Vertical spring & maximum length

[stunner5000pt]

Homework Statement

A spring with spring constant k has an unstretched length of L. A mass m is hung vertically from the spring. The mass is pulled down a distance x and given an initial velocity upward of v.
Determine the maximum length that the spring reaches in terms of the above variables.
Assume the spring has mass zero.

Relevant Equations

Conservation of energy
$$\Delta E_{k} + \Delta E_{g} + \Delta E_{s} = 0$$

Included a diagram as well... forgive me... I cannot seem to 'uninvert' the attached picture

final velocity is zero
if we set the lowest point that the mass reaches as zero, then the final height is zero
let H be the 'extra' length that the spring reaches over and above the initial stretch

$$\Delta E_{k} + \Delta E_{g} + \Delta E_{s} = 0$$
$$\frac{1}{2} m (v_{2}^2 - v_{1}^2 ) + mg (h_{2} - h_{1}) + \frac{1}{2} k ( (x + h_{f} )^2 - x^2) = 0$$
using the things that are zero above

$$-\frac{1}{2} v^2 - mg h_{f} + \frac{1}{2} k ( (x + h_{f} )^2 - x^2) = 0$$

$$-\frac{1}{2} v^2 - mg h_{f} + \frac{1}{2} k ( 2x h_{f} + h_{f}^2 ) = 0$$

$$\frac{k}{2} h_{f}^2 + h_{x} ( -mg + kx) - \frac{1}{2} v^2 = 0$$

ok at this point it's getting a bit messy as this requires to go into a quadratic formula but, is this correct so far?

Thank you for help in advance!

 

[PeroK]

That looks good. I agree that the final expression will be messy.

Don't forget that the spring extends when first loaded with the mass.

 

[PeroK]

$$\frac{k}{2} h_{f}^2 + h_{x} ( -mg + kx) - \frac{1}{2} v^2 = 0$$

You seem to have lost an $m$ from your KE term!

 

[stunner5000pt]

That looks good. I agree that the final expression will be messy.

Don't forget that the spring extends when first loaded with the mass.

You seem to have lost an $m$ from your KE term!

Whoops! thanks for catching this! Something seemed off initially when i wrote it down too
Just want to clarify.. did I not include the spring extending the mass when it first loaded?

 

[PeroK]

Whoops! thanks for catching this! Something seemed off initially when i wrote it down too
Just want to clarify.. did I not include the spring extending the mass when it first loaded?

The question is slightly ambiguous. The spring has natural length $L$. If we hang the mass from the spring, then it will stretch to $L + x_0$, where $kx_0 = mg$, and that will be the equilibrium point.

When it says the mass is "pulled down a distance $x$", I would interpret that as a further extension to $L + x_0 + x$. But, it could mean that $x$ is intended to be what I've called $x_0$. Or, it could be that the extension is now simply $L + x$, where $x > x_0$.

In any case, I don't like calling an initial extension something like $x$, as we'd want to use $x$ as the variable in our equations. This seems to be something some problems setters do. It annoys me because it forces us to use a new variable.

Finally, it seems odd to pull something down and give it an upward velocity. Why not a downward velocity? That seems more natural.

It feels to me like one of these problems where someone who doesn't really understand physics just cobbles together some random ideas. Where did it come from?

 

[haruspex]

The question is slightly ambiguous.

I would agree if it said the spring is pulled down by x, but it says the mass is pulled down. To me, that means this is in addition to the extension caused by the weight of the mass.

 

[Mister T]

Don't forget that the spring extends when first loaded with the mass.

So it's not clear to me from the problem statement whether or not this extension is included when "the mass is pulled down a distance x".