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Springs work and energy

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring of constant 18 N/cm is compressed a distance 7cm by a 0.5 kg mass, then released. It skids over a frictional surface of length 1.6m with coefficient of friction 0.17, then compresses the second spring of constant 5 N/cm. the acceleration of gravity is 9.8m/s^2


    2. Relevant equations
    PE=1/2kx^2 --- potential energy = 1/2 kinetic length ^2
    KE=1/2mv^2 --- kinetic energy = 1/2 mass velocity ^2
    W=Fd ---work = force distance



    3. The attempt at a solution
    i converted all the cm to m.. then plugged them into EP=1/2kx^2.. so
    PE=1/2(.18)(0.07)^2
    PE=0.000441J
    but this already didn't look right, however, I still used it in W=Fd.
    W = 0.000441*1.6
    W = 0.000705600
    now this REALLY isn't looking right, and I really have NO idea what I'm doing, except getting random numbers. As much as I love physics (really, I use to want to be a physicist) i have never really understood what I'm doing as far as these harder problems go. If anyone could help me, that would be amazing. thanks!


    Mathieu
     
  2. jcsd
  3. Dec 1, 2009 #2
    Are you solving for energy lost by friction from the initial position (first compressed spring) to the final position (second compressed spring)? Is the mass connected to the two springs, or is the mass launched off the first spring and into the second spring? In you're attempt at a solution, you seem to substitute PE for F in the work equation. When dealing with energy, I think you use:
    W = [tex]\Delta[/tex]KE = -[tex]\Delta[/tex]PE = [tex]\int[/tex][tex]\vec{F}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex], because [tex]\vec{F_{spring}}[/tex]= -k[tex]\vec{x}[/tex]. If the mass is launched, then the initial spring energy is converted into an initial kinetic energy of the mass (-[tex]\Delta[/tex][tex]SE_{i}[/tex] = [tex]\Delta[/tex][tex]KE_{i}[/tex]). Some of that initial kinetic energy is lost as frictional energy, and the mass collides with the second spring with a final kinetic energy ([tex]KE_{i}[/tex] - FE = [tex]KE_{f}[/tex]). The final kinetic energy is converted into spring energy, and the mass stops (-[tex]\Delta[/tex][tex]KE_{f}[/tex]= [tex]\Delta[/tex][tex]SE_{f}[/tex]). I'm not sure what you're solving for, but I think this is how you would solve for energy lost. Feel free to correct me.
     
  4. Dec 1, 2009 #3
    I'm sorry..i didn't post the actual question...i'm so dumb haha its:

    How far will the second spring compress in order to bring the mass to a stop? answer in units of cm.

    and the mass isn't connected, the first spring launches it across the surface to the second spring
     
  5. Dec 1, 2009 #4

    rl.bhat

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    Homework Helper

    Hi Mathieu, welcome to PF.
    Check your conversions. 1 cm = 0.01 m.
    Then 18 N/cm = ....... N/m?
     
  6. Dec 1, 2009 #5
    0.18 N/m...correct?
     
  7. Dec 2, 2009 #6

    rl.bhat

    User Avatar
    Homework Helper

    The spring used in the problem requires 18 N force to stretch 1 cm. How much force is needed to stretch the same spring through 1 m?
     
  8. Dec 2, 2009 #7
    ki = 1800 N/m
    kf = 500 N/m
    xi = 0.07 m
    m = 0.5 kg
    r = 1.6 m
    [tex]\mu_{k}[/tex] = 0.17
    g = 9.81 m/s2

    -[tex]\Delta[/tex][tex]SE_{i}[/tex] = [tex]\Delta[/tex][tex]KE_{i}[/tex]
    -0.5kixi = 0.5mvi^2

    [tex]KE_{i}[/tex] - FE = [tex]KE_{f}[/tex]
    0.5mvi^2 - [tex]\mu_{k}[/tex]mgr = 0.5mvf^2

    -[tex]\Delta[/tex][tex]KE_{f}[/tex]= [tex]\Delta[/tex][tex]SE_{f}[/tex]
    -0.5mvf^2 = 0.5kfxf

    Substituting the KEs for PEs into the middle equation, we get
    -0.5kixi - [tex]\mu_{k}[/tex]mgr = -0.5kfxf
    xf = (kixi + 2[tex]\mu_{k}[/tex]mgr)/kf
    xf is how far the second spring will compress in order to bring the mass to a stop, and is in m, so convert to cm.
     
  9. Dec 2, 2009 #8
    Hey everyone, I'd thought I'd let everyone know that I did figure this out, our teacher conveniently forgot to teach us the equations for work done by non-conservative forces, so after learning that, it just randomly occurred to me how to solve it.

    The short version is this, take PE of the first spring, make that me initial energy in the Work done by non-conservative forces equation (Eo-Wnc=E) and take PE of the second spring and make that the final energy, so basically PE1 - Wnc = Pe2

    I explain in more detail here:

    PE=1/2kv^2, and Eo-Wnc=E
    (Wnc is work of nonconservative forces, Eo is starting energy and E is...final energy)

    Because of the law of conservation of energy, I thought I could do this:

    PE = E

    Or rather, because there are two springs:

    PE1 = Eo
    E = PE2

    So what I did was this:

    PE1=1/2k1x2
    PE1=1/2(1800)(0.07)2
    PE = 4.41J

    PE = Eo

    Eo-Wnc = E

    PE - Wnc = E

    PE - (mgμk*d) = E (Because W=Fd and F is = to Nμk (normal force mu k)

    4.41 - (0.5*9.8*0.17*1.67) = PE2

    3.01889 = PE2

    PE = 1/2 kx2

    3.01889 = 1/2(500)x2

    2(3.01889) = 500x2

    0.123088 = x2

    X = 0.109888852937866

    now because I converted everything to meters and the answer has to be in centimeters, the answer would be 10.9888852937866cm.

    And I'm proud to say I figured it all out by myself ^^ hahaha i feel smart now :)





    THANK YOU EVERY FOR YOUR HELP!!!!!

    I really appreciate it!


    Mathieu
     
    Last edited: Dec 2, 2009
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