1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring AP physics Question

  1. Nov 9, 2004 #1
    My Teacher gave me an interesting problem in AP Physics which seemed simple at first but has me going insane. The problem was what would happen to the potential energy contained in a spring if 0.2 kg was added to the original 0.2 kg on the spring, which stretched the spring 5 cm. The mathematical answer is that the potential energy would quadruple. The problem I have with the answer is this: when you first add the 0.2 kg mass, the spring stretches 5 cm from its original state, but according to the answer, if you add the next 0.2 kg mass, the length must once again be increased by 5 cm. This is found using the equation Fs = (1/2)k(x^2). But as the spring lengthens, it takes more work to increase the length by the same amount again, so shouldn't another 0.2 kg increase the length of the spring by less than 5 cm? If this is so then the actual answer should only be that the potential energy is doubled. One example I tried to use against my teacher was this: Say that I stand on a spring scale. The scale should read my weight because of the work that gravity does on the scale by pulling me down onto the scale. Now if a person who weighs the exact same as I do stands on the scale with me, it should obviously read double my weight because gravity is now doing double the work. Now if the potential energy in the spring quadrupled then it would have the ability to do twice the work that gravity was doing, therefore launching my friend and I through the ceiling. Basically I'm extremely confused and maybe my logic is completely incorrect. Thank you if you read all of this...
     
  2. jcsd
  3. Nov 9, 2004 #2

    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 is your system?
    what forces are acting on the objects in your system?
    what is the total net force of your system?
     
  4. Nov 9, 2004 #3
    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...
     
  5. Nov 9, 2004 #4
    Try and find a formula online that connects mass with potential energy of the spring
     
  6. Nov 9, 2004 #5
    You have
    [tex]
    F=-kx
    [/tex]
    [tex]
    -mg=-kx
    [/tex]
    [tex]
    PE = -k x^2
    [/tex]

    Can't put PE in terms of mass?
     
  7. Nov 9, 2004 #6

    Doc Al

    User Avatar

    Staff: Mentor

    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?
     
  8. Nov 9, 2004 #7
    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.
     
  9. Nov 9, 2004 #8

    Doc Al

    User Avatar

    Staff: Mentor

    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.

    Sure it is. It's twice the force---just enough to give twice the stretch.
     
  10. Nov 9, 2004 #9
    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...
     
  11. Nov 10, 2004 #10

    Doc Al

    User Avatar

    Staff: Mentor

    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.
     
  12. Nov 10, 2004 #11
    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.
     
  13. Nov 10, 2004 #12
    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.
     
  14. Nov 10, 2004 #13
    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
     
  15. Nov 10, 2004 #14

    Doc Al

    User Avatar

    Staff: Mentor

    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.)
     
  16. Nov 10, 2004 #15
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Spring AP physics Question
Loading...