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Homework Help: Am I underthinking this work problem?

  1. May 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass m is dropped from rest above a relaxed spring of stiffness k a distance D. Find the position from where it was released where the mass attains its maximum velocity and find that maximum velocity

    2. Relevant equations
    [tex]W_{net}=\Delta E[/tex]

    3. The attempt at a solution
    ok... so obviously the mass is going to be at maximum velocity when it as distance D from the place of drop right? I stated that we can bascially ignore the spring in the problem because at the maximum velocty, the mass will JUST hit the spring and the spring wont have any impact or do any work on the mass.

    so here is what i did;

    [tex]W_{net}=\Delta E[/tex]

    There is no work in this problem so,

    [tex]0=\Delta GPE + \Delta KE [/tex]
    [tex]0 = \frac{1}{2}mv_f^2-mgD [/tex]
    [tex]\frac{1}{2}mv_f^2 = mgD [/tex]
    [tex] v_f = \sqrt{2gD} [/tex]

    am i missing something crucial?
  2. jcsd
  3. May 17, 2015 #2
    ugh, also can someone tell me why everything I type in latex is centered? I use the same code for mathhelpforum and it works fine ther. here is the code I used for the last latex equation i wrote:

    [ tex]v_f=\sqrt{2gD} [ /tex]

    whats wrong with that above code that that makes everything centered?
  4. May 17, 2015 #3
    Consider a limiting case of small d and small k. That is, the weight has basically no time to accelerate before hitting a very weak spring.
  5. May 17, 2015 #4


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    Given that the first thing it asks for is the position the mass was dropped from, I suggest D is not what one would assume. Perhaps it is the maximum distance the mass falls?
    Either way, as Insightful says, you are wrong about the max speed being where it first contacts the spring. Think about acceleration.
  6. May 17, 2015 #5
    well the longer time it falls for the higher its velocity is correct? D is the distance from the position of release to the top of the spring (so just before the mass would contact the spring) so wouldnt that be the position where the velocity would be maximum?
  7. May 17, 2015 #6


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    No. See my edited reply.
  8. May 17, 2015 #7
    yeah, im not understanding where I'm going wrong. the mass is in free fall.... so the acceleration is 9.8. the acceleration is going to be 9.8 until it hits the spring where it's kinetic energy is transferred to elastic potential energy. do you mind telling me how the mass is not at its max velocity right before it hits the spring?
  9. May 17, 2015 #8


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    Suppose it has hit the spring and depressed it a small distance, so the spring force is still quite small. Which way is it accelerating at this point?
  10. May 17, 2015 #9
    oh... ok. it is still accelerating downward. so my answer only work for springs that have a pretty large k value correct? (when the spring force is greater than mg).
  11. May 17, 2015 #10
    how do i find when the spring force is greater than the gravitational force?

    through a side F=ma problem, found that:

    [tex] k =\frac{m(a+g)}{x} [/tex]


    [tex] a = \frac{mg-kx}{m} [/tex]

    is that helpful in solving the priblem?
  12. May 17, 2015 #11


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    Yes. By that, when is a zero?
  13. May 17, 2015 #12
    alright. i fixed my answer a bit. doing an F=ma problem I did the follwing with the mass as the system and the y going down...

    [tex] F_g - F_{el} = ma [/tex]
    we get [tex] a=\frac{mg-kx}{m} [/tex]

    we want to find the point at which a turns from positive to negative (becuase thats when acceleration changed directions)

    so [tex] mg-kx > 0[/tex]
    is true as long as
    [tex] x< \frac{mg}{k} [/tex] correct? where x is the length compression.

    now we do the entire problem again:

    now I think that the distance where velocity is greatest is the following:

    [tex] D+\frac{mg}{k} [/tex]

    system: mass, earth, and spring:

    work = change in energy
    [tex] 0 = \Delta GPE + \Delta EPE + \Delta KE [/tex]
    [tex] = -mg(D+\frac{mg}{k}) + \frac{m^2g^2}{2k}+ \frac{1}{2}mv^2 [/tex]
    [tex] v =\sqrt{-2\bigg{(} \frac{-mg(D+\frac{mg}{k}) + \frac{m^2g^2}{2k}}{m}\bigg)} [/tex]
    Last edited: May 17, 2015
  14. May 17, 2015 #13
    is my above solution correct?
  15. May 17, 2015 #14


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    You have some sign issues (is the change in GPE positive or negative?), and mg/2 is not an energy, it' a force.
  16. May 17, 2015 #15
    wait wait. dont correct me. i see an error.
  17. May 17, 2015 #16
    i think i fixed it now.
  18. May 17, 2015 #17
    Check your final rearrangement to get v.
  19. May 17, 2015 #18
    my bad. there should be a negative inside the radical right?
  20. May 17, 2015 #19
    Why is 1/2mv^2 under the radical?
  21. May 17, 2015 #20
    oh thats an accident as well. i forgot to take it out... yeah i make way too many stupid mistakes
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