Simple energy question, but getting different results with different methods

[defetey]

Homework Statement

A linear elastic spring is $$0.15m$$ long. When the upper end is held in the hand and a $$0.5kg$$ mass is suspended from the lower end, its length becomes $$0.22m$$. What is the spring constant?

Homework Equations

$$E_{E}=\frac{1}{2}kx^{2}$$

$$\vec{F_{spring}}=kx$$

$$\vec{F_{G}}=mg$$

$$E_{G}=mgh$$

The Attempt at a Solution

First method:

Since the block is at rest, $$\vec{F}_{net}=0$$

$$\vec{F}_{G}-\vec{F}_{spring}=0$$

$$(0.5)(9.8)=k(0.07)$$

$$k=70N/m$$Second method:

If we set the gravitational reference to where the block stops, it will be $$0.07m$$ up originally (with no elastic energy since it is not stretching the spring yet). Then when it is at the bottom, it will be $$0$$, and so all that energy will be elastic energy.

$$E_{G}=E_{E}$$

$$(0.5)(9.8)(0.07)=\frac{1}{2}k(0.07)^{2}$$

$$k=140N/m$$

I know the first method is right, because with $$k=70N/m$$, $$\vec{F}=kx$$ will be $$=4.9N$$, which is must be to cancel out the gravitational force and have it at rest. But shouldn't the second method also yield the same answer?

 

[collinsmark]

First method:

Since the block is at rest, $$\vec{F}_{net}=0$$

$$\vec{F}_{G}-\vec{F}_{spring}=0$$

$$(0.5)(9.8)=k(0.07)$$

$$k=70N/m$$

Your first method looks good to me!

Second method:

If we set the gravitational reference to where the block stops, it will be $$0.07m$$ up originally (with no elastic energy since it is not stretching the spring yet). Then when it is at the bottom, it will be $$0$$, and so all that energy will be elastic energy.
[...]
But shouldn't the second method also yield the same answer?

No. Ask yourself what would happen if you attached a weight to a spring, attached one side of a spring to your left hand (and you hold your left hand completely still from this point forward), and held the mass in your right hand such that the spring remains completely unstretched. Then release the mass. What happens?

Does the mass fall to an immediate state of static equilibrium and stay there?

It certainly does not! Instead it falls some distance (twice as much as the distance you used above), and then rises back up to where it was initially released, and this whole process repeats. You've created simple harmonic motion. The oscillation will continue indefinitely; well, at least until friction causes the oscillations to fade away, or until some other external force comes into play and stops the oscillations (and either situation involves work and energy).

 

[berkeman]

Thanks collinsmark, this one had me confused!