# Mass on a spring

1. Feb 10, 2008

### blalien

[SOLVED] Mass on a spring

1. The problem statement, all variables and given/known data
This problem is from Gregory’s Classical Mechanics

A light spring of natural length a is placed on a horizontal floor in the upright position. When a block of mass M is resting in equilibrium on top of the spring, the compression of the spring is a/15. The block is now lifted to a height 3a/2 above the floor and released from rest. Find the compression of the spring when the block first comes to rest.

2. Relevant equations
Mgh = the potential energy of the mass at a height h above the floor
1/2kx^2 = the potential energy of the spring, since we are assuming that the spring obeys Hooke's Law

3. The attempt at a solution

It doesn’t help that the book never defines “natural length” or “compression.” But anyway, this is my logic:

Assume that, on the horizontal floor, V = 0. So, you have two systems and two equations:

Mass starts at rest on spring -> Mass compresses spring by a/15
Mga = Mg(14a/15)+1/2k(a/15)^2

Mass starts at 3a/2 above floor -> Mass compresses spring by x
Mg(3a/2) = Mg(a-x)+1/2kx^2

Then you solve for k and x. Unfortunately, this yields the wrong answer, which is apparently x = a/3. I don’t want to be told exactly how to do the problem. But, could you please just point to the faulty part in my logic? I would really appreciate the help.

Thanks!

2. Feb 10, 2008

### Staff: Mentor

Natural length = uncompressed/unstretched length; compression = distance the spring is compressed (away from its uncompressed position).

Use the given information to determine the spring constant.

Measure spring energy from the unstretched position. When the spring is pulled up to 3a/2, what is its spring potential energy? (Not zero!)

3. Feb 10, 2008

### blalien

Oh, I see. So, the mass and the spring are lifted up to the height 3a/2. I was assuming that the mass alone was lifted up then dropped, but I guess your way makes more sense. Then the second equation is actually:

Mg(3a/2)+1/2k(a/2)^2 = Mg(a-x)+1/2kx^2

That still doesn't yield the right answer, though.

So, when there is no mass on the spring, the spring's equilibrium position is clearly at height h=a. Therefore, the spring energy is 1/2k(h-a)^2, where h is the height of the top of the spring.

When the mass is on the spring, does this remain true? Is the equilibrium position still at h=a, or is it at h=14a/15? Or do we have to redefine the spring energy of the system?

4. Feb 10, 2008

### Staff: Mentor

Now that I reread the problem, I think your interpretation makes more sense!

First find the spring constant using the initial information as I pointed out earlier, then apply the second equation you had before:
This should work. (Sorry about that!)

5. Feb 10, 2008

### blalien

Well yeah, therein lies the problem.

So we start with Mga = Mg(14a/15)+1/2k(a/15)^2. Solving for k gives k = 30Mg/a

Then we have Mg(3a/2) = Mg(a-x)+1/2kx^2
Mg(3a/2) = Mg(a-x)+1/2(30Mg/a)x^2

Then, solving for x gives x = a(sqrt(31)+1)/30. The solution given in the back of the book is, "Spring is compressed by a/3." Now, the book's solutions have been wrong before, but x = a/3 does sound more intuitive. So one of those equations is incorrect. The question is finding the error.

6. Feb 10, 2008

### Staff: Mentor

To find the spring constant, use a force equation, not an energy one. You are told that a certain force (Mg) compresses the spring by a certain amount (a/15) . Use Hooke's law to find k.

7. Feb 10, 2008

### blalien

Ah, there we go. Thanks so much!