Taken from the 2011 F = MA Exam (USAPHO Qualifiers)

A spring has an equilibrium length of 2.0 meters and a spring constant of 10 newtons/meter. Alice is pulling on one end of the spring with a force of 3.0 newtons. Bob is pulling on the opposite end of the spring with a force of 3.0 newtons, in the opposite direction. What is the resulting length of the spring?

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## Comments

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TopNewestBy the way, please do not use the CMC tag in any new discussions. That's specifically for Cody's Math Contest.

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Sorry, I accidentally included it. Thanks for letting me know.

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Thanks, you beat me to it.

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First Law of Newton: If there's a force on something, it'll accelerate. Since this spring isn't moving, there's no force on it. If there wasn't any Bob pulling from the other end, Alice will drag the spring with her all the way home (and if has no mass, it'll be totally upstretched too). Picture Sally hooking one end to her bike and driving home. It won't be stretched at all.

Now how do springs behave? If you want to elongate a spring (with constant 10 N/m) by one meter, you'll have to hold both ends and pull, and keep pulling till it's 1 m longer and hold it that way. At that point, you'll be exerting 10 N force with each hand. The forces on the hands have to equal, otherwise, the spring will accelerate. Or you could tie one end to a wall and pull. Or in this case, a Bob guy who'll halfway for you. The advantage you get if you have Bob is that while stretching, he'll do half the work for you. However, once done, both exert exactly the same force; they have to. Bob is no longer necessary. A wall will do just as well.

For this problem, the constant is 10 N/m => 0.1 m/N. Since the force on the ends is 3 newtons, the elongations is 0.3 m, from which we conclude that the final length is 2.3 m.

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OH, okay. Thanks! So what would happen if Alice was pulling with 4 N and Bob was pulling with 3 N? The spring would obviously accelerate, but how much would it stretch by?

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Visualise this: a train engine hooked to a compartment and pulling it forward. They're connected to each other by those massive springs. Now, will the force on both ends of the spring be the same? By the way, the spring is much lighter compared to the engine and compartment.

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But now, let's look at it this way. If the train is accelerating with a, then the spring should have the same acceleration. So the spring has an acceleration a. Now, forget the train exists. There is black everywhere. Just plain scary empty void. And in the middle is the spring. See nothing else. The force on it should be mass times acceleration. And since its mass is negligible, so is the force.

So does that mean the spring has nearly zero force on it? Well, yes. Since, there are two forces on it, one forward and one backward, they should almost perfectly cancel. My point here is, when you are applying second law, the only mass to be taken into account is the spring, don't bring the train into the equation. It doesn't matter if it's there or not. The fact that forces on both ends is nearly equal is the next part. Before that, we deduce that force is zero. Then, since there are two forces, we say that they are nearly equal?

If you sketch out the free-body diagrams and solve the equations, you'll get the result. But try to see why this is so counter-intuitive. It might help a lot more in convincing you than all the mathematics in the world.

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Correct?

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For the train problem though, the difference can't be too large. Because if it was so, the spring would just shoot ahead while the engine and the compartment slowly go forward, which is as ridiculous to visualise as it is impossible.

For the present problem, it is precisely zero since forces can't be applied on massless objects.

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Hopefully this is right: the total force on the spring is \(6\) Newtons, so by Hooke's law we have \[F = kd\] or \[6 \text{ N} = 10 \textstyle\frac{\text{N}}{\text{m}} \cdot d\] where \(d\) is the additional length of the spring. Thus, \(d = 0.6 \text{m},\) and the spring stretches to \(\boxed{2.6 \text{m}}.\)

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That's what I thought too. However, the answer key says it's 2.3m and, unfortunately, there are no solutions.

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kx = 6 , ................where k and x are spring constant and extension in spring.

x = 0.6

So, resultant length = 2.6m

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This was my answer too, but, like I said below, the answer is actually 2.3 meters.

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Are you sure about the authenticity of the key???

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The spring can be thought of as two springs in series where the net spring constant is 10 N/m.

1/knet = 1/k1 + 1/k2

Since k1 = k2,

knet = 20 N/m

Therefore the extension of each spring is

3 N = 20 N/m * x

x = 0.15 m

The total extension is then 0.3 m, resulting to a net length of 2.3 m.

Not sure about this though.

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Are you sure springs can be divided up in that way? I guess my main question is why don't they have the same spring coefficient.

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The equation you have used is for springs in series and not them in parallel..Also, in the given situation, the two portions are in series configuration.

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I edited my post. Sorry for the typo. I meant that the springs are in series.

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Nevermind, I read up on springs in parallel and in series. Thanks!

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ans : 3m

Explanation : When you pull the spring on the two ends, the centre of the spring remains stationary. Now the whole sring acts like two springs in series, each with constants 2k.( in series 1/(k1) + 1/(k2) = 1/k) .. Thus for Bob and Alice, now, its like pulling springs of force constant 2

k= 23 = 6 N/m.so, 3 = 6 * x

Totally, the extension is 0.5 + 0.5 = 1 m.

Thus the new length is 2 + 1 = 3 m.

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Well, first off, the answer is actually 2.3 meters. Also, I don't think springs work like that; that is, I don't think you can just divide the spring into two smaller springs with equal spring coefficients.

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Yes...it is meaningful to view the system as two springs in series. And, I am not dividing the spring into two with equal coefficients. The two springs are in series and their effective constant is k. So, by the equation , the new coefficient is 2k.

The mistake I did was to wrongly take in the value of k from the question as 3N/m instead of 10N/m.(you can see that in my comment above)So, the corrected equation is :

3 = (2*10) * x

x = 0.15 m

Total extension = 0.3 m

Thus new length = 2.3 m.

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