# Spring AP physics Question

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Gfoxboy said:

think about what ks means -- spring constant * displacement

deriving the energy equation 1/2 k s^2 -- how does U(E) rise as s changes? think about how force works, and see if you can figure out why you don't go skyrocketing through the ceiling.

what forces are acting on the objects in your system?
what is the total net force of your system?

I went through the math and I understand mathematically why the answer is that the potential energy quadruples. What I still don't understand is why the potential energy can be twice as much as the force required to hold up the weights... obviously the recoiling force of the spring would be equal to the force of gravity pulling down on the two 0.2 kg weights. But according to the potential energy equation the total energy is four times the amount of potential energy from just one 0.2 kg weight. I just don't see why it requires four times as much energy to hold up only twice the weight...

Try and find a formula online that connects mass with potential energy of the spring

You have
$$F=-kx$$
$$-mg=-kx$$
$$PE = -k x^2$$

Can't put PE in terms of mass?

Doc Al
Mentor
Gfoxboy said:
I went through the math and I understand mathematically why the answer is that the potential energy quadruples. What I still don't understand is why the potential energy can be twice as much as the force required to hold up the weights... obviously the recoiling force of the spring would be equal to the force of gravity pulling down on the two 0.2 kg weights. But according to the potential energy equation the total energy is four times the amount of potential energy from just one 0.2 kg weight. I just don't see why it requires four times as much energy to hold up only twice the weight...
It's not the spring potential energy that holds up the weight, it's the tension (force) in the spring. (Force and energy are two different things!) The amount of spring potential energy that gets stored in the spring depends on how much work was done (by gravity) to stretch the spring.

When the first 0.2 kg weight was put on the spring, it stretched 5 cm. How much work was done in stretching the spring? As the spring stretched, the force (given by kx) went from 0 to kX (X = 5 cm). The work done is average force (kX/2) times displacement (X) = kX^2/2.

When the second 0.2 kg was added, the spring stretched from 5 cm to 10 cm. The force on the spring during this stretch is not just mg = (.2)g, but twice that: mg = (.4) g. So now the force goes from kX to 2kX. The work done during this second stretch is average force (3/2 kX) times displacement (X) = 3/2 kX^2. Thus the total work done (and energy stored in the spring) is 2kX^2.

Think about it: the more that spring is stretched, the harder it gets to stretch it another cm. (The added weight is just the additional force needed, not the total force.)

Does this help?

Well if the longer a spring gets the harder it is to stretch then doesn't adding the same weight again not double the length? The first 0.2 kg stretched it 5 cm, but the second 0.2 kg should stretch it less because it became harder to stretch it that additional distance. Both 0.2k kilogram masses exert the same force, but when added together the force isn't enough to double the distance.

Doc Al
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Gfoxboy said:
The first 0.2 kg stretched it 5 cm, but the second 0.2 kg should stretch it less because it became harder to stretch it that additional distance.
It is harder to stretch the additional distance: It's three times harder (on average) to stretch the spring the second 5 cm! But it only takes twice the weight to create twice the (maximum) stretch.

Both 0.2k kilogram masses exert the same force, but when added together the force isn't enough to double the distance.
Sure it is. It's twice the force---just enough to give twice the stretch.

Wait so the first weight applies a force of X and the second weight applies a force of 3x...?? I feel like I'm missing something very simple here...

Doc Al
Mentor
Gfoxboy said:
Wait so the first weight applies a force of X and the second weight applies a force of 3x...?? I feel like I'm missing something very simple here...
No, assuming equilibrium the first mass applies a force of mg = kX, while the first + second masses together apply a force of 2mg = 2kX, which is twice as much.

However, the force that the spring exerts while being stretched varies. The average force that the spring exerts while being stretched is 3 times greater for the second X of stretching (from X to 2X) than for the first X of stretch (from 0 to X). Thus the work done on the spring is 3 times greater during the stretch from X to 2X. That's why the energy stored in the spring when stretched to 2X is four times the energy stored when only stretched X.

I guess I may never make sense out of this. It seems as though more potential energy is being created than work that is done by gravity. Thanks for all your help.

Oh only one thing I want to clear up. Work obviously equals force times distance so, how can the work done on the spring be 3 times greater from x to 2x if the force is only doubled? This is the only reason I don't understand this problem. It seems like doubling the weight should only double the force, therefore the distance would be smaller the second time and not actually reach 2x. If you could elaborate more on that I may be able to understand this.

Work = integral( -kx ) [a to b], where a is spring length at equlibrium and b is the length it is streched to. Basically, integrating the force equation the length you strech it yields the work done to stretch it

Doc Al
Mentor
Gfoxboy said:
Oh only one thing I want to clear up. Work obviously equals force times distance so, how can the work done on the spring be 3 times greater from x to 2x if the force is only doubled?
Yes, work = Force x distance... But the force is not constant. So you need to use the average force to calculate the work. The maximum force from x to 2x is at the point 2x; that maximum force is twice the force at point x. But the average force is 3 times greater between x to 2x, compared to 0 to x, so the work is three times greater. See my posts above for details.

(Of course, if you familiar with integration, the work work is the integral of -Fdx = kx dx ==> 1/2 k x^2.)

AHA! Thank you both. Finally it makes sense. I feel like I've been hiding under a rock or something. I didn't differentiate between average and instantaneous forces. Finally I understand thank you for all your help and patience.