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 $μ_{k}$ . When the block $A$ is sliding on the table, the tension in the string is:

\frac{m_{1} m_{2} (1 + μ_{k}) g}{m_{1} + m_{2}}

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\frac{m_{1} m_{2} (1 - μ_{k}) g}{m_{1} + m_{2}}

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\frac{(m_{2} + μ_{k} m_{1}) g}{(m_{1} + m_{2})}

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\frac{(m_{2} - μ_{k} m_{1}) g}{(m_{1} + m_{2})}

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Solution

The correct option is A $\frac{m_{1} m_{2} (1 + μ_{k}) g}{m_{1} + m_{2}}$
The blocks $m_{1}$ and $m_{2}$ will move with combined acceleration a:
From F.B.D. of block $m_{1}$
$T - f_{1} = m_{1} a$ ...(i)
as the block $m_{1}$ is sliding, kinetic friction will be acting:
$T - μ_{k} N = m_{1} a$ ...(ii)
$N = m g$ ...(iii)
From F.B.D. of block $m_{2}$
$m_{2} g - T = m_{2} a$ ...(iv)
adding (ii) and (iv)
$a = \frac{m_{2} g - μ_{k} N}{m_{1} + m_{2}}$ ...(v)
$T = m_{2} g - m_{2} a = m_{2} (g - \frac{m_{2} g - μ_{k} N}{m_{1} + m_{2}})$
$= m_{2} (\frac{m_{1} g + μ_{k} m_{1} g}{m_{1} + m_{2}})$
$= m_{1} m_{2} (\frac{1 + μ_{k}}{m_{1} + m_{2}} g)$
hence correct answer is option A

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Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be: