Physics question - object released at equillibrium position of spring

[shadowysun]

Hello everybody,

I am having a lot of trouble with the following question and would appreciate it if someone showed me the correct way to do it. Thanks in advance for your help! :)

Homework Statement

A 0.20 kg ball attached to a vertical spring (k = 28 N/M) is released from rest from the unstretched equilibrium position of the spring. Determine how far the ball falls, under negligible air resitance, before being brought to a momentary stop by the spring.

So,
m = 0.20 kg
k = 28 N/M
x = ?

Homework Equations

Fspring = kx
Fg = mg
EE = 1/2 kx^2
EG = mg (delta h)

The Attempt at a Solution

My friend and I both attempted to solve this problem. We used different methods and obtained different answers, but neither of us know which answer is right and why the other is wrong.

Here is the 1st way we did it:
When the ball is brought to a momentary stop, Fnet = 0, so
Fg = Fspring
mg = kx
x = mg/k

And here is the 2nd way:
Since energy is conserved,
Ebefore = Eafter
EG lost = EE gained
mg(delta h) = 1/2 kx^2

delta h = x (since the ball falls the same distance that the spring stretches), therefore
mgx = 1/2 kx^2
x = 2mg/k

But this is twice as large as the solution obtained using the first method. Am I overlooking something in one of the methods? Which method is correct?

Again, thanks in advance!

 

[Doc Al]

Here is the 1st way we did it:
When the ball is brought to a momentary stop, Fnet = 0, so

Fnet = 0 means that the acceleration is zero, not the speed.

 

[rl.bhat]

x = mg/k is true only during stretched equilibrium position of the spring. Therefore 2nd method is correct.

 

[shadowysun]

Omg. I can't believe I made such a dumb mistake >_<. Thank you so much for your help!