MHB Physics Spring Compression Equation

[Dethrone]

Suppose a mass, $m$, is dropped from a height $L$ relative to the top of a uncompressed spring, which is on the surface of the floor. Calculate the compression, $\Delta x$ of the spring, given the mass.

Okay, so I completely made up this question, but here is my confusion:

Which equation is appropriate to solve this?

$$mgL=\frac{1}{2}k(\Delta x)^2-mg \Delta x$$

OR

$$mgL=\frac{1}{2}k\Delta x^2$$

 

[MarkFL]

Do you have any information regarding the stiffness of the spring? Are you considering that the weight of the mass is still working on the spring as it's being compressed?

 

[Dethrone]

All needed values are given, including $k$. I'm not sure whether or not the mass is still working on it...the question (from my test) was simply the effect of the mass dropping on an uncompressed spring, and finding its compression/deflection, or whatever you want to call it :D

 

[MarkFL]

All needed values are given, including $k$. I'm not sure whether or not the mass is still working on it...the question (from my test) was simply the effect of the mass dropping on an uncompressed spring, and finding its compression/deflection, or whatever you want to call it :D

Well, obviously energy considerations are the way to go here, as you are doing. The initial gravitational potential energy, will be converted into the potential energy of the spring.

I would say your first equation is correct, since it accounts for the additional gravitational energy lost during compression.

 

[topsquark]

Which equation is appropriate to solve this?

$$mgL=\frac{1}{2}k(\Delta x)^2-mg \Delta x$$

Looks good to me. Typically we set an origin for the spring potential energy to be at the top of the uncompressed spring as the spring has no PE there and choose a positive direction in the direction of the spring compression. So you can actually get rid of the deltas there.

-Dan