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Homework Help: Spring (conservation energy) with friction

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass 1.6 kg is attached to a horizontal spring that has a force constant 900 N/m. The spring is compressed 2.0 cm and is then released from rest.

    a) A constant friction force of 3.8 N retards the block's motion from the moment it is released. At what position x of the block is its speed a maximum?

    K=900 N/m

    2. Relevant equations

    PE=.5Kx^2 (elastic potential energy function from the spring)
    PE[a]+KE[a]=PE+KE, where [a] is the initial position of the question, and is where the position of x.

    3. The attempt at a solution

    Since [a] is the initial position, then the equation that I wrote above will get rid of KE[a], since there is no Kinetic energy involved (It is at rest). Furthermore, at , it will get rid of PE because when finding the speed is max, PE is zero and KE is at the max.

    That being said, what throws me off is the friction that slows down the block after it is released.

    So my next guess was since we got rid of KE[a] and PE, I put the equations accordingly like this:


    But I dont know where to apply the Friction given.
  2. jcsd
  3. Mar 15, 2009 #2


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    Welcome to PF!

    Hi neotriz! Welcome to PF! :smile:

    The work-energy theorem states that work done = energy lost,

    so PE[a]+KE[a]=PE+KE + work done :wink:
  4. Mar 15, 2009 #3
    Re: Welcome to PF!

    Thanks :smile:

    But I still dont get it.
    If it that was the case, how would I find the velocity at ?

  5. Mar 15, 2009 #4
    Also, the answer must leave in cm unit. So I guess I have dont have to change the unit when it is compressing 2cm, right?
  6. Mar 15, 2009 #5


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    Just call it v
    No, wrong!

    Change into m for formulas like 1/2 kx2, and change back again at the end, or the squares can get you into serious trouble. :frown:
  7. Mar 15, 2009 #6
    I dont understand

    When I do the calculation, I have everything except the displacements (the one it is asking me to find) and the max velocity. How would I find the latter one?

    One more thing: Do we assume at x, PE is zero?
  8. Mar 15, 2009 #7


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    You can set PE = 0 anywhere … PE is always relative

    but it makes the maths a lot easier if you set it at 0 at the uncompressed length :wink:
    Show us how far you've got with your calculation :smile:
  9. Mar 15, 2009 #8
    Here's what I have:

    PE[a]=KE+Work done by friction (correct)?


    .5K(x)^2 = .5mv^2 + Work done

    .5(900N/m)(2.00 X 1-^-2 m)^2 = .5(1.6)v^2 + 3.8N(s)

    NOTE: I changed 2.00 cm to 2.00 X 10^-2

    Through alebraic manipulations, I get:

    .18 = .8v^2 + 3.8(s), where I guess (s) is what I am trying to find
  10. Mar 15, 2009 #9


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    No, you need .5Kx2 (with different x) on both sides

    and your s is related to x
  11. Mar 15, 2009 #10
    So that means
    what I wrote before:
    PE[a]=KE+Work done

    "The Work Done" is included the Friction's and the Spring's at x?
  12. Mar 15, 2009 #11


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    perhaps I should have been clearer …

    the work done in that equation is only the work done by friction

    (technically, PE is work done, but when it's a conservative force, such as gravity or a spring, we call it PE instead)
  13. Mar 15, 2009 #12
    So where is the other .5Kx^2 from the otherside is coming from?

    I'm really sorry, I'm new at this physics thinking
  14. Mar 15, 2009 #13


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    One side is the initial energy of the spring, the other side is the energy of the spring at a general time
  15. Mar 15, 2009 #14
    But I thought at x, there is no PE, meaning KE is at max.
  16. Mar 16, 2009 #15


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    uhh? :confused:

    x is at max KE, so at min PE, not zero PE

    you can set PE = 0 at any position you like

    but if you want to use PE = 1/2 kx2, then x is measured relative to the uncompressed length, and the PE will therefore be 0 at the uncompressed length.
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