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Calculating elongation of the spring Consider a spring of spring constant 100 N/m.. it is suspended vertically and a block of 1kg is attached to it.. calculate the elongation of the spring from the equilibrium position.

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I did it in two methods and i get different answers.. a) using net force at equilibrum is zero at the final position the net force is zero. and there are two forces acting hence -kx + mg = 0 so x =mg/k ----- (1) b) Considering energy conservation Total Energy before = Total energy after T.E before = PEg = mgh (Say h is the height above the ground just when block released) T.E after = PEg + PEspring = mg (h-x) + 1/2 Kx^2 mgh = mgh - mgx +!/2 kx^2 mgx = 1/2kx^2 mg = 1/2kx x = 2mg/k see i get double the answer :(!!
In b), you were finding x at the lowest point if the spring is released from some height. In a) x is elongation when the block reach equilibrium, not oscillating anymore.
Ohhhhhh.. stupid .. stupid I AM!!!!!.. HTANK YOU!!!!

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No probs :)
is it not possible to find the elongation a) part which you mentioned using energy conservation?
maybe i have to use the k.E also.. right??
correct.. i have to consider KE .. cause it ll have some K.E at that point!!!!!!
Hmm, I don't know if it's possible. Because if you drop the block from some height, it'll stop only if the energy is dissipated by friction. With the friction into account, conservation of mechanical energy couldn't be applied.
yea thanks .. !!

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