A block A of mass m 1 rests on a horizontal table. A light string connected to it passes over a frictionless…

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

Solution

From the figure,
\(\begin{aligned}
& m_2 g-T=m_2 a \ldots \text { (i) } \\
& T-\mu_k m_1 g=m_1 a \ldots \text { (ii) } \\
& \text {multiply eqn (i) with } \mathrm{m}_1 \text { and eqn (ii) with } \mathrm{m}_2 \\
& m_1 m_2 g-T m_1=m_1 m_2 a \ldots \text { (iii) } \\
& T m_2-\mu_k m_1 m_2 g=m_1 m_2 a \ldots \text { (iv) } \\
& \text{From eqn (iii) and (iv), we get,} \\
& m_1 m_2 g-T m_1=T m_2-\mu_k m_1 m_2 g \\
& m_1 m_2 g\left(1+\mu_k\right)=T\left(m_1+m_2\right) \\
& \therefore T=\frac{m_1 m_2 g\left(1+\mu_k\right)}{m_1+m_2}
\end{aligned}\)

Asked in: NEET 2015 (Phase 1)

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