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Tension in a String

A block A of ...

Question

A block A of mass m₁ rests on a horizontal table. A lights 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 . When the block A is sliding on the table, the tension in the string is :

A

m₁m₂(1 - μ_k)g/(m₁ + m₂)

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B

(m₂ + μ_km₁)g/(m₁ + m₂)

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C

(m₂ - μ_km₁)g/(m₁ + m₂)

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D

m₁m₂(1 + μ_k)g/(m₁ + m₂)

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Solution

The correct option is D
m₁m₂(1 + μ_k)g/(m₁ + m₂)

See figure alongside

Let T be the tension in the string.

Let a be the acceleration of the combination. We have,

$m_{2} g - T = m_{2} a$ …(1)

for block B.

And

$T - μ_{k} m_{1} g = m_{1} a$ …(2)

for block A.

Adding equation (1) and (2) we get,

$(m_{2} - μ_{k} m_{1}) g = (m_{1} + m_{2}) a$
$∴ a = \frac{(m_{2} - μ_{k} m_{1}) g}{(m_{1} + m_{2})}$ …(3)

From equation (2) and (3) we get,

$T = μ_{k} m_{1} g + m_{1} a$

$= μ_{k} m_{1} g + m_{1} g [\frac{m_{2} - μ_{k} m_{1}}{(m_{1} + m_{2})}] = m_{1} g [μ k \frac{(m_{2} - μ_{k} m_{1})}{(m_{1} + m_{2})}] = m_{1} g [\frac{μ_{k} m_{1} + μ_{k} m_{2} + m_{2} - μ_{1} m_{1}}{(m_{1} + m_{2})}] = m_{1} g [\frac{m_{2} (1 + μ_{k})}{(m_{1} + m_{2})}] = \frac{m_{1} m_{2} (1 + μ_{k}) g}{(m_{1} + m_{2})}$

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