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Mass on Incline with Spring Attached

  1. Jun 19, 2013 #1
    1. The problem statement, all variables and given/known data

    swhKirX.png

    2. Relevant equations

    Forces and/or energy

    3. The attempt at a solution

    I honestly don't know where to start. First I tried kx = mgsin(tan-1 (3/4))) and got the answer E, which is wrong. Then I tried using energy, mgh = 1/2kx^2 which resulted in an answer that wasn't on the list. Any ideas on how to begin?
     
  2. jcsd
  3. Jun 19, 2013 #2

    rude man

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    At any point after letting go of the mass,
    energy stored in spring + potential energy + kinetic energy = constant. (Why?)
     
  4. Jun 19, 2013 #3
    Because total mechanical energy of the system is conserved?

    Edit: nvm I figured it out!
     
    Last edited: Jun 19, 2013
  5. Jun 19, 2013 #4

    CAF123

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    Why is ##kx = mg \sin \theta## wrong here?, ##\theta## the inclination of ramp.
     
  6. Jun 19, 2013 #5

    rude man

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    Well, it violates the law of conservation of energy.

    If you slowly let the mass slip until it was stationary then you'd be right, but that is not what's happening here.
     
  7. Jun 19, 2013 #6

    CAF123

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    Perhaps I misinterpreted something: when the mass is released, it obeys ##mg\sin\theta - kx = ma_x##So if they were ever equal, then the block would then move at constant speed.

    The energy approach would indeed seem more appropriate.
     
  8. Jun 19, 2013 #7

    rude man

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    Indeed it is.

    The point of where the mass momentarily stops moving is not the point of zero acceleration. It is the point of zero velocity.
     
  9. Jun 20, 2013 #8

    CAF123

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    Yes, but what does the acceleration vectors look like in such a case? It can't be like the vectors on a pendulum, so the acc. vectors will lie along direction of motion.
     
  10. Jun 20, 2013 #9

    rude man

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    Let s = distance along ramp starting from the point where the spring is relaxed, where s = 0:
    Then form F = ma:
    -ks + mg sin(θ) = md2s/dt2
    Solve this differential equation for s, take ds/dt and then d2s/dt2 gives you your acceleration. You will find s(t) and its derivatives are non-decaying oscillations.

    You can also solve the given problem this way instead of invoking energy arguments: solve for ds/dt and set ds/dt = second zero (the first is obviously when you let go of the mass). Then solve for T, the time of the second zero of ds/dt, and then s(T).
     
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