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PhysicsW.E.P.MiscellaneousHard2 minQB

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A body of mass $m = 10^{- 2} kg$ is moving in a medium and experiences a frictional force $F = - k v^{2} .$ Its initial speed is $v_{0} = 10 m s^{- 1} .$ After $10 s$ its kinetic energy is $\frac{1}{8} m v_{0}^{2},$ then value of $k$ will be:-

Options:

Answer:

Solution:

After $10 sec,$ $\frac{1}{2} m v^{2} = \frac{1}{8} m v_{0}^{2}$
$\Rightarrow v = \frac{v_{0}}{2} = 5 m s^{- 1}$
Acceleration, $a = - \frac{k v^{2}}{m}$
$\Rightarrow \frac{d v}{d t} = - \frac{k v^{2}}{m}$
$\Rightarrow \int_{10}^{5} \frac{d v}{v^{2}} = - \frac{k}{m} \int_{0}^{10} d t$
$- \frac{1}{5} + \frac{1}{10} = \frac{- k}{10^{- 2}} \times 10$
$k = 10^{- 4} kg m^{- 1}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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