A body of mass m = 10–2 kg is moving in a medium and experiences a… | JEE Main 2017

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Inertia, First & Second Laws

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 $\frac{1}{8}mv_0^2$ , the value of k will be:

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Answer: (d)

KE $= \frac{1}{2}mv^2 = \frac{1}{8}mv_0^2 \Rightarrow v = \frac{v_0}{2} = 5\text{ m/s}$ .
$m\frac{dv}{dt} = -kv^2 \Rightarrow \frac{1}{v} - \frac{1}{v_0} = \frac{kt}{m}$
$\frac{1}{5} - \frac{1}{10} = \frac{k \times 10}{10^{-2}} \Rightarrow 0.1 = 1000k$
$k = 10^{-4}$ ; units of $F/v^2$ are $\text{kg m}^{-1}$ .
So the answer is (d) $10^{-4}\text{ kg m}^{-1}$ .

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