Question

A body of mass $m = 1\, kg$ is moving in a medium and experiences a fractions force $F = - kv$, where v is the speed of the body. The initial speed is $v_0 = 10 \, ms^{-1}$ and after 10s, its energy becomes half of initial energy. Then, the value of $k$ is

• A. $10 \ln \sqrt{2}$
• B. $\ln \sqrt{2}$
• C. $\frac{\ln 2}{20}$
• D. $10 \ln 2$
• E. $\ln 2$

Answer 1

Answer: (C)

$\because$ According to the question,
$\frac{1}{2} m v_{f}^{2}=\frac{1}{2} \times \frac{1}{2} m v_{i}^{2}$
$\left(v_{i}, v_{f}=\right.$ initial and final speeds of the body)
or $v_{f}^{2}=\frac{v_{i}^{2}}{2}$ or $v_{f}=\frac{10}{\sqrt{2}}$
Given, $f=-k v$
or $m a=-k v \Rightarrow \frac{m d v}{d t}=-k v$
$\Rightarrow \int_\limits{10}^{10 / \sqrt{2}} \frac{1}{v} d v=-\int_{0}^{10} k d t (\because m=1\, kg )$
$\Rightarrow (\ln v)_{10}^{10 / \sqrt{2}}=-k(10)$
$\Rightarrow \ln \frac{10}{\sqrt{2}}-\ln 10=-k(10)$
$\Rightarrow k=\frac{1}{10} \ln \left(\frac{10}{10 / \sqrt{2}}\right)=\frac{1}{10} \ln \sqrt{2}=\frac{\ln 2}{20}$