# Mass Dropped on Vertical Spring

## Homework Statement

A mass, m, is dropped from a height, h, on an initially uncompressed spring of length L.

(a) Determine the amount by which the spring is compressed before the mass comes to rest.
(b) Determine the mass's maximum speed.

## Homework Equations

U(e)=.5k*x^2, U(g)=mgh, KE=.5mv^2

I think I am supposed to do this all symbolically, since I don't have k.

## The Attempt at a Solution

For part (a):
So, take the bottom of compression to be the 0 for gravitational potential energy. Then, we can say that the mass is dropped from a distance d above the top of the unstretched spring, where d=h-L. If the spring compresses by a distance, x, then let us call L-x the 0 for gravitational potential energy. I end up with this conservation of energy equation:
E0=Ef... so: mg(h)= .5k*x^2+ mg(L-x)
This simplifies to: 0=.5k*x^2-mgx-mg(L-h)

I can use the quadratic equation to solve this, but is that too complicated for this type of question? Which solution to that equation would I use, the + or the - section?

For part (b), I believe I find the new equilibrium position of the mass (x=mg/k) and then plug that into the conservation of energy as the x, solving for v at that point. Correct?

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Yes, solving the quadratic equation would be the correct way to solve this. As for whether it's too complicated or not, if you were taught to solve quadratic equations, I guess it's reasonable to expect you to use that knowledge in solving problems.

It seem to me that since you've taken L-x as the length of the spring after compression, x would be positive.

I agree with your method of solving part (b).

Ok... we weren't explicitly taught to solve the quadratic, but I see no way around it. Is this a standard sort of problem, which would then include that formula?

And I was trying to figure out whether the choice of x as positive means I choose the positive root (well, not the positive root but the one that involves addition).

And for (b), I just realize that I actually need to find all of that in terms of a 0 for gravitational PE. Should I set that 0 at the new equilibrium point?

Is this a standard sort of problem, which would then include that formula?
That I don't know about, sorry. But it seems like a perfectly valid problem to me, to which a solution can be found.

And I was trying to figure out whether the choice of x as positive means I choose the positive root (well, not the positive root but the one that involves addition).
I tried solving the quadratic equation. One of the two solutions for x is negative, provided h>L. And btw, there is a small algebraic mistake in your final expression for the quadratic equation.

And for (b), I just realize that I actually need to find all of that in terms of a 0 for gravitational PE. Should I set that 0 at the new equilibrium point?
It doesn't matter where you choose to have zero gravitational PE. As long as you apply energy conservation correctly, you will get the solution.