A body of mass m= 10^{-2} ;kg is moving | vedclass

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Question

Let ${V_f}$ is the final speed of the body.

From questions,

$\begin{gathered}
  \frac{1}{2}mV_f^2 = \frac{1}{8}mV_0^{2\,}\,\,\,\,\,\,

Vedclass · Accepted Answer

$10^{-4}$ $kg m^{-1}$

Vedclass · Answer

Let ${V_f}$ is the final speed of the body.

From questions,

$\begin{gathered}
  \frac{1}{2}mV_f^2 = \frac{1}{8}mV_0^{2\,}\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,{V_f} = \frac{{{V_0}}}{2} = 5m/s \hfill \
  F = m\left( {\frac{{dV}}{{dt}}} ight) =  - k{V^2}\,\,	herefore \left( {{{10}^{ - 2}}} ight)\frac{{dV}}{{dt}} =  - k{V^2} \hfill \
  \int\limits_{10}^5 {\frac{{dV}}{{{V^2}}} =  - 100K\int\limits_0^{10} {dt} }  \hfill \
  \frac{1}{5} - \frac{1}{{10}} = 100k\left( {10} ight)\,\,\,\,or,\,\,K = {10^{ - 4}}\,kg{m^{ - 1}} \hfill \ 
\end{gathered} $

A body of mass $m= 10^{-2} \;kg$ is moving in a medium and experiences a frictional force $F= -kv^2$ Its intial speed is $v_0= 10$ $ms^{-1}$ If, after $10\ s$, its energy is $\frac{1}{8}$ $mv_0^2$ the value of $k$ will be

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