Gregory’s Classical Mechanics Mass on a spring

[blalien]

[SOLVED] Mass on a spring

Homework Statement

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.

Homework 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

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!

 

[Doc Al]

It doesn’t help that the book never defines “natural length” or “compression.”

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

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

Use the given information to determine the spring constant.

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

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

 

[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?

 

[Doc Al]

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.

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:

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

This should work. (Sorry about that!)

 

[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.

 

[Doc Al]

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

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.

 

[blalien]

Ah, there we go. Thanks so much!