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Conservation of Energy + Hooke's Law

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    A 2.70kg mass is pushed against a horizontal spring of force constant 25.0 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring. When the spring has been compressed enough to store 12.0 J of potential energy, the mass is suddenly released from rest. What is the greatest speed the mass reaches?

    2. Relevant equations
    Conservation of Energy
    Hooke's Law (F = -kx)

    3. The attempt at a solution
    As soon as they put a spring in the conservation of energy problems, I become confused. First, I don't understand why it's 1/2kx^2 (which doesn't apply to this problem I don't think), and second, I have a hard time determining if it's positive or negative. Here is what I come up with for this problem, there are more parts to it and I was hoping that you guys could help me work through my spring-a-phobia.

    Edit: I see a mistake already, fixing the image I posted.
     
    Last edited: Oct 29, 2011
  2. jcsd
  3. Oct 29, 2011 #2
    First, it should have been 25N/cm so that's really 2500 N/m.

    Then:
    [tex]12.0 = \frac{1}{2}(2.7)v^{2}[/tex]
    [tex]v = 2.98[/tex]

    That seems correct to me. The second part is to determine when this speed occurs. I'm not sure how to do this exactly. Intuition tells me that it happens right as the block leaves the spring, or when the spring is at equilibrium. I don't know hot to express that mathematically though. If I say that x is zero in the conservation equation I get that 0 = KE which is clearly not the case.

    This is my answer for the second part:
    Am I on the right track so far?
     
    Last edited: Oct 29, 2011
  4. Oct 29, 2011 #3

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    Hi QuarkCharmer! :smile:

    A compressed spring holds the potential to release energy.
    In neutral position it obviously does not hold energy, therefore a compressed spring holds positive energy.

    Why is it [itex] {1\over 2}kx^2[/itex]?
    Well:
    [tex]W=\int Fdx = \int kxdx = {1\over 2}k x^2[/tex]

    And yes, your result for v is right.

    When the (perfect massless) spring is back in its equilibrium point, it has transferred all of its energy to the mass and the mass.
    (The same integral formula shows that all stored power has been returned at this point.)
    Therefore the spring won't accelerate in opposite direction.
    Hooke's Law is observed, because there is no mass fixed to the spring anymore, so F=ma has to be zero.
     
    Last edited: Oct 29, 2011
  5. Oct 29, 2011 #4
    Thank you ILS,

    Wouldn't the spring continue to oscillate after the mass disconnects? Also, wouldn't the integral contain a negative kx, since F = -kx, such that:
    [tex]W_{elastic/spring} = \int -kx dx = -\frac{1}{2}kx^{2}[/tex]

    The next part want's to know what the greatest acceleration of the mass is. Here is how I worked it out:

    Since I assume the acceleration of the mass stops when the mass leaves the spring, I needed to determine the length the spring was compressed. I did this via:

    [tex]\frac{1}{2}kx^{2}=12[/tex]
    [tex]\frac{1}{2}(2500)x^{2}=12[/tex]

    I put k as 2500, since the desired units are N/m and I was given 25N/cm, so:
    [tex]\frac{25N}{cm}\frac{100cm}{1m} = \frac{2500N}{m}[/tex]

    This tells me that x is 0.0980 m (seems too small...)

    Then I used kinematics to find the accleration:
    [tex]V_{f}^{2} = V_{i}^{2}+2aΔx[/tex]
    [tex]2.98^{2} = 0^{2}+2a(0.0980)[/tex]
    [tex]a = 45.4 m/s^{2}[/tex]
    Which is incorrect?
     
  6. Oct 29, 2011 #5

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    I suddenly realize what your problem with the minus sign is.

    The magnitude of elastic force is F=kx.
    The vector of the elastic force is [itex]\vec F = -k \vec x[/itex].

    I can imagine this can be confusing.
    I usually only use the magnitude F=kx.
    When relevant we need to realize of course that its direction is opposite to the deviation.


    When the mass disconnects, the remaining mass attached to the spring is m=0.
    If the spring would get to some position x, its elastic force would be F=kx.
    Using F=ma the resulting acceleration a would be infinite!

    Suffice to say: no, it does not oscillate after the mass disconnects (assuming a perfect massless spring).


    Looks good.


    No, this formula is not applicable.
    It only applies when the acceleration "a" is constant, which it isn't.

    You should use F=kx to find the maximum acceleration.
     
  7. Oct 29, 2011 #6
    Understood. So, "yes" it would still oscillate, but not in these imaginary perfect condition physics problems.


    I understand the error with using kinematics there! (and the part about the negative in Hooke's Law, thank you)

    [tex]F=ma[/tex]
    [tex]F=kx[/tex]
    [tex]kx=ma[/tex]
    [tex](2500)(0.0980)=(2.7)a[/tex]
    [tex]a_{max} = 90.7 m/s^{2}[/tex]

    That seems outrageously high? So I assume I am applying this wrong as well?
     
  8. Oct 29, 2011 #7

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    No, you applied it correctly.
    It means you can successfully shoot your mass upward (it will overcome gravity). :smile:
     
  9. Oct 29, 2011 #8
    Just seems like a large number for acceleration with a spring! That's nearly 300ft/s, or about 1/2 the speed of an average hand gun round. That's some spring they have there...


    The last part of this practice question want's to know when this max acceleration occurs.

    That's going to be when the spring reaches equilibrium correct?
     
  10. Oct 29, 2011 #9

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    No, to the contrary.
    Are you mixing acceleration and velocity?

    Remember that you calculated a speed v=2.98 m/s!

    Note that a=F/m.
    When is F at its maximum?
     
  11. Oct 29, 2011 #10
    Is this the idea then?

    15xls2x.jpg
     
  12. Oct 29, 2011 #11

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    Depends on what you have on the x-axis.
    Is it x or t?
    It should be x.
     
  13. Oct 29, 2011 #12
    I was thinking of it's acceleration vs time, but vs distance might be more appropriate yes. Either way, that would make the max acceleration occur right when the spring starts to expand back to it's equilibrium state correct?
     
  14. Oct 29, 2011 #13

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    Good! :smile:

    FYI, acceleration vs time is a cosine.
     
  15. Oct 29, 2011 #14
    Because that is modeled by a simple harmonic motion right?
     
  16. Oct 29, 2011 #15

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  17. Oct 29, 2011 #16
    Great! That's the section I am starting now. Trying to work ahead a few weeks in homework/subjects because I plan to be super busy with another class next week. I just finished all of the homework involving springs and such using knowledge gained in this thread. Thank you!

    It's Pendulum time now!
     
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