A block of mass m = 10,kg rests on a horizont | vedclass

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Question

$f =\mu  (M + m) g$

$\begin{gathered}
  a = \frac{f}{{M + m}} = \frac{{\mu \left( {M + m} ight)g}}{{(M + m)}} = \mu g \hfill \
&nbsp

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

$f =\mu  (M + m) g$

$\begin{gathered}
  a = \frac{f}{{M + m}} = \frac{{\mu \left( {M + m} ight)g}}{{(M + m)}} = \mu g \hfill \
   = 0.05 	imes 10 = 0.5\,m{s^{ - 2}} \hfill \
  {V_0} = \frac{{Initial\,momentum\,}}{{\left( {M + m} ight)}} = \frac{{0.05V}}{{10.05}} \hfill \
  m = 50g\,\,\,\,\,\,\,\,\,\,\,\,M = 10\,kg \hfill \ 
\end{gathered} $

${\left( {\frac{{0.05v}}{{10.05}}} ight)^2} = 2 	imes 0.5 	imes 2$

Solving we get $v = 201\sqrt 2 $ object falling from height $H.$

$\begin{gathered}
  \frac{V}{{10}} = \sqrt {2gh}  \hfill \
  \frac{{201\sqrt 2 }}{{10}} = \sqrt {2 	imes 10 	imes H}  \hfill \ 
\end{gathered} $

$H = 40\; m = 0.04\; km$

$\begin{gathered}
  {v^2} - {u^2} = 2as \hfill \
  0 - {u^2} = 2as \hfill \
  {u^2} = 2as \hfill \ 
\end{gathered} $

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