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Three blocks of masses $m_{1} = 1 k g, m_{2} = 1.5 k g$ and $m_{3} = 2 k g$ are in contact with each other on a frictionless surface as shown in fig. with an acceleration of $4 m / s^{2}$
Then find the resultant force on each block.

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$4 N, 6 N, 8 N$

$4 N, 6 N, 10 N$

$5 N, 6 N, 8 N$

$4 N, 4.5 N, 6 N$

answer is A.

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Detailed Solution

For $m_{1}$

$F - N_{1} = m_{1} a = 1 \times 4 \Rightarrow F - N_{1} = 4 N$

For $m_{2}$

$N_{1} - N_{2} = m_{2} a = 1.5 \times 4 = 6 \Rightarrow N_{1} - N_{2} = 6 N$

For $m_{3}$

$N_{2} = m_{3} a = 2 \times 4 \Rightarrow N_{2} = 8 N$

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