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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

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

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

1 Answer

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$
Hence a is the correct answer.


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

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