Question

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Q12. A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is ${\mathrm{\mu }}_{\mathrm{k}}$. When the block A is sliding on the table, the tension in the string is :-

$$\begin{gathered} \left(1\right)\frac{\left({\mathrm{m}}_2-{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\right)\mathrm{g}}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}\left(2\right)\frac{{\displaystyle {\mathrm{m}}_1{\mathrm{m}}_2\left(1+{\mathrm{\mu }}_{\mathrm{k}}\right)\mathrm{g}}}{{\displaystyle \left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}} \\ \left(3\right)\frac{{\displaystyle {\mathrm{m}}_1{\mathrm{m}}_2\left(1-{\mathrm{\mu }}_{\mathrm{k}}\right)\mathrm{g}}}{{\displaystyle \left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}}\left(4\right)\frac{{\displaystyle \left({\mathrm{m}}_2+{\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\right)\mathrm{g}}}{{\displaystyle \left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}} \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} Fromthefreebodydiagramof{m}_2 \\ {m}_{2}g-T=m_{2}a...1 \\ Fromthefreebodydiagramof{m}_1 \\ T-\mu N=m_{1}a \\ T-\mu m_{1}g=m_{1}a...2 \\ addingeqn1and2 \\ {m}_{2}g-T+T-\mu m_{1}g=m_{2}a+m_{1}a \\ {m}_{2}g-\mu m_{1}g=m_{2}a+m_{1}a \\ a=\frac{(m_2-\mu m_1)g}{m_2+m_1} \\ {m}_{2}g-T=m_{2}a \\ {m}_{2}g-m_{2}a=T \\ T=m_{2}g-m_2\times \frac{(m_2-\mu m_1)g}{m_2+m_1} \\ T=m_{2}g(1-\frac{(m_2-\mu m_1)}{m_2+m_1} \\ T=m_{2}g(\frac{m_2+m_1-m_2+\mu m_1}{m_1+m_2}) \\ T=m_2{m}_{1}g\frac{(1+\mu )}{(m_1+m_2)} \\ Regards \end{gathered}$$