Spring PE

[Pisiform]

The springs have a spring constant k and a natural length L0. The material used to make the springs has a Young's modulus of 2 x 1011 Pa. In the first experiment a mass m is suspended from a spring. The mass stretches the spring to a new length L, called the equilibrium length.

The mass in the first experiment is pulled down a distance A from its equilibrium position and then released from rest. The mass will then oscillate with simple harmonic motion. As the mass moves up and down, energy is dissipated due to factors such as air resistance and internal heating of the spring. The mass will no longer oscillate when the total energy dissipated equals:

A - kL^2/2
B - kA^2/2
C - k(L + A)^2/2
D - kL0^2/2

The answer is of course B which I found out by process of elimination. To my understanding, none of the answer choices were the correct answer since the spring is vertical and so its total PE would have been:

mg = k(L-Lo)
L-Lo = mg/k

PE = 0.5 k [mg/k + A]^2

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[Gengar]

Why would you use L-L0+A for the displacement? It's oscillating from L to a distance A back to L and then to -A. It isn't oscillating back to L0.

 

[Pisiform]

Why would you use L-L0+A for the displacement? It's oscillating from L to a distance A back to L and then to -A. It isn't oscillating back to L0.

I know, but that would not correspond to its total energy. If this was a horizontal spring I would have agreed with you. F = kx or in 0.5kx^2 , x is the displacement of the mass from the unstrectched length L0 from the sprinh if we need total energy

 

[milski]

In the first experiment PEf = 1/2 * k * (L-L0)^2
After stretching PEi = 1/2 * k * (L-L0+A)^2
PEi-PEf = k/2 * ((L-L0+A)^2-(L-L0)^2) which does not seem to be any of the answers. Strange...

Edit: Disregard this post - it does not account for the change in potential energy of the mass due to lowering it.

 

[Ssina]

the first mass shouldn't matter, it is like making the spring out of a heavier metal to it's equilibrium is stretched further.... so only distance A matter.

PE very initial is 1/2 K (L-L0)^2... because it was stretched
and at then end, when it loses the extra PE that was applied to stretch it to A (which from the original point is L0+L+A) is still the same... so it won't lose the PE gained by the mass, it will only lose the energy applied to stretch it

 

[Gengar]

In the first experiment PEf = 1/2 * k * (L-L0)^2
After stretching PEi = 1/2 * k * (L-L0+A)^2
PEi-PEf = k/2 * ((L-L0+A)^2-(L-L0)^2) which does not seem to be any of the answers. Strange...

Maybe I'm just stupid but I don't see how this is right. I was under the impression that the x in .5kx^2 is always the displacement from the system's resting position, which in this case, is A. How does the fact that it's a vertical spring change that?

 

[milski]

Maybe I'm just stupid but I don't see how this is right. I was under the impression that the x in .5kx^2 is always the displacement from the system's resting position, which in this case, is A. How does the fact that it's a vertical spring change that?

But I don't think that the L position is really a resting position - it's just an equilibrium positions when some force applied to it.

In other words, you're trying to say that extending the spring by A will change its energy by a constant amount both when you extend it from L0 to L0+A and from L to L+A.

On the other hand, my math does not seem to account for the decrease of potential energy from lowering the mass itself. So I'll have to redo that.

 

[milski]

Ok, that works out. Let dL = L-L0, PEi is the potential energy with the spring stretched to dL+A and PEf is the spring at rest with the mass hanging from it.
PEf = k/2 * dL^2
PEi = k/2 * (dL + A)^2 - mgA (we lose some potential energy by lowering the mass)
From mg = k * dL (gravity acting on the mass is same size opposite direction as the spring force on it) we can get:
PEi = k/2 * (dL + A)^2 - mgA = k/2 * (dL + A)^2 - k * dL * A =
= k(dL^2/2 + dL * A + A^2/2 - dL * A) = k/2 * (L^2-A^2)
The we have PEf - PEi = k/2 * dL^2 - k/2 * (L^2 - A^2) = k/2 * A^2 which is answer B.

That was a nice question! We also proved that we can treat a spring with a mass hanging from it the same as a relaxed spring with a different neutral position and the same k coefficient.

 

[BerkReviewTeach]

I know, but that would not correspond to its total energy. If this was a horizontal spring I would have agreed with you. F = kx or in 0.5kx^2 , x is the displacement of the mass from the unstrectched length L0 from the sprinh if we need total energy

I agree with Pisiform here. The question was oversimplified when they assumed horizontal rules for a vertical spring.

Vertical springs are annoying because there's change in mgh to consider. They have a greater resting displacement at equilibrium than if the spring were horizontal, so the stretching and compressing of the spring is asymmetric. Whether it's horizontal or vertical, the equilibrium position is still defined as having 0 potential energy and we assume that when it's at equilibrium the system has no spring potential energy. But the displacement is different when it's up versus down.

When we stretch a spring from equilibrium, we are giving the system PE. In the case of a vertical spring, it is harder to pull it down from equilibrium (in terms of spring resistance) than to push it up (because it's already been stretch a little from what it wants to be--i.e., what it would be if it were horizontal), but this asymmetry is offset by the fact that pushing it up means we are working against gravity and that takes energy too. In an ideal system, pushing the mass up by 1 cm requires the same amount of energy as pulling it down by 1 cm. Up means we do a little work against the spring and a little work against gravity. Down means we do a lot of work against the spring, but we get some energy back from gravity. Again, we assume the two balance each other out in an ideal system to make our calculation easier.

So I agree with Pisiform that choice B is the best answer to choose on this question, but it's not a correct answer. The potential energy of the system at the start is 0.5ka^2 + mg(delta)h, which means that's how much energy must be lost during the dampening.

 

[milski]

I agree with Pisiform here. The question was oversimplified when they assumed horizontal rules for a vertical spring.

Vertical springs are annoying because there's change in mgh to consider. They have a greater resting displacement at equilibrium than if the spring were horizontal, so the stretching and compressing of the spring is asymmetric. Whether it's horizontal or vertical, the equilibrium position is still defined as having 0 potential energy and we assume that when it's at equilibrium the system has no spring potential energy. But the displacement is different when it's up versus down.

When we stretch a spring from equilibrium, we are giving the system PE. In the case of a vertical spring, it is harder to pull it down from equilibrium (in terms of spring resistance) than to push it up (because it's already been stretch a little from what it wants to be--i.e., what it would be if it were horizontal), but this asymmetry is offset by the fact that pushing it up means we are working against gravity and that takes energy too. In an ideal system, pushing the mass up by 1 cm requires the same amount of energy as pulling it down by 1 cm. Up means we do a little work against the spring and a little work against gravity. Down means we do a lot of work against the spring, but we get some energy back from gravity. Again, we assume the two balance each other out in an ideal system to make our calculation easier.

So I agree with Pisiform that choice B is the best answer to choose on this question, but it's not a correct answer. The potential energy of the system at the start is 0.5ka^2 + mg(delta)h, which means that's how much energy must be lost during the dampening.

Pisiform is correct about the displacements in the potential energy formulas that he/she is using. But if you account for the potential energy from gravity, you get exactly answer B. See post #8 for the math.

 

[MrNeuro]

Ok, that works out. Let dL = L-L0, PEi is the potential energy with the spring stretched to dL+A and PEf is the spring at rest with the mass hanging from it.
PEf = k/2 * dL^2
PEi = k/2 * (dL + A)^2 - mgA (we lose some potential energy by lowering the mass)
From mg = k * dL (gravity acting on the mass is same size opposite direction as the spring force on it) we can get:
PEi = k/2 * (dL + A)^2 - mgA = k/2 * (dL + A)^2 - k * dL * A =
= k(dL^2/2 + dL * A + A^2/2 - dL * A) = k/2 * (L^2-A^2)
The we have PEf - PEi = k/2 * dL^2 - k/2 * (L^2 - A^2) = k/2 * A^2 which is answer B.

That was a nice question! We also proved that we can treat a spring with a mass hanging from it the same as a relaxed spring with a different neutral position and the same k coefficient.

two questions
1) why do you have PEf? are you doing PEf - PEi isn't that just work?
2) isn't the displacement just A not L - L0 + A. making the PE= .5 k A^2 - mgA

 

[milski]

two questions
1) why do you have PEf? are you doing PEf - PEi isn't that just work?
2) isn't the displacement just A not L - L0 + A. making the PE= .5 k A^2 - mgA

1) Because the spring already has some potential energy when it's being extended by the weight in experiment one. Yes, PEi-PEf is the work done by the force extending the spring from L to L+A. You should be able to calculate the work and get the same result.
2) The k/2*x^2 formula for potential energy of a sprint is correct only when the displacement x is from the neutral position that spring. That is when there are no forces applied to the spring at all. When the spring is extended to L, there is m*g force applied to it, so that is not its neutral position.

 

[ljc]

I guess I'm a bit rusty here, but can someone remind me (in words, not math) what causes the mass to cease oscillating? Why does the difference in potential energies (PEf-PEi, or whatever is in the math above) correspond to a halt in oscillation?

 

[milski]

I guess I'm a bit rusty here, but can someone remind me (in words, not math) what causes the mass to cease oscillating? Why does the difference in potential energies (PEf-PEi, or whatever is in the math above) correspond to a halt in oscillation?

The oscillation up and down requires energy, which was added when the weight was pulled down. In an ideal system, the oscillation will go on forever. In the real world, according to the problem, "energy is dissipated due to factors such as air resistance and internal heating of the spring." Once all that energy is dissipated the movement will stop.

 

[ljc]

So essentially what this question is asking is "How much energy was added to the system?"

Got it, thanks!

 

[milski]

So essentially what this question is asking is "How much energy was added to the system?"

Got it, thanks!

Yes.

I have no idea what's with the Young's modulus and why it was given. Is it normal for MCAT questions to have extra information in them?

 

[MT Headed]

Yes.

I have no idea what's with the Young's modulus and why it was given. Is it normal for MCAT questions to have extra information in them?

I can't think of any in the discretes, but it is a poular tactic to give a big passage with a ton of information and then ask something really simple about the data. Sometimes it is as simple a pulling one number out of a table of chemical data, though it is usually a trend "according to the data, adding more water to the reaction makes it run (a) faster (b) slower (c) hotter" etc.

Review books are notorious for adding unnecessary information to questions. It gets kind of annoying after a while.