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Byju's Answer

Standard XII

Physics

Second Law of Motion

Three blocks ...

Question

Three blocks $m_{1}, m_{2}$ and $m_{3}$ of masses $8 kg, 3 kg$ and $1 kg$ are placed in contact on a smooth surface. Forces $F_{1} = 140 N$ and $F_{2} = 20 N$ are acting on blocks $m_{1}$ and $m_{3}$ respectively as shown. The reaction between blocks $m_{2}$ and $m_{3}$ is

2.5 N

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7.5 N

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22.5 N

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30 N

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Solution

The correct option is D $30 N$
Considering blocks as single mass
$\Rightarrow 140 - 20 = 12 a$
$a = 10 {m/s}^{2}$ which is towards right direction
Now FBD of $^{'} m_{3}^{'}$
Let $T$ be the reaction force on the block .
Acceleration of $1 kg$ block will also be towards right
$T - 20 = 1 (10)$
$T = 30 N$

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Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. A force of $40 N$ pulls the system. Then calculate the value of $T$ (in $N$ ) if $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ .