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 2 years ago
#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
 2 years ago
#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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shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.0Good @chand . Would you mind explaining to @goutham1995 ?

chand
 2 years ago
Best ResponseYou've already chosen the best response.1sure. 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

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.0 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.

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.0The 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. :)

goutham1995
 2 years ago
Best ResponseYou've already chosen the best response.0lol..thats what i did..but why will it be wrong? if we dont consider energy, how else can we do it?

shivam_bhalla
 2 years ago
Best ResponseYou've already chosen the best response.0@VincentLyon.Fr , can you help us here please ??

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.0Energy 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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