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Spring oscillation kinematics question.

  1. Dec 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A massless spring with sprint constant k is vertically mounted so that bottom end is firmly attached to the ground, and the top end free. A ball with mass m falls vertically down on the top end of the spring, becoming attached, so that the ball oscillates vertically on the spring. What equation describes the acceleration a of the ball when it is at a height y above the original position of the top end of the spring? Let down be negative, and neglect air resistance; g is the magnitude of the acceleration of free fall.


    2. Relevant equations
    F = -kx
    vf^2 = vi^2 + 2ax
    Ei = Ef
    KE = 1/2 mv^2
    PE of spring = 1/2 kx^2

    3. The attempt at a solution
    For this problem, I tried to work backwards given the height y. And I used the kinematics formula and plugged in the intial speed at the point when the ball is oscillated back at the equilibrium of the spring and the final speed being at height y.
    Then I worked with the F= -kx, letting the force equal to the weight mg, and used 1/2 kx^2 = 1/2 mv^2. then I solved for velocity and plugged that in the equation to find the final velocity at y.

    I think I went somewhere wrong in here because none of the answers matched.

    Btw, this a number 16, from the 2008 F = ma exam.

    And if anyone has enough time, can someone please explain to me the best way of studying for this exam?
    Would it be a good idea to read the entire mechanics part in the book Fundamentals of Physics? Because I feel I don't really grasp all the material and understand everything.

    thanks.
     
  2. jcsd
  3. Dec 25, 2011 #2
    Accelerations come from forces, what forces are acting on the ball?
     
  4. Dec 25, 2011 #3
    gravity, forces from the spring?
     
  5. Dec 25, 2011 #4

    SammyS

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    When the ball is at a height y above the original position of the top end of the spring, is the spring in contact with the ball ?
     
  6. Dec 25, 2011 #5
    no, it says y above the original position of the top end of the spring.
     
  7. Dec 25, 2011 #6

    SammyS

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    OK, then at that position, what force is acting on the ball ?
     
  8. Dec 25, 2011 #7
    gravitational force?
     
  9. Dec 26, 2011 #8
    The question does say "A ball with mass m falls vertically down on the top end of the spring, becoming attached" which makes me think that the mass is in some way "stuck" to the spring which would mean that the spring does exert a force on the ball once it has passed it's equilibrium point.
     
  10. Dec 26, 2011 #9
    it does.... it attaches to the spring and goes past the equilibrium point on the spring until the spring goes up again which launches the ball.

    so does anyone have any idea about how to do this.
     
  11. Dec 26, 2011 #10

    SammyS

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    Good point! I missed that.

    Well that makes sense, because then it says it oscillates. I must have missed that too.

    @ Ishida52134,

    This means that the ball is not launched.

    So above the equilibrium point, the spring is being stretched by the ball.
     
  12. Dec 26, 2011 #11
    okay so how do you do it.....
     
  13. Dec 26, 2011 #12

    SammyS

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    When the ball is a height, y, above the original position of the top end of the spring, what is the force exerted on the ball by the spring? ... What is the force exerted on the ball by gravity?

    What is the direction of each of these forces ?
     
  14. Dec 26, 2011 #13
    spring: ky
    gravity: -mg
    following standard coordinate axes.
     
  15. Dec 26, 2011 #14
    What is the net force (sum of the forces) on the ball? What does this net force equal to according to Newton's 2nd Law?
     
  16. Dec 26, 2011 #15
    ky - mg = ma?

    can u just tell me how to do it instead of asking me questions over a whole week. I honestly don't want to spend 1 week figuring out how to do one problem.
     
  17. Dec 26, 2011 #16

    SammyS

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    When y is positive, assuming that is above the spring's equilibrium position, then the spring exerts a downward force on the ball. That force should be negative.
     
  18. Dec 26, 2011 #17
    so how do you do the problem.......
     
  19. Dec 26, 2011 #18

    SammyS

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    Force exerted by spring: -ky

    Force exerted by gravity: -mg

    Call Mr. Newton to find the acceleration.
     
  20. Dec 26, 2011 #19
    -ky - mg = ma? ..........
     
  21. Dec 26, 2011 #20
    Just divide both sides to solve for a
    a = -ky/m - g
    The original question asked "what equation describes the acceleration when the ball is y is above the original position of the top end of the spring?" which is almost like asking find a(y), acceleration as a function of y.
     
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