Work Energy Power

JEE Main level, single correct, medium

122 of 183

Question 122 of 183

A body of mass m=10−2kgm={10}^{-2}kg is moving in a medium and experiences a frictional force F=−kv2.F=-k{v}^{2}. Its initial speed is v0=10ms−1.{v}_{0}=10m{s}^{-1}. After 10s10s its kinetic energy is 18mv02,\dfrac{1}{8}m{v}_{0}^{2}, then value of kk will be:-

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Answer: (D) 10−4kgm−1{10}^{-4}kg{m}^{-1}

After 10sec,10 \text{sec,} 12mv2=18mv02\dfrac{1}{2}m{v}^{2}=\dfrac{1}{8}m{v}_{0}^{2}
⇒v=v02=5ms−1\Rightarrow v=\dfrac{{v}_{0}}{2}=5m{s}^{-1}
Acceleration, a=−kv2ma=-\dfrac{k{v}^{2}}{m}
⇒dvdt=−kv2m\Rightarrow \dfrac{dv}{dt}=-\dfrac{k{v}^{2}}{m}
⇒∫105dvv2=−km∫010dt\Rightarrow \int _{10}^{5}\dfrac{dv}{{v}^{2}}=-\dfrac{k}{m}\int _{0}^{10}dt
−15+110=−k10−2×10-\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{-k}{{10}^{-2}}\times 10
k=10−4kgm−1k={10}^{-4}kg{m}^{-1}

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