Discussion Forum

A body of mass m = 10^-2 kg is moving in a medium and experiences a frictional force F= -kv₂ . Its initial is v₀ =10 ms^-1 . If , after 10 s. its energy is  $\frac{1}{8}mv_{0}^{2}$ , the value of k will be

Answer:

Explanation:

Given , force  F= -kv² 

    $\therefore$   Acceleration, a= $\frac{-k}{m}v^{2}$

  or      $\frac{\text{d}v}{\text{d}t}= \frac{-k}{m}v^{2}\Rightarrow \frac{dv}{v^{2}}=-\frac{k}{m}dt$

  Now with limits , we have

         $\int_{10}^{v} \frac{dv}{v^{2}}= -\frac{k}{m}\int_{0}^{t}dt $

$\Rightarrow \left(- \frac{1}{v}\right)_{10}^{v}= -\frac{k}{m}t\Rightarrow \frac{1}{v} = 0.1+\frac{kt}{m}$

$\Rightarrow v= \frac{1}{0.1+\frac{kt}{m}}= \frac{1}{0.1+1000k}$

 $\Rightarrow \frac{1}{2}\times m\times v^{2}=\frac{1}{8}mv_{0}^{2}$

                 $\Rightarrow v=\frac{v_{0}}{2}=5$

$\Rightarrow  \frac{1}{0.1+1000k}=5\Rightarrow 1=0.5+5000k$

$\Rightarrow  k=\frac{0.5}{5000}\Rightarrow k=10^{-4}kg/m$