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Platform on springs

  1. Aug 14, 2005 #1
    A platform of mass 0.8 kg is supported on four springs (only two springs are shown in the picture, but there are four altogether). A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s.

    The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6 cm below its original position.

    -->What is the effective spring constant of all four springs taken together?

    i did the following: F = kx

    so F/x = k

    i'm thinking the only force contributing to this is dropping (force) of the modeling clay on the platform. or does the weight (gravity) of the platform contribute to any of this?
     
  2. jcsd
  3. Aug 14, 2005 #2

    Astronuc

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    Presumably, the mass of the platform is considered in the initial position of the springs.

    The additional mass of the clay displaces the springs further.

    At rest, this becomes a static problem.

    If one is considering the oscillation, then one would have to consider the combined mass.
     
  4. Aug 14, 2005 #3

    Fermat

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    If the table has come to rest, then the velocity with which the clay struck the table and any resulting oscillations are irrlevant.
    What you did is pefectly correct. Also, the weight of the platform doesn't get involved since we are told it stopped (6 cm) below its original position - so it's only the 6 cm compression that has to be considered and all of that is provided by the (static force) of the weight of the clay on the platform/spring combination.
     
  5. Aug 14, 2005 #4

    HallsofIvy

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    The only thing that is relevant here is that you are told the mass of the clay and that the table comes to rest 6 cm. below its original height. In other words, compressing the spring 6 cm= 0.06 m will support a force of 0.8g = 0.8(9.81) Newtons.

    The fact that clay was originally dropped from a height affects the oscillation but not the final equilibrium position. The mass (and weight) of the table itself determines the original height of the table. It is the added weight of the clay that causes the 6 cm compression.
     
  6. Aug 14, 2005 #5
    ok, the figured out the k = 98.1 N/m,

    next the problem asks:

    With what amplitude does the platform oscillate immediately after the clay hits the platform?

    i first used the momentum conservation, and found the velocity of the clay+platform system is equal to .36

    since total E = .5kA^2

    i'm not sure here, but i set initial E = final E like the following:

    initial KE + initial PE = final KE + final PE
    .5mv^2 + 0 = .5(m of clay + M of platform).36^2 + .5kx^2

    then the final KE equal to .5kA^2, it got me the incorrect answer, but is there anything wrong with this though?
     
  7. Aug 14, 2005 #6

    HallsofIvy

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    At maximum compression (the amplitude of the oscillation), the speed of the clay and table is 0, not ".36" (that's the speed of clay and table immediately after impact which is not relevant).
     
  8. Aug 14, 2005 #7
    well, the book has pretty much the same example, it says to find the final speed of the clay and platform system, find kinetic energy of the system after collision then set this KE to .5kA^2, tried that example myself, it didnt work out ...
     
  9. Aug 15, 2005 #8

    HallsofIvy

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    But you didn't do that, did you? You had ".5mv^2 + 0 = .5(m of clay + M of platform).36^2 + .5kx^2" That ".4kx^2" is the ".5k A^2" but you haven't set that equal to the kinetic energy after collision. The kinetic energy of the clay, platform combination immediately after collision should be exactly the same as the kinetic energy of the clay alone just before collision- that's what I was using.
     
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