Mass Dropped on Vertical Spring

[schaefera]

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?

 

[omoplata]

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).

 

[schaefera]

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?

 

[omoplata]

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.

 

[rwisz]

Your approach to part (a) is correct. Using the conservation of energy equation, you can solve for the compression distance, x. Since the mass is dropped from a height above the spring, the initial gravitational potential energy is equal to mgh, where h is the initial height. The final gravitational potential energy is equal to mg(L-x), where L-x is the new equilibrium position of the mass on the compressed spring. The spring potential energy is equal to 0.5kx^2, where k is the spring constant.

To solve for x, you will need to use the quadratic formula, as you have correctly stated. You will get two solutions, but only one of them will be physically meaningful. The negative solution will result in a negative compression distance, which is not possible. So, you should use the positive solution.

For part (b), you are correct in finding the new equilibrium position of the mass, x=mg/k. This is the point at which the spring force and the gravitational force are balanced, and the mass will come to rest. To find the maximum speed, you can use the conservation of energy equation again, with the initial potential energy being mgh and the final kinetic energy being 0.5mv^2. Solving for v will give you the maximum speed of the mass.