Question

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

• A. $\frac{m_1m_2(1+\mu_k)g}{(m_1+m_2)}$
• B. $\frac{m_1m_2(1-\mu_k)g}{(m_1+m_2)}$
• C. $\frac{(m_1+\mu_k m_1)g}{(m_1+m_2)}$
• D. $\frac{(m_2-\mu_k m_1)g}{(m_1+m_2)}$

Answer 1

Answer: (A)

$\frac{m_1m_2(1+\mu_k)g}{(m_1+m_2)}$

Answer 2

${m_2g -T = m_2a}$
${T - \mu_k \,\,\, m_1g = m_1a}$
${ \Rightarrow a = \frac {(m_2 -\mu_km_1)_g}{m_1 + m_2}}$
For the block of mass $'m_2'$
$m_2g -T = m_2\bigg[\frac {m-2 - \mu_k}{m_1 + m_2}\bigg]g$
$m_2g - T = m_2{\bigg[\frac{m_2 -\mu_km-1}{m_1+m_2}\bigg]} m_2g =m_2g{\bigg[\frac{m_2 -\mu_km-1}{m_1+M_2}\bigg]}$
$\Rightarrow {T = \frac{m_1m_2(1+\mu_k)_g}{m_1 +m_2}}$