What is the maximum distance a block will fall on a vertically hanging spring?

[shawli]

Homework Statement

A spring with a spring constant of 450 N/m hangs vertically. You attach a 2.2 kg block to it and allow the mass to fall. What is the maximum distance the block will fall before it begins moving upward?

Homework Equations

I'm not exactly sure how the conservation of energy equation would look for this. So, when the mass pulls the spring down to its maximum displacement, there is only elastic potential energy in the isolated spring system. Now I'm not sure what other point to use this information in reference to.

If it's the equilibrium position that I'm also supposed to look at, then I'm not sure what forms of energy are present at that point.

The Attempt at a Solution

 

[shawli]

Wait nevermind ... I googled it. The equilibrium position no longer equals x = 0 for a vertical spring, and the formula becomes h = mg/k.
Whoops lol.

 

[pgardn]

Wait nevermind ... I googled it. The equilibrium position no longer equals x = 0 for a vertical spring, and the formula becomes h = mg/k.
Whoops lol.

Does the problem make sense though?

 

[shawli]

Erm, from what I understand... A mass of 2.2kg is attached to a spring while it is at the equilibrium position. Gravity makes the spring stretch downwards until it reaches a maximum point where there is only elastic potential energy, and then it starts to move back up again.

I'm still a bit confused about what energy is acting where in terms of the "isolated system".

And this formula, h = mg/k, actually isn't giving me the correct answer. ...So I guess it doesn't make sense lol. :/

 

[shawli]

Okay I think I got it!

At the equilibrium position, only gravitational potential energy is present since the mass-spring system has not moved yet at that instant and since the spring is vertical.

The mass pulls the spring down to its maximum displacement, and at that point only elastic potential energy is present.

So:

Eg = Ee
m * g * h = 0.5 * k * x^2

At the equilibrium position, the height of the mass should be equal to the maximum displacement after the mass falls. So, x = h:

m * g * x = 0.5 * k * x^2
0 = 0.5*450*x^2 - 2.2*9.8*x
x= 0 and 0.095m

Woot right answer :). Is my reasoning all correct?

 

[pgardn]

Maybe this will help with the concepts.

Woot? Am I helping an Aggie?

\m/ that's a hookem horns sign... :)

Your reasoning using energy looks correct without doing the math myself.