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# In the arrangement shown mass m is released from rest with the spring unstretched . The force constant of spring is K. The maximum extension in the spring

$(a)\;x_m=\frac{2mg}{k} \quad (b)\;x_m=\frac{mg}{k} \quad (c)\;x_m=\frac{mg}{2k}\quad (d)\;x_m=\frac{k}{2 mg}$

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## 1 Answer

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When the spring extends to maximum extension $x_0$
The potential energy of spring= gravitational potential of block
$mgx_m=\large\frac{1}{2}$$k x_m^2$
$x_m=\large\frac{2mg}{k}$
Hence a is the correct answer.

answered Jul 18, 2013 by
edited Feb 10, 2014 by meena.p

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