- Home
- Standard 11
- Physics
A block $A$ of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block $B$ of mass $m_2$ is suspended. The coefficient of kinetic friction between the block and the table is $\mu _k.$ When the block $A$ is sliding on the table, the tension in the string is
$\frac{{\left( {{m_2} + {\mu _k}{m_1}} \right)g}}{{{m_1} + {m_2}}}$
$\;\frac{{\left( {{m_2} - {\mu _k}{m_1}} \right)g}}{{{m_1} + {m_2}}}$
$\;\frac{{{m_1}{m_2}\left( {1 + {\mu _k}} \right)g}}{{{m_1} + {m_2}}}$
$\;\frac{{{m_1}{m_2}\left( {1 - {\mu _k}} \right)g}}{{{m_1} + {m_2}}}$
Solution
$\begin{array}{l}
{m_2}g – T = {m_2}a\\
T – \mu k\,{m_1}g = {m_1}a\\
\Rightarrow \,a = \frac{{\left( {{m_2} – \mu km} \right)g}}{{{m_1} + {m_2}}}\\
For\,the\,block\,of\,mass\,'m{'_2}\\
{m_2}g – T = {m_2}\left[ {\frac{{m – 2 – {\mu _k}}}{{{m_1} + {m_2}}}} \right]
\end{array}$
$\begin{array}{l}
g{m_2}g – T = {m_2}\left[ {\frac{{{m_2} – {\mu _k}m – 1}}{{{m_1} + {m_2}}}} \right]\\
{m_2}g = {m_2}g\left[ {\frac{{{m_2} – {\mu _k}m – 1}}{{{m_1} + {M_2}}}} \right]
\end{array}$
