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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Laws of Motion

NEET 2015 · Physics · Laws of Motion

A block A of mass m _ 1 rests on a horizontal table. A light string...

A block A\text{A} of mass m1{m}_{1} 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\text{B} of mass m2{m}_{2} is suspended. The coefficient of kinetic friction between the block and the table is μk{\mu}_{k}. When the block B\text{B} is sliding on the table, the tension in string is:

Options

  1. A\dfrac{\left({m}_{2} + {\mu}_{k}{m}_{1}\right)g}{\left({m}_{1} + {m}_{2}\right)}
  2. B\dfrac{\left({m}_{2} - {\mu}_{k}{m}_{1}\right)g}{\left({m}_{1} + {m}_{2}\right)}
  3. C\dfrac{{m}_{1}{m}_{2}\left(1 + {\mu}_{k}\right)g}{\left({m}_{1} + {m}_{2}\right)}
  4. D\dfrac{{m}_{1}{m}_{2}\left(1 - {\mu}_{k}\right)g}{\left({m}_{1} + {m}_{2}\right)}

Correct answer

C. \dfrac{{m}_{1}{m}_{2}\left(1 + {\mu}_{k}\right)g}{\left({m}_{1} + {m}_{2}\right)}

Step-by-step solution

From the figure, m2g−T=m2a… (i) T−μkm1g=m1a… (ii) multiply eqn (i) with  m1 and eqn (ii) with m2m1m2g−Tm1=m1m2a… (iii) Tm2−μkm1m2g=m1m2a… (iv) From eqn (iii) and (iv), we get,m1m2g−Tm1=Tm2−μkm1m2gm1m2g(1+μk)=T(m1+m2)∴T=m1m2g(1+μk)m1+m2\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)\text{ 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})$
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