-k{v}^2$. Its initial speed is $v_0=10m{s}^{-1}$. If, after time $10s$, 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

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

Step 1. Given data

Mass of the body, $m={10}^{-2}kg$

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

Initial speed, $v_0=10m{s}^{-1}$

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

Step 2. Finding the value of $k$

Final kinetic energy

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

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

We know that,

Force, $F=m\frac{dv}{dt}$

$-k{v}^2={10}^{-2}\frac{dv}{dt}$

${\int }_{v_0}^{\frac{v_0}{2}}\frac{dv}{v^2}={\int }_0^t\frac{-kdt}{{10}^{-2}}$

${\left[-\frac{1}{v_0}\right]}_{v_0}^{\frac{v_0}{2}}=-\frac{kt}{m}$

$\frac{1}{v_0}-\frac{2}{v_0}=-\frac{kt}{m}$

$k=\frac{m}{v_{0}t}$

$k=\frac{{10}^{-4}}{10\times 10}$

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

Hence, the correct option is $\left(C\right)$.