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Work by a spring

  1. Mar 6, 2017 #1
    I think I solved this correctly, but it puzzles me why they made two separate questions for the first two points.

    1. The problem statement, all variables and given/known data

    A block of known mass m is attached to the free end of a spring, on a flat horizontal surface without friction. The spring is compressed by a known distance x, and then released. The elastic constant of the spring is known K. Calculate:

    a. The energy stored in the spring while it's compressed.
    b. The Work done by the spring to extend.
    c. The speed of the block when the spring passes by its rest position point.

    2. Relevant equations
    Energy stored in spring = 0.5 * K * x2

    Sum of non-conservative forces Work = Mechanical Energy at the end - Mechanical Energy at the start
    Kinetic energy = 0.5 * m * v2

    3. The attempt at a solution

    So, what bothers me here are the points a. and b.

    The energy stored in the spring should be:
    0.5 * K * x2

    The Work done by the spring to extend should be...? I'm assuming they refer to the work done from when the spring is released until it goes back to its rest position. So it should regularly be, again:

    0.5*K * x2

    ...shouldn't it?

    Third point should be simply:

    0 = ME_end - ME_start => ME_start = ME_end =>
    => 0.5* k * x2 + 0.5*m*v02 = 0.5* k * x02 + 0.5*m*vf2

    So vf = sqrt (k * x2 / m)
     
  2. jcsd
  3. Mar 6, 2017 #2

    gneill

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    Staff: Mentor

    Suppose that they meant when the spring reaches its maximum extension? Would that make the answers to (a) and (b) different?
     
  4. Mar 7, 2017 #3
    I would guess so, yea. As far as I'm aware, from a graphic method the Work by a spring is done by the area under the curve, which becomes a trapezoid, where the area then results as:

    0.5 * K * (Xfinal2 - Xinitial2)

    Which would agree with the situation here: if one of the two "X" of the aforewritten equation is = 0, the formula becomes the usual 0.5 * K * X2

    The only issue might be the sign of the equation, so maybe point b. is = -a. ...?
     
  5. Mar 7, 2017 #4

    gneill

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    Staff: Mentor

    Think about the force acting on the block due to the spring. In particular, consider the direction of that force at various points along the path that the block takes while the spring is extending. What does that do to the work calculation along the path?
     
  6. Mar 7, 2017 #5
    If the spring is compressed from right to left, then the spring should do a positive Work as it's de-compressing from left to right; then, once it reaches it rest position, as it's extending further to the right it should start doing negative Work.

    Assuming the spring "counter-extends" to the right by a distance X equal to the compression distance, that would mean that the overall Work to go from a position "-X" to a position "+X" (putting X0 at its rest position) would be net 0 (+something -something), similar to the Work done by gravity if an object goes down by -h and then up by +h, because in one case we have Force and motion going the same way (+Work), while in the second part of the situation we have Force going one way (to the left for the spring, down for gravity) while the motion goes in the opposite direction (to the right for the spring, up for gravity).

    Assuming what I just said is correct, I still don't see why a. and b. would be different, if by "when the spring extends" refers to when the spring reaches its rest position again. On the contrary, if the problem meant "when the spring has counter-extended to its maximum value, before compressing back again", then my answer would have been net Work = 0.
     
  7. Mar 7, 2017 #6

    gneill

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    Staff: Mentor

    So in one case the answers would be the same and in the other case they would be different, right?
     
  8. Mar 7, 2017 #7
    lol yes, I would say so :D
    That's because in one case the spring releases all its stored energy to return to its rest position, so its work should be equal to the energy spent.

    In the other case the spring releases its stored energy to return to its rest position, but the energy that the spring transferred to the block (as kinetic energy) will cause the block to pull the spring further, thus causing the spring to do a negative Work that has the same value as the positive Work it did before.
     
  9. Mar 7, 2017 #8

    gneill

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    Staff: Mentor

    So, in the absence of an explicit definition of "extend" in the problem statement, I'd suggest that you cover both cases in your answer to (b).
     
  10. Mar 7, 2017 #9
    Ok, but are my answers correct? :D
    To recap:
    - If until rest position, Work = energy stored = 0.5 * k * x2
    - If until maximum counter-extension, Work = 0.5 * k * x2 - 0.5 * k * x2 = 0
    (Or, alternatively, Work = 0.5 * k * (x2 - (-x)2)
     
  11. Mar 7, 2017 #10

    gneill

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    Staff: Mentor

    Right. Although you might want to drop the "counter" in "counter-extension": It's not a recognized term. "Maximum extension" and "maximum compression" are more commonly used terms.
     
  12. Mar 7, 2017 #11
    Perfect. Thank you very much.
     
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