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Spring-energy question

  1. Jun 23, 2011 #1

    Femme_physics

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

    http://img198.imageshack.us/img198/9417/springenergy.jpg [Broken]

    A)

    A body moves on a horizontal surface at a constant velocity of V = 2 m/s as described in pic A. In its way it hits point B at the spring in its horizontal position, whose in relaxed position. The mass of the body is m = 1 kg and the spring's constant is k = 80 N/m

    Calculate how much will the spring shorten as a result from the body hitting it presuming the surface is frictionless.

    B)

    The body is released from the spring at point B, at an initial velocity of V = 2 m/s as described in pic B. At what horizontal distance from point B will the body move to after being released from the spring, if the kinetic friction coeffecient between the body and the plain is μ = 0.25. The mass of the body is idential to caluse A.

    You must do the calculations in 2 ways:

    1) By considering work and energy
    2) By Newton's second law (sigma F = ma)

    2. Relevant equations



    3. The attempt at a solution

    http://img705.imageshack.us/img705/8687/mghinaction1.jpg [Broken]

    http://img8.imageshack.us/img8/3046/mghinaction2.jpg [Broken]

    ( I don't have the solution manual answer to that, but I think this is correct )

    I'm getting a bit stuck at clause B. I wanna draw a free body diagram, but I'm not sure what I should use for force. Should I use

    F = 80 x 0.224

    Or should I discover my force doing sum of all forces on x = ma?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 23, 2011 #2

    I like Serena

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    Hi femme-physics! :smile:

    When your problem "starts", the spring is fully extended in its relaxed position.
    Since the mass is not attached, the spring does not take part in this problem anymore.

    So what other forces do you have, and how is friction calculated?

    In the other spring-problem, you had to do the same thing.
    Remember how you did it?
     
  4. Jun 23, 2011 #3

    Femme_physics

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    Hi ILS :)

    Since in clause A there's no friction I can just ignore it and use conservation of mechanical energy (which I did do) to get the answer. I got the right answer didn't I?

    Hmm you're right. I hope that works?

    http://img585.imageshack.us/img585/3302/felw.jpg [Broken]

    http://img4.imageshack.us/img4/7868/fel2.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Jun 23, 2011 #4

    I like Serena

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    Yes. And you did! :smile:


    Well, there are 4 things I would nit-pick about.
    But yes, this works, and you have the right answer. :)
     
    Last edited: Jun 23, 2011
  6. Jun 23, 2011 #5

    Femme_physics

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    w00t! Thanks! ;)

    I just didn't understand if they wanted me to do BOTH clauses with both methods getting to the same result, or just do each clause in a different method (not that I know any other way to do it)

    But what would you nit pick about?
     
  7. Jun 23, 2011 #6

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    :D

    Indeed there is no other way to do it!


    In order of importance (most important first):
    1. You have a formula in which you use Fs Δx, but this should be a Δx.
      That is the first formula in the second scan.
      (Otherwise there would be no purpose in calculating a.)
    2. You calculate Fk and use μk, but this should be Fs and μs.
      (This would confuse your students, since they won't be able to find Fk.)
    3. In the same formula where you use Fs Δx, you have swapped vo and vf. Good that you swapped the numbers too when you plugged them in.
      (Again, this would confuse your students.)
    4. In your free body diagram you have included Fel and the spring, but we concluded that they play no role in this problem. And then you left Fel out in your forces sum!
      (Your students may be confused no longer since they might have given up! :wink:)

    Still, you did a good job! I think you're starting to "get" this energy-and-work stuff! :smile:
     
    Last edited by a moderator: May 5, 2017
  8. Jun 23, 2011 #7

    Femme_physics

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    thanks for the encouragement ;)

    You're right. I initially plugged "a" but then I saw that in my other thread I plugged Fs but it wasn't corrected so I wasn't sure what to think. I'm glad you straightened me out :)

    Oh, how peculiar. Yet, the question gives us only the coeffecient of Fk, and we calculate Fs using that Fk coeffecient?!?

    Swapped? This is the formula:

    theformula.jpg

    I plugged in 0 for Vf (the distance it stops), and 2 for Vo (its initial velocity). I don't see where's the mistake here?

    LOL -- Don't worry, I haven't taken up on explaining spring-energy problems, yet.

    Does it really play no role? Did they really gave this figure out for no apparent reason but to confuse the students?
     
  9. Jun 23, 2011 #8

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    In the other thread it was correct, since you used a different formula. ;)
    A formula that included the mass m in the kinetic energy.


    Looking at your problem statement, the only k that I see is the spring constant k.
    But this k is totally unrelated to friction.
    As you can see, this would cause confusion.

    And actually you have a friction coefficient which you named μ, and not μk.



    Hmm, not sure what I was thinking, because you're right! :redface:


    Yes, it really plays no role.
    The problem maker actually made it easier (depending on your point of view).

    The real problem is a mass that runs into a spring and is bounced back.
    In the problem statement this was split into 2 separate problems.
    The first where friction is neglected, and a second where the spring does not play a role.

    So they broke the real problem down for you. Wasn't that nice of them? :wink:
     
  10. Jun 23, 2011 #9
    I think you got it, but what did you mean by Fel? What's "el"? I have never seen anyone write the distance-velocity-acceleration formula like that...

    It took me a while to recognize that you had it in the work-energy theorem.
     
  11. Jun 23, 2011 #10

    Femme_physics

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    Oh, right.

    Well they named it that way. But that did say: "f the kinetic friction coeffecient between the body and the plain is μ = 0.25."


    w00t! :D

    I guess it was! But I'm really surprised they gave me a figure I'm not supposed to use. But so be it!
     
  12. Jun 23, 2011 #11

    Femme_physics

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    "Force elastic" -- that's how they write spring-force it in our solution manual :shy:I just copied the notation.
     
  13. Jun 23, 2011 #12

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    Huh? :uhh:
    Aren't you saying the same thing as I did?

    Note that this is the second part of the sentence, that says after the spring became irrelevant (my interpretation of the first part of the sentence).
     
  14. Jun 23, 2011 #13
    Now try to figure out the initial compression in question 2), that's a challenge for you!
     
  15. Jun 23, 2011 #14

    Femme_physics

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    I don't get it, I am given the kinetic coeffecient of friction but in my solution I use the static one? How do I even know that the static one and kinetic one match?

    @ flyingpig - One step at a time heh!
     
  16. Jun 23, 2011 #15

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    Ahaaaaaa!!! :biggrin:
    I finally pinpointed your use of the subscripts "s" versus "k".
    They're abbreviations for kinetic and static!
    (Usually I use the subscript "f" for friction. :wink:)

    In all the problems you posted I think it has always been about kinetic friction, hasn't it?

    Use of Fk would be ok then! :blushing:

    But you should replace all references to Fs by Fk then.
    The work done by static friction would be negligible (it only "works" in, say, the first millimeter, which makes its "work" very small compared to the work of the kinetic friction).
     
  17. Jun 23, 2011 #16

    Femme_physics

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    Ahhhhh! :biggrin:

    That clears it. So, static friction is negligible I see.

    Done and done ;) thanks master!
     
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