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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! :)

1. The problem statement, all variables and given/known data

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 = ?

2. Relevant equations

Fspring = kx

Fg = mg

EE = 1/2 kx^2

EG = mg (delta h)

3. 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!

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# Homework Help: Physics question - object released at equillibrium position of spring

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