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Physics Question on work, energy and power

A body of mass m=10−2 kgm=10^{-2} \, kg is moving in a medium and experiences a frictional force F=−kv2F=-kv^2. Its initial speed is v0=10 ms−1v_0=10 \, ms^{-1}. If, after 10 s10\,s, its energy is 18mv02\frac{1}{8} mv_0{^{2}}, the value of kk will be :

A

10−3kg m−110^{-3} kg \, m^{-1}

B

10−3kg s−110^{-3} kg \, s^{-1}

C

10−4kg m−110^{-4} kg \, m^{-1}

D

10−1kg m−1s−110^{-1} kg \, m^{-1} s^{-1}

Answer

10−4kg m−110^{-4} kg \, m^{-1}

Explanation

Solution

kfki=18mv0212mv02=14\frac{k_{f}}{k_{i}}=\frac{\frac{1}{8} m v_{0}^{2}}{\frac{1}{2} m v_{0}^{2}}=\frac{1}{4}
vfvi=12\frac{v_{f}}{v_{i}}=\frac{1}{2}
vf=v02v_{f}=\frac{v_{0}}{2}
−kv2=mdvdt-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}
[−1v]v0v02=−kmt0\left[-\frac{1}{v}\right]_{v_{0}}^{\frac{v_{0}}{2}}=\frac{-k}{m} t_{0}
1v0−2v0=−kmt0\frac{1}{v_{0}}-\frac{2}{v_{0}}=-\frac{k}{m} t_{0}
−1v0=−kmt0-\frac{1}{v_{0}}=-\frac{k}{m} t_{0}
k=mv0t0k=\frac{m}{v_{0} t_{0}}
=10−210×10=\frac{10^{-2}}{10 \times 10}
=10−4 kg m−1=10^{-4}\, kg \,m ^{-1}