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A block of mass M is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be
A
Mg/2k
B
Mg/k
C
2Mg/k
D
4Mg/k
Detailed Solution
The block is released from rest and comes momentarily to rest again at the maximum extension x, so its kinetic energy is zero at both positions.
Loss of gravitational potential energy = gain in elastic potential energy of the spring.
$Mgx = \frac{1}{2}kx^2$
$x = \frac{2Mg}{k}$
(This is twice the equilibrium extension Mg/k, because the block overshoots the equilibrium position and oscillates about it.)
Loss of gravitational potential energy = gain in elastic potential energy of the spring.
$Mgx = \frac{1}{2}kx^2$
$x = \frac{2Mg}{k}$
(This is twice the equilibrium extension Mg/k, because the block overshoots the equilibrium position and oscillates about it.)