Verified Solution • 2025–26 Session moderate

QJEEM17Phba02: JEE Main 2017 Physics - Class 11 JEEM17Phba02 Solution: A body of mass m = 10 –2 kg is moving in a medium and…

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 18mv02{1 \over 8}mv_0^2 , the value of k will be:

Multiple Choice Options

  • A 10-1 kg m-1 s-1
  • B 10-3 kg m-1
  • C 10-3 kg s-1
  • D 10-4 kg m-1

Step-by-Step Verified Solution

According to the question, final kinetic energy = 18mv02{1 \over 8}mv_0^2

Let final speed of the body = Vf

So final kinetic energy = 12mvf2{1 \over 2}mv_f^2

According to question,

12mvf2{1 \over 2}mv_f^2 = 18mv02{1 \over 8}mv_0^2

⇒vf=v02\Rightarrow {v_f} = {{{v_0}} \over 2} = 102{{10} \over 2} = 5 m/s

Given that, F = –kv2

⇒\Rightarrow m(dvdt)m\left( {{{dv} \over {dt}}} \right) =−kv2= - k{v^2}

⇒10−2(dvdt)=−kv2\Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}

⇒∫105dvv2=−100k∫010dt\Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt}

⇒15−110=100k×10\Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10

⇒k=10−4kg m−1\Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}

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