Question

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

Answer 1

Answer: (C)

$\frac{k_{f}}{k_{i}}=\frac{\frac{1}{8} m v_{0}^{2}}{\frac{1}{2} m v_{0}^{2}}=\frac{1}{4}$
$\frac{v_{f}}{v_{i}}=\frac{1}{2}$
$v_{f}=\frac{v_{0}}{2}$
$-k v^{2}=\frac{m d v}{d t}$
$\int_\limits{v_{0}}^{\frac{v_{0}}{2}} \frac{d v}{v^{2}}=\int_\limits{0}^{t_{0}} \frac{-k d t}{m}$
$\left[-\frac{1}{v}\right]_{v_{0}}^{\frac{v_{0}}{2}}=\frac{-k}{m} t_{0}$
$\frac{1}{v_{0}}-\frac{2}{v_{0}}=-\frac{k}{m} t_{0}$
$-\frac{1}{v_{0}}=-\frac{k}{m} t_{0}$
$k=\frac{m}{v_{0} t_{0}}$
$=\frac{10^{-2}}{10 \times 10}$
$=10^{-4}\, kg \,m ^{-1}$