IIT JEE , AIEEE Forum - Physics , Mechanics

shamips

a mass hangs on spring of spring constant k. What will be the maximum elongation as given by mg=kx or mgx=1/2kxsq. Why two answers differ? please explain the basic concept.

shamips

Nobody is able to clarify properly. The mass is hanging just like we weigh on a spring balance.No oscillations. what will be the energy stored in spring? what will be the work done by mass and what will be the maximum elongation?

If I use mg=kx =>x=mg/k then I calculate Es=0.5kx**2 =0.5k(mg/k)**2, am I correct?

By the way I say what will be the elongation,instead of max.elongation does it mean differently ?

biki

the second one is correct

shashidhars

dear biki,
study SHM(work power energy wud also do) to clear dis doubt.
d body wud hav some KE(=work done by spring force) at x = mg/k.vich wud take it to x=2*mg/k.

ramyani

From FBD it seems mg = kx

krishna.gopal

Second one is correct. The first one assumes that maximum elongation will come when force is zero. But this is wrong. Because even when force is zero velocity is finite and more elongation is possible. The limit of elongation is given by energy conservation which is second equation

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