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  4. A body of mass m=10-2 kg is moving in a medium and experiences a frictional force F=-kv2. Its initial speed is v0=10 ms-1. If, after 10 s, its energy is (1/8) mv02, the value of k will be :

Q. 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 :

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Solution:

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

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