Why Do Different Methods of Loading Affect Spring Calculations?

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confusseed
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can someone explain to me how springs work?
I already know the equation of elastic pe= 1/2kx^2 but I need more explanation!
how come when a mass is dropped from a hanging spring I can multiply the mass by 9.8m/s^2 to get the force, then divide the answer by the spring's constant (20n/cm) then divide by two and get the right answer every time but when it is gently let down, I get the right answer without dividing by two and how come these methods do not work when the spring is on the ground and the mass is being dropped from a height?
 
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Your making me confused! I think you should try expressing yourslef more clearly, and people will be better equipted to help you!

Ok, for the P.E. I do not know how in-depth you go into this stuff. PE = -INT(F), that is, the potential energy is ALWAYS the negative integral of the force. Since the force for a spring is -kx, it is apparent where the PE equation comes from. We expect it to be positive all the time, because for all displacements (stretching of the spring) it must have a positive potential, as it stands to gain kinetic energy. Also note for x=o the potential is zeo, as we expect.

I think you want to find the displacement of a mass hanging on a spring. Well, first note that there are two forces acting on the mass. Gravitational force acting down, and th spring acting upwards. Therefore:
-kx = mg

mg is the force due to gravity, and -kx the force due to the spring. We equate them because when the block comes to rest, the forces must be equal.

Solving for x:
x = -mg/k

The negative sign just tells us it is displaced downwards.
I don't know what else you want.