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#Simple yet not so simple qn 1) an ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to it's lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the string is a)4Mg/k b)2Mg/k c)Mg/k d)Mg/2k

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the answer is 2mg/k
Good @chand . Would you mind explaining to @goutham1995 ?

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sure. when a mass attached to a spring is suddenly released, energy is conserved. in this case, spring potential energy is equal to the potential energy of spring. 1/2 kx^2 = mgx and thus x = 2mg/k
- System is conservative (ME is conserved). - In both situation, mass is at rest, hence KE = 0 and ΔKE = 0 - Gravitational PE is converted in Elastic PE hence the equation derived by chand.
The common mistake which people make in such question is F = mg= kx So x = mg/k This is the mistake I wanted to point out. :)
lol..thats what i did..but why will it be wrong? if we dont consider energy, how else can we do it?
@Vincent-Lyon.Fr , can you help us here please ??
Energy makes it much easier, but you can also solve it 'the hard way' using SHM equations. I will post this solution, but have no time right now.

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