Laws of Motion - Result Question 40

40. A block $A$ of mass $m_1$ 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 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

[2015]

(a) $\frac{(m_2-\mu km_1) g}{(m_1+m_2)}$

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

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

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

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Answer:

Correct Answer: 40. (b)

Solution:

  1. (b) For the motion of both the blocks

$m_1 a=T-\mu_k m_1 g$

$m_2 g-T=m_2 a$

$a=\frac{m_2 g-\mu_k m_1 g}{m_1+m_2}$

$m_2 g-T=(m_2)(\frac{m_2 g-\mu_k m_1 g}{m_1+m_2})$

solving we get tension in the string

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