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Second Law of Motion

Two blocks of...

Question

Two blocks of masses $3 k g$ and $2 k g$ are placed in contact with each other on a frictionless table. Find the force on the common cross-sectional area of contact if a force of $5 N$ is applied on (i)bigger block (ii)Smaller block

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Solution

Given
$F = 5 N$
$m_{1} = 3 k g$
$m_{2} = 2 k g$
$F = m a$
$F = (m_{1} + m_{2}) a$
$5 = (3 + 2) a$
$\frac{5}{(3 + 2)} = a$
$a = 1 m s^{- 2}$
For Bigger Block,
$F = m a$
$F = 3 \times 1$
$F = 3 N$
For Smaller Block,
$F = m a$
$F = 2 \times 1$
$F = 2 N$

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