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A problem involving Newton's second law.

  1. Apr 15, 2013 #1
    1. The problem statement, all variables and given/known data
    see this link http://i45.tinypic.com/10gwhw2.jpg


    2. Relevant equations

    F=kx, F=ma

    3. The attempt at a solution

    ok first I find the total length of the pulley system,
    2Sa+Sb=L, differentiating with respect to time i get,
    2Va + VB=0 and thus 2Aa+Ab =0
    So Aa =-Ab/2

    next I first analyze block A and do a force balance to get,
    -Ma*g - k*x + 2T = Ma*Aa

    following this i analyze Block B doing a force balance again,

    T-Mb*g = Mb*Ab

    so to eliminate one of the variables I input Aa=-Ab/2 into the equation describing block A.

    I get -Ma*g-k*x+2T+Ma(-Ab/2)

    then i get -2(Ma*g-k*X+2T)=Ab

    so now i substitute this in Block B with real values

    T-98.1=10(19.62-400+2T)
    so T = 195.04
    then Ab = 19.62-400+2(195.04) = 9.7 m/s^2

    So now I have (delta)y = 0.5 m A=9.7
    using the following equation i can solve for t, Δy=(1/2)at^(2)
    doing so i get 0.32s
    thus v = at = 9.7(0.32) =3.104 m/s

    However i do not have a solution for this exercise and unsure if my attempt is correct or not. Any help is greatly appreciated thanks!
     
  2. jcsd
  3. Apr 15, 2013 #2

    gneill

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    Staff: Mentor

    Since the length of the spring changes with time, the force will vary with time, too. That means the accelerations will not be constant, so you can't just apply the constant acceleration kinematic equations. Instead, you'll end up with a differential equation to solve.

    Have you considered using a conservation of energy approach? Muuuuch simpler.
     
  4. Apr 15, 2013 #3
    Ok so I assume my working is correct up to the point with which i calculated T.

    so my new attempt is as follows:

    W=(1/2)*m*vf^(2) - (1/2)*m*vi^(2)
    F*D= (1/2)*m*vf^(2) - (1/2)*m*vi^(2)

    so F*D = (-Ma*g-k*x+2T)*Δy = (1/2)*mvf^(2)

    subbing in values and solving for vf i get 0.98m/s^2

    I'm guessing however this is incorrect because the force that the spring exerts is not constant?
     
  5. Apr 15, 2013 #4

    gneill

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    Staff: Mentor

    Yup. This approach must yield a differential equation. The spring force is not constant, so the accelerations are not constant and you cannot apply the constant acceleration kinematic formulae.

    Try an energy conservation approach.
     
  6. Apr 15, 2013 #5
    Ok my next attempt is as follows

    (1/2)ky^(2) = Mb*vf^(2)

    so subbing in values to solve for vf i get 4.472m/s.
     
  7. Apr 15, 2013 #6
    actually maybe this is the equation i need?

    (1/2)ky^(2) +Ma*g*y=(1/2)Mb*Vf^(2)
    and solving i get 4.686 m/s
     
  8. Apr 15, 2013 #7

    gneill

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    Suppose you set the individual gravitational PE's of the two masses to zero when the system is at equilibrium. Also set the spring's PE to zero at equilibrium. (You can do this because PE is a relative measure).

    Now, whenever the system is at the equilibrium position, those PE's will be zero.

    How much additional energy will be stored in the spring when mass B is displaced by 0.5m and held there? What's the total KE at that time? (Remember, the masses are stationary). So what's the total energy PE+KE at that time?
     
  9. Apr 15, 2013 #8
    PEa = Ma*Aa*y and KEa = 0
    PEb= Mb*Ab*y and KEb =0

    I can't use g as acceleration right? because the forces acting on the objects interfere with its normal functioning? or am i way off the mark :(
     
    Last edited: Apr 15, 2013
  10. Apr 15, 2013 #9

    gneill

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    Staff: Mentor

    Okay, not what I asked. What's the change in PE stored in the spring when mass MB is displaced by 0.5m?
    You won't need ANY accelerations or forces using the conservation of energy method.
     
  11. Apr 15, 2013 #10
    ok so that would be -kx
     
  12. Apr 15, 2013 #11

    gneill

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    Staff: Mentor

    No, What's the formula for the energy stored in a spring?
     
  13. Apr 15, 2013 #12
    sorry, (1/2)ky^(2)
     
  14. Apr 15, 2013 #13

    gneill

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    So... What's the additional energy stored in the spring when mass B is displaced?
     
  15. Apr 15, 2013 #14
    (1/2)Mbvf^(1/2)
     
  16. Apr 15, 2013 #15

    gneill

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    Staff: Mentor

    No. There's no motion at this point. Mass B is displaced and held stationary at its new position. But the spring has been stretched. What energy does SPRING gain?
     
  17. Apr 15, 2013 #16
    potential energy
     
  18. Apr 15, 2013 #17

    gneill

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    Staff: Mentor

    Make an effort. Calculate the value!
     
  19. Apr 15, 2013 #18
    Ok I'm thinking, (1/2)k(xf^(2)+xi^(2))=1/2(Mb)vf^(2)

    subbing in values i get

    6.32 m/s?
     
  20. Apr 15, 2013 #19

    gneill

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    Staff: Mentor

    How do you justify the equation above? (What does it represent?)

    What about mass A? Won't it be moving too?

    The reason I ask you to calculate the potential energy added to the spring when the displacement is carried out is that it represents the energy added to the overall system, and will stay with the system when it returns to the equilibrium position. When the system returns to the equilibrium position all the gravitational potential energy changes due to displacement disappear -- everything is back in its original position. What's left is the new energy that was stored in the spring, which has then been released as KE since the spring, too, is back in its original position.

    So why not calculate that spring energy? How much is the spring extended when MB is displaced by 0.5m? How much energy does that put into the spring, in Joules?
     
  21. Apr 15, 2013 #20
    " How much is the spring extended when MB is displaced by 0.5m?"
    the spring is extended 0.5m.

    "How much energy does that put into the spring, in Joules?"
    (1/2)k(y)^(2) = (1/2)*800*(0.5)^2 = 100 J
     
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