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Spring forces, etc

  1. Jun 16, 2008 #1
    Spring forces, etc.... :(

    1. The problem statement, all variables and given/known data

    Doing past exams before a real exam and i completely missed this whole section of lectures. lol

    A Block of mass 0.800kg is given an intitial velocity v = 1.2m/s to the left and collides with a light spring of force constant k = 50.0N/m

    2. Relevant equations

    If the surface is frictionless, calculate the maximum compression of the spring after the collision.

    3. The attempt at a solution

    I have absolutely no idea how to begin this questions. any help with these sorts of questions or this one in particular would be great
  2. jcsd
  3. Jun 16, 2008 #2
    Hey alex9050. Welcome to PF.

    First off, try think what the block will be doing when the spring is fully compressed. At what speed will it be traveling?
  4. Jun 16, 2008 #3

    scroll to the bottom... theres almost exactly the same problem except with different numbers... read it carefully because i believe they made an error by putting -250 instead of -500 at the bottom... just plug in your numbers and try and understand the work energy laws
  5. Jun 16, 2008 #4
    Sorry, what are you trying to look for in your original question?
  6. Jun 16, 2008 #5
    Im not really sure what the question is... but just based on the i would start by finding the kinetic energy of the block, since we know the surface is frictionless, think about where all that energy has to go when it hits the spring.
  7. Jun 16, 2008 #6
    we're trying to calculate the maximum compression of the spring made by the block before the block goes in the other direction

    and ed, i do understand the block will be stationary for a split second, but i have no idea on the formula to use to answer the question :$
    Last edited: Jun 16, 2008
  8. Jun 16, 2008 #7
    and this is a work done question.
  9. Jun 16, 2008 #8


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    Equate the kinetic energy in the block to the potential energy in the spring at maximum compression, since then the block will have no kinetic energy, and the energy must have gone somewhere.
  10. Jun 16, 2008 #9
    so that means that its 1/2kx^2 = 0 ? then just rearrange and solve ? because the KE = .5mv^2 = 0 because velocity is 0 at the point of maximum compression ?
  11. Jun 16, 2008 #10
    the equations you will want to use are

    Ek= (1/2)(m)(v^2 - u^2) where u is the initial velocity
    Eel= (1/2)(constant)(Δx^2) where Δx is how far the spring is compressed.

    see if you can figure out where to go from there.
  12. Jun 16, 2008 #11


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    Nooooo. (1/2)*kx^2 is equal to the INITIAL kinetic energy. At the point where v=0 all of the initial kinetic energy has been transferred to potential energy in the spring.
  13. Jun 16, 2008 #12
    yes the velocity will be 0 at the point of maximum compression. but thats not what you want to use. you want to find the kinetic energy before it compresses the spring. so use the .5mv^2 before it gets to the spring and set things equal. sorry for the last post, i didnt see you had the formulas. but find the kinetic energy before it ever gets to the spring.
  14. Jun 16, 2008 #13
    wait wait, working it out now.
  15. Jun 16, 2008 #14
    i got 0.152 so 0.152M?
  16. Jun 16, 2008 #15
    it would be positive because the compression in the same direction as the velocity you use.
  17. Jun 16, 2008 #16


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  18. Jun 16, 2008 #17
    that is what i got too.
  19. Jun 17, 2008 #18


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    MUST be right then. :)
  20. Jun 17, 2008 #19
    so i ended up as .5*.8*(1.2)^2 = .5*50 *x^2 does that sound right ?

    equaling 0.152m
  21. Jun 17, 2008 #20


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    I think we are all agreed that that's right.
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