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Physics Question on work, energy and power

A body of mass m=1 kgm = 1\, kg is moving in a medium and experiences a fractions force F=−kvF = - kv, where v is the speed of the body. The initial speed is v0=10 ms−1v_0 = 10 \, ms^{-1} and after 10s, its energy becomes half of initial energy. Then, the value of kk is

A

10ln⁡210 \ln \sqrt{2}

B

ln⁡2\ln \sqrt{2}

C

ln⁡220\frac{\ln 2}{20}

D

10ln⁡210 \ln 2

Answer

ln⁡220\frac{\ln 2}{20}

Explanation

Solution

∵\because According to the question,
12mvf2=12×12mvi2\frac{1}{2} m v_{f}^{2}=\frac{1}{2} \times \frac{1}{2} m v_{i}^{2}
(vi,vf=\left(v_{i}, v_{f}=\right. initial and final speeds of the body)
or vf2=vi22v_{f}^{2}=\frac{v_{i}^{2}}{2} or vf=102v_{f}=\frac{10}{\sqrt{2}}
Given, f=−kvf=-k v
or ma=−kv⇒mdvdt=−kvm 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 )
⇒(ln⁡v)1010/2=−k(10)\Rightarrow(\ln v)_{10}^{10 / \sqrt{2}}=-k(10)
⇒ln⁡102−ln⁡10=−k(10)\Rightarrow \ln \frac{10}{\sqrt{2}}-\ln 10=-k(10)
⇒k=110ln⁡(1010/2)=110ln⁡2=ln⁡220\Rightarrow k=\frac{1}{10} \ln \left(\frac{10}{10 / \sqrt{2}}\right)=\frac{1}{10} \ln \sqrt{2}=\frac{\ln 2}{20}