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Spring Force

Two identical...

Question

Two identical blocks A and B, each of mass 'm' resting on a smooth horizontal surface, are inter connected by spring of stiffness 'K'. If the block B is acted on by a horizontal force 'F' and the elongation of the spring is 'e', the relative acceleration between the blocks is equal to

A

\frac{F}{2 m}

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B

\frac{F - K e}{m}

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C

\frac{F - 2 K e}{m}

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D

\frac{K e}{m}

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Solution

The correct option is D $\frac{F - 2 K e}{m}$
Block A has force only due to the spring.
$a_{1} = k e / m$
Block B has the additional force $F$ as well.
$∴ F - k e = m a_{2} \Rightarrow a_{2} = \frac{F - k e}{m}$
$∴ r e l a t i v e a c c e l e r a t i o n = a_{2} - a_{1} = \frac{F - 2 k e}{m}$

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