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NEET 2015 (AIPMT, Cancelled Paper) · Friction Basics

Friction Basics question from NEET 2015 (AIPMT, Cancelled Paper)

Q26NEET 2015 · AIPMT · CANCELLED PAPERMedium

Friction Basics

A block A of mass m1 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 m2 is suspended. The coefficient of kinetic friction between the block and the table is μ\muk. When the block A is sliding on the table, the tension in the string is

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Answer: (a)

  1. Block B: m2g−T=m2am_2 g - T = m_2 a; block A: T−μkm1g=m1aT - \mu_k m_1 g = m_1 a.
  2. Adding: a=(m2−μkm1)gm1+m2a = \dfrac{(m_2 - \mu_k m_1)g}{m_1 + m_2}
  3. T=m2(g−a)=m2g m1+μkm1m1+m2T = m_2(g - a) = m_2 g\,\dfrac{m_1 + \mu_k m_1}{m_1 + m_2}
  4. So the answer is (a) T=m1m2(1+μk)gm1+m2T = \dfrac{m_1 m_2(1+\mu_k)g}{m_1+m_2}.
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