Spring Mechanical Energy

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Homework Statement


A 150 kg mass is attached to a vertical spring, stretching it by 15.0 cm. This same spring is fastened to the floor and compressed by 25.0 cm. A 5.00 kg mass is then placed on the spring. If the spring is released and the mass is launched straight up into the air by the spring, how high will the mass go above the spring's equilibrium position. Assume that the mass is released at the spring's equilibrium position.


Homework Equations


Initial Mechanical Energy = Final Mechanical Energy
KE(i) + Gravitational PE(i) + Spring PE(i) = KE(f) + Gravitational PE(f) + Spring PE(f)


The Attempt at a Solution



Using the first part of the experiment to find the spring constant:

m = 150 kg
xi = 0 m
xf = .15 m
g = 9.8 m/s^2

Gravitational PE (i) = Spring PE(f)

mgh = .5kx^2

[2(150 kg)(9.8 m/s^2)(.15 m)]/(.15 m)^2 = K

19600 = K

For the second part:

m = 5 kg
xi = 0 m
xf = .25 m
g = 9.8 m/s^2

Gravitational PE (i) + Spring PE (i) = Gravitational PE (f)

mgh + .5kx^2 = mgh

(5 kg)(9.8 m/s^2)(.25 m) + .5(19600)(.25)^2 = (5 kg)(9.8 m/s^2)(h final)

12.75 m = h

Does this attempt at the solution seem right?
 
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Answers and Replies

  • #2
collinsmark
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Using the first part of the experiment to find the spring constant:

m = 150 kg
xi = 0 m
xf = .15 m
g = 9.8 m/s^2

Gravitational PE (i) = Spring PE(f)

mgh = .5kx^2

[2(150 kg)(9.8 m/s^2)(.15 m)]/(.15 m)^2 = K

19600 = K
Be careful with this approach. I don't think it's right (see below). It depends on how you interpret the part of the problem statement that says, "A 150 kg mass is attached to a vertical spring, stretching it by 15.0 cm."

Your approach is correct, if you interpret the statement as follows: "A spring is hung from the ceiling, in equilibrium, and a 150 kg mass is attached to it, (such that the spring itself is still unstretched) and then the mass is released. The mass will oscillate up and down, reaching a maximum height equal to its original position, and a minimum height of 15 cm below its original position."

If the above is how you interpret the problem statement, your approach is correct. But that's not how I interpret the problem statement.

I interpret the problem statement as saying, "A spring is hung from the ceiling, and a 150 kg mass is attached to it. The mass is lowered slowly such that the spring-mass system reaches a state of static equilibrium. At this point, the spring has stretched by 15 cm."

Both lead to very different results.

If you interpret the problems statement like I do, then you can't use conservation of kinetic and potential energy to find the spring constant (some of the energy is lost as heat, when the spring is gradually lowered). But you can use F = kx to find the spring constant.

[Edit: By that I mean that in the latter interpretation, some energy is lost to some external force as the spring is gradually lowered; or the energy is lost as heat due to air resistance as the oscillations gradually fade away.]
For the second part:

m = 5 kg
xi = 0 m
xf = .25 m
g = 9.8 m/s^2

Gravitational PE (i) + Spring PE (i) = Gravitational PE (f)

mgh + .5kx^2 = mgh

(5 kg)(9.8 m/s^2)(.25 m) + .5(19600)(.25)^2 = (5 kg)(9.8 m/s^2)(h final)

12.75 m = h
Be careful regarding +/- signs. Or in other words, be careful about what things go on which side of the equation. I think there's a mistake in there somewhere.
 
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  • #3
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Ah, thank you for the reply collinsmark. I'm still pretty stuck. I set the GPE = 0 when the mass was stretched down .15 m. Is there some sort of sign error? Should it be -mgh = .5kx^2?
 
  • #4
collinsmark
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Ah, thank you for the reply collinsmark. I'm still pretty stuck. I set the GPE = 0 when the mass was stretched down .15 m. Is there some sort of sign error? Should it be -mgh = .5kx^2?
Again, it's all in the interpretation of what is meant by, "A 150 kg mass is attached to a vertical spring, stretching it by 15.0 cm." Does that mean the spring/mass system is in static equilibrium at 15 cm, or does that mean the spring/mass system is oscillating up and down with a maximum peak-to-peak amplitude of 15 cm (i.e. oscillating [via simple harmonic motion] between 0 cm and -15 cm)?

You are approaching the problem under the interpretation that the mass/spring system will oscillate with a 15 cm peak-to-peak amplitude (via simple harmonic motion). If that's how you are supposed to interpret it, than your approach is okay. [Edit: for the first part, that is.]

But I don't interpret the problem statement that way. If I had to guess I would say that the problem statement means that the system is in static equilibrium when the mass/spring is stretched by 15 cm. But you can't use conservation of energy for that (because not all forces involved will be conservative). But you can use another formula in this case to find the spring constant: Hooke's law.
F = kx.
 
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  • #5
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The approach I wanted to take on was of static equilibrium and I think I understand what you mean now. If I was to then use Hooke's law and Solve for F and x would it just be:

F = m*a
= (150 kg) (9.8 m/s^2)
= 1470 N

x = .15 m

k = (1470 N) / (.15 m)
= 9800
 
  • #6
collinsmark
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The approach I wanted to take on was of static equilibrium and I think I understand what you mean now. If I was to then use Hooke's law and Solve for F and x would it just be:

F = m*a
= (150 kg) (9.8 m/s^2)
= 1470 N

x = .15 m

k = (1470 N) / (.15 m)
= 9800
Yes, that right! :approve:

I know this whole spring stuff can be confusing, but this scenario is something worth considering. Convincing yourself of this may be worth the effort, and getting a good handle on this can help you with similar problems in the future.

If you were to hold a mass on top of an uncompressed spring and let it go, the mass would drop down some distance. But it wouldn't stop there! Instead, the spring would immediately push it back up to where it started! And the process would continue indefinitely, or until air resistance caused the oscillation to die out, or until somebody stepped in and manually stopped the system from oscillating.

A point worth noting, is that the mass would have maximum kinetic energy at the very point where static equilibrium would occur if the oscillations died away. In other words, for an ideal system, the mass would fall right past the point of equilibrium before it started coming back the other way. And for an ideal system, this point of static equilibrium is the midpoint of the oscillation. In other words, if the static equilibrium point is a distance h below the starting point (where the spring by itself is uncompressed), the mass will fall by 2h before coming back up.

We can use conservation of energy on this system if we ignore air resistance and put up with the oscillations. At the highest point in the oscillation, all the energy is in the form of gravitational energy. At the lowest point, all the energy is contained in the spring (there is no kinetic energy in either of these two points). Since the distance traversed from top to bottom is 2h, we can say,

mg(2h) = (1/2)k(2h)2

2mgh = 2kh2

mg = kh

And we get Hooke's Law! :cool:

(Again, I don't expect you to put this on your homework. It's easier to just start with Hooke's law from the beginning. The reason I explained the above was just to show that conservation of energy is really not violated, even though it might seem so if you're not careful.)
 
  • #7
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Thanks a lot for the insight! I always welcome and love explanations and I appreciate that you showed me this.

As for the part when the spring gets compressed and the mass is launched into the air, is it strange that I have GPE(i) and SPE(i) = GPE(f)?

When the spring is initially compressed, there is a spring energy that must be taken into account and because there is a change in height (0 m - .25m) that shows how much the spring was pressed down, there also has to be initial GPE. When the final GPE is in the air at max height (which is what I'm trying to solve), would I need to add .25 m to the height so that it is literally:

(5 kg)(9.8 m/s^2)(.25 m - 0 m) + .5(9800)(.25 m - 0m)^2 = (5 kg)(9.8 m/s^2)(h + .25)

Then again, you mention something about my signs being wrong and using the wrong equation on one side so I might be making the same mistake all over again.
 
  • #8
collinsmark
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(5 kg)(9.8 m/s^2)(.25 m - 0 m) + .5(9800)(.25 m - 0m)^2 = (5 kg)(9.8 m/s^2)(h + .25)
There still one or two things not quite right with the above. It has to do with the reference height for gravitational potential energy.

It turns out that you get to choose reference height for gravitational potential energy (just to be clear, the reference height is the height that the gravitational potential energy is zero -- and you get to define where that is). The reference height could be almost anything, as long as you are consistent, and don't change it. It could be the ceiling, the floor, center of the Earth (well, the center of the Earth might make things a little complicated, but you could still do it!), anything. 'Just pick a height and stick with it.

I would pick the spring's totally-uncompressed equilibrium position as the reference height. But that's just me. You can pick whatever you want.

Looking at the left side of your equation, there is a change of height 0.25 m below this reference point (as you have pointed out). But is the height above or below the reference point, and what does that say about the sign of the 0.25 m? (Is it positive or negative?)

And on the right side of the equation, you have yet another 0.25 m in your gravitational potential energy term. But don't change gravitational reference heights in the middle of an equation. Stick with a reference height that you have already chosen, or redo both sides of the equation such that there is only one, single gravitational potential reference height.
 
  • #9
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Ah, should I have taken the direction into consideration then? If so, then it would just be (-.25 m - 0m) for the initial height and for the final height, (h - .25 m), where h represents the final height without the height compression (since it is asking how high will the mass go above the spring's equilibrium position)
 
  • #10
collinsmark
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Ah, should I have taken the direction into consideration then? If so, then it would just be (-.25 m - 0m) for the initial height
Yes, that's true if you have chosen your reference height for gravitational potential energy to be the spring's uncompressed height. In other words, if you have chosen your reference height such that the gravitational potential energy is zero when the spring is in its totally-uncompressed equilibrium position, then yes. :approve:

(And by the way, you did exactly as I would do. :smile:)
and for the final height, (h - .25 m), where h represents the final height without the height compression (since it is asking how high will the mass go above the spring's equilibrium position)
You've seem to have changed the reference height for the gravitational potential energy :frown:! If your reference height is the spring's totally-uncompressed equilibrium height, then how far is h above this reference? :wink:

(And by the way, when the spring is compressed by 0.25 m it is not in its uncompressed equilibrium position. Just for clarity, I am defining the "totally-uncompressed equilibrium position" as the height of the spring all by itself without any mass or anything on it, for this particular case.)

[Edit: And I think I'm defining the equilibrium height (in this case) the same way that the problem statement is.]

[Second edit: And just in case there is still confusion about this, what I am defining as the spring's "totally-uncompressed equilibrium height" in this part, is different than the equilibrium position of the spring/mass system in the first part.]
 
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  • #11
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So then the height, h, would just be that (h + .25 m) instead, since it opposes the initial height, when the spring was compressed and the height was negative.

My problem was understanding the whole "set your own GPE = 0 and stay consistent with it" sort of deal. My professor mentioned that as well and I just never seemed to grasp that as well. So in setting the GPE(i) = 0 at the totally uncompressed equilibrium height, that would be completely acceptable so long as I take into consideration the change in height in the GPE(f), when the spring is compressed or stretched.
 
  • #12
collinsmark
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h = h. That's my point. Don't go with h = (h + .25 m), because that doesn't make any sense. Define h however your want, but then stick with it.

[Edit: If you really need to redefine it later, at least do the same thing to both sides of the equation. -- Draw a free body diagram if you need to. It might help.]
 
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  • #13
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I finally understand why (h +.25 m) is wrong. We had set the Equilibrium to be 0 when there was no compression. The change in height initially due to compression was (0-.25 m). However, in order to find the final height, it would simply be (0 + h). In setting the final height to (h + .25 m) I'm changing the equilibrium to be 0 at -.25 m, which is not consistent at all with what I initially set the equilibrium to. Thanks a bunch collinsmark, sorry you had to put up with this, haha
 

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