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Conceptual doubt about elastic potential energy of a system

  1. Feb 5, 2010 #1
    Good afternoon,

    First of all, this is not a homework question, but I'm not sure whether it should be posted on the Homework section.

    My problem arose when I tried to think about what happens when a spring does work in two blocks.
    First, consider two blocks (1 and 2) on a frictionless horizontal plane, and a spring. The two blocks are attached to each side of the spring, and the spring is initially on its equilibrium position.
    Then, the two blocks are pulled away from each other by two external forces opposite to each other (F1 and F2), with the same magnitude, parallel to the spring, so that the spring is decompressed. So, the spring applies a restoring elastic force on each block. Let's call the restoring forces Fe1 (opposite to F1) and Fe2 (opposite to F2).
    The displacement of blocks 1 and 2 are respectively x1 and x2. So, the spring is decompressed by a length of x = (x1 + x2) (which makes Fe1 = Fe2 = -k(x1 + x2)).
    It means that the system stores a potential energy defined by:
    [tex]\Delta U=\frac{1}{2}k(x_1+x_2)^2[/tex]
    Right?
    But, then, a problem arose in the reasoning:
    Individually, each restoring force does negative work on each block. For block 1, the work done by the restoring force is:
    [tex]W_{Fe1}=-\frac{1}{2}kx_1^2[/tex]
    For block 2:
    [tex]W_{Fe2}=-\frac{1}{2}kx_2^2[/tex]
    But the sum of the works by the conservative forces equals the negative of the variation of potential energy of the system.
    So, the sum of these works, [tex]W=W_{Fe1}+W_{Fe2}[/tex], should give:
    [tex]W=W_{Fe1}+W_{Fe2}=-\Delta U[/tex]
    But, in this case, the above is not true. What is wrong with this reasoning?

    Thank you in advance.
     
    Last edited: Feb 5, 2010
  2. jcsd
  3. Feb 5, 2010 #2
    Your two forces must be equal and opposite, right?? otherwise the spring and two blocks will go accelerating off in one direction or the other.

    If the two forces are equal and opposite, your example seems the same as having a spring attached to a fixed wall and pulling a mass with that force f to extend the spring...

    and your energy is the 1/2kx2 as you supposed....
     
  4. Feb 6, 2010 #3
    Yes, they are equal and opposite.

    Good. So, it could be analyzed as if one of the blocks was fixed and the other block did the entire displacement (x1 + x2).
    But what is wrong with the rest of the reasoning? For me, the sum of the works done by the conservative forces (Fe1 = Fe2) on each block should equal the negative of the expression for the potential energy, but it apparently doesn't.
     
  5. Feb 6, 2010 #4

    rcgldr

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    Homework Helper

    There's some missing information here. Are the blocks massless? Is the force applied to those blocks constant or does it increase over time? If blocks have mass, and the spring is massless, and the force on each block is constant, then the blocks are accelerating outwards until

    | k (x1 + x2) | = | F1 | = | F2 |

    At this point the blocks start decelerating, but continue to move outwards because of their momentum.
     
  6. Feb 6, 2010 #5
    Let me confirm the missing information (I thought of this when elaborating the example, but forgot to inform):
    - The blocks have mass (which may be called m1 and m2, respectively);
    - The spring is massless;
    - The applied forces are constant.
    Then, the information you gave is valid. When the restoring forces equal the applied forces, the blocks will start decelerating.

    EDIT: Back to the original problem (that the sum of the works by the conservative forces should equal −ΔU):
    Maybe the work done by the elastic force on each block is not as I calculated?


    Thank you in advance.
     
    Last edited: Feb 6, 2010
  7. Feb 6, 2010 #6

    Doc Al

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    Staff: Mentor

    Right.
    These work calculations are not correct. They assume that the force on each block is -kx, but it's really -k(x1 + x2).

    Take the simple case where x1 = x2 = x. The spring PE is 1/2k(2x)2 = 2kx2.

    The force on each block is -2kx, so the work done on each block is -1/2(2k)x2 = -kx2. So the total work done on both blocks equals -2kx2, as expected.
     
  8. Feb 6, 2010 #7
    OK, I had already noticed that. I wasn't thinking straight when I wrote that. The force is -k(x1 + x2).

    OK, I understood what you said. Knowing that the displacement of each block is x and F = -2kx, I could also find this work integrating F from 0 to x (I did this, and the result was correct, W = -kx2). Also, I can find the total work done by the spring by integrating -kx from 0 to 2x (since the elongation of the spring, which I called x, varies from 0 to 2x and I consider it as if one of the blocks was fixed).

    Now, let's see how it would stay if I assumed x1 to be different from x2.
    If I call x the displacement of one block, integrating F = -kx from 0 to (x1 + x2) will give the correct expression for the work done by the spring on the two blocks: W = -(1/2)k(x1 + x2)2. However, I'm having trouble trying to find out what the expression for work on each block would be in this case. (Of course, if x1 was to be different from x2, both external forces would have to be different too.)

    Thank you in advance.
     
    Last edited: Feb 6, 2010
  9. Feb 6, 2010 #8

    diazona

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    The trick is that, when you try to calculate the work done on one of the blocks, the x in F = -kx is not the same as the x in [itex]W = \int F\mathrm{d}x[/itex]. The former is the total elongation of the spring (x1 + x2), but the latter is the distance over which the force acts, i.e. the distance over which the block moves (just x1). It's not often that you run into a situation where the two distances are different, which is why [itex]W = \frac{1}{2}kx^2[/itex] usually works.
     
  10. Feb 6, 2010 #9

    Doc Al

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    x1 and x2 will be different if the masses of the blocks are different. As long as you pull with equal force on each end, the center of mass of the system will not move. Use that to figure out how x1 relates to x2, then use that to figure out the work done on each block. Hint: Since the center of mass is fixed, you can treat the spring as composed of two smaller springs connected together. Figure out each smaller spring's effective spring constant.
    Not necessarily. Keep the forces the same.
     
  11. Feb 6, 2010 #10
    OK, thank you for the help.
    I haven't read about center of mass yet; when I do, I will try to work a solution.
     
  12. Feb 7, 2010 #11
    Yes, you are right. In this case, the displacement of each block is x; but, as a function of the displacement of each block, the elongation of the spring is 2x. In other words, as you said, here, the total elongation of the spring doesn't equal the displacement of each block.
     
  13. Feb 7, 2010 #12
    It will make spring constant of smaller springs equal to 2k..
     
  14. Feb 8, 2010 #13
    OK, I've read about center of mass, and tried to work a solution. What I've got so far is a relation between x1 and x2 based on the masses of the blocks.
    Look what I did:
    First, I choose a referential in m1 (which I consider to be punctual) for measuring the center of mass. I call "l" the natural length of the spring.
    The expression for finding the X coordinate of the center of mass (Xcm) when the spring is in the natural length would be:
    [tex]X_{cm}=m_1(0) + m_2l[/tex]
    [tex]X_{cm}= m_2l[/tex]
    When the forces are applied, and each block moves respectively x1 and x2, the expression for the X coordinate of the center of mass becomes:
    [tex]X_{cm}=m_1(-x_1) + m_2(l+x_2)[/tex] (block 1 moved -x1 with respect to the chosen origin.)
    [tex]X_{cm}=-m_1x_1 + m_2(l+x_2)[/tex]
    As the forces are equal and opposite, the center of mass doesn't move. Then, the first expression equals the last one:
    [tex]m_2l=-m_1x_1 + m_2(l+x_2)[/tex]
    [tex]m_2l-m_2(l+x_2)=-m_1x_1 \Rightarrow m_2(-x_2)=-m_1x_1[/tex]
    Then, I find a relation between the displacement of the blocks and their masses:
    [tex]m_2x_2=m_1x_1[/tex]
    Is this correct?
    If it is correct, it's interesting because it indicates that, the bigger the mass, the smaller the displacement will be.
    So, if the elastic force is F = -k(x1 + x2), it could be expressed in terms of just x1 or x2:
    [tex]F=-kx_1\left(1 + \frac{m_1}{m_2}\right)[/tex]
    [tex]F=-kx_2\left(1 + \frac{m_2}{m_1}\right)[/tex]
    Right? It seems right, because if masses are equal, the ratios above (m1/m2 and m2/m1) are 1, so the F = -2kx1 = -2kx2, and x1 = x2, as you said earlier.
     
    Last edited: Feb 8, 2010
  15. Feb 19, 2010 #14
    There is another interesting consideration about this topic.
    As in the last post, the force as a function of the displacement of the two blocks is:
    [tex]F=-k(x_1+x_2)[/tex]
    When the blocks are released after having decompressed the spring, the elastic forces will produce acceleration on each block.
    Then, the acceleration on block 1 is:
    [tex]a_1=-\frac{k}{m_1}(x_1+x_2)[/tex]
    And the acceleration on block 2 is:
    [tex]a_2=-\frac{k}{m_2}(x_1+x_2)[/tex]
    As the two accelerations are opposite to each other, the relative acceleration aR of the blocks relative to each other will be:
    aR = a1 + a2
    [tex]a_R=a_1+a_2=-\frac{k}{m_1}(x_1+x_2)-\frac{k}{m_2}(x_1+x_2)[/tex]
    [tex]a_R=-k(x_1+x_2)\left(\frac{1}{m_1}+\frac{1}{m_2}\right)[/tex]
    As F=-k(x1+x2):
    [tex]a_R=\frac{F}{\mu}[/tex]
    where [tex]\mu=\frac{1}{\frac{1}{m_1}+\frac{1}{m_2}\right}[/tex].
     
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