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Maximum distance a spring will stretch

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.070m. A 0.200kg lump of putty is dropped from rest onto the frame from a height of 30.0cm

    Find the maximum distance the frame moves downward from its initial position


    2. Relevant equations
    SEp+GEp+Ek = SEp+GEp+Ek

    mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2

    initial momentum = final momentum


    3. The attempt at a solution
    To find velocity of the putty just before it hits the frame:
    Ep = Ek
    mgh = (0.5)mv2
    (0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
    (v)(v) = 5.886
    v = 2.426 m/s

    To find the velocity of the putty and the frame after the putty hits:
    initial momentum = final momentum
    mv + mv = mv + mv
    (0.200)(2.426) + 0 = (0.200+0.150)v
    v = 1.386 m/s

    To find k of the spring:
    GEp = SEp
    mgh = (0.5)kx2
    (0.15)(9.81)(0.07) = (0.5)(k)(0.07)2
    k = 42 N/m

    To find the distance the frame moves downward:
    mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (1.5)(k)(x)2
    (0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(42)(0.070)2 = (0) + (0) + (1.5)(42)(h)2
    (3.4335h - 0.240345) + (0.439) = 21h2
    0 = 21h2 - 3.4335h - 0.199
    h = 0.20887 or -0.04537
    h - 0.07 = 0.1389 or -0.1153
    I figure the height (because of my reference) must be negative, so it must be -0.1153m

    Is this correct? I got a bit confused about h on the last equation ...
     
  2. jcsd
  3. Nov 4, 2011 #2

    PhanthomJay

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    You should have used m = 0.2, but since mass cancels, it makes no difference
    this looks good.
    no, you are off by a multiple. The frame is initially at rest in its equilibrium position at 0.07 m from the unstretched position of the spring.
    use the positive value of h, not the negative. Draw a sketch. Otherwise, your equation is correct after you change the k value.
     
    Last edited: Nov 5, 2011
  4. Nov 5, 2011 #3
    I am confused by this. Shouldn't I be able to use the equation

    SEp+GEp+Ek = SEp+GEp+Ek

    Where the left side of the equation is when the frame is raised and has no weight on the spring and the right side of the equation is when the frame is at rest 0.07 m from the un-stretched position of the frame. So Ek can be disregarded because the frame is at rest before and after so i would just use the equation

    SEp+GEp = SEp+GEp

    Since I am saying there is no weight on the spring before, and I am putting my height reference point at where the frame is at rest after, the equation is

    GEp = GEp
    mgh = (0.5)kx2

    I am still missing where I messed up
     
  5. Nov 5, 2011 #4

    PhanthomJay

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    The frame is at rest at 0.07 meters below the unstretched spring position because you slowly lowered it with your hand, applying a force until it comes to a slow stop. By applying a force, mechanical energy is NOT conserved, since you have done non conservative work. Only if you let the mass drop from the unstretched spring position would mgh =1/2 kx^2, and then x would not be 0.07 m.
     
  6. Nov 6, 2011 #5
    It doesn't say that the frame is slowly lowered by your hand. Couldn't you assume that it is simply dropped and that there are no external forces because it should end up 0.07m lower at rest whether you use your hand or not. If you didn't use your hand, i understand that it would drop lower that 0.07m but it would rebound upwards (so it would have kinetic energy and mgh would not equal 1/2kx^2), correct?

    Also, I cannot think of another way to solve for k of the spring without using the conservation of energy. What other way is there in this situation?
     
  7. Nov 6, 2011 #6

    PhanthomJay

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    It is implied. Otherwise, for an ideal spring, the frame would be bouncing up and down like a yo-yo, forever.
    No, it will not be at rest at 0.07 m. In fact, it will have its max speed at that point, although the acceleration at that point will be 0.
    yes, to 0.14 m
    It would rebound upwards, and mechanical energy would always be conserved (PE_s +PE_g + KE = constant at any point
    Hooke's Law!
     
  8. Nov 8, 2011 #7
    My second attempt at the solution:
    To find velocity of the putty just before it hits the frame:
    Ep = Ek
    mgh = (0.5)mv2
    (0.150)(9.81)(0.30) = (0.5)(0.150)(v)(v)
    (v)(v) = 5.886
    v = 2.426 m/s

    To find the velocity of the putty and the frame after the putty hits:
    initial momentum = final momentum
    mv + mv = mv + mv
    (0.200)(2.426) + 0 = (0.200+0.150)v
    v = 1.386 m/s

    To find k of the spring:
    Fg = Fs
    mg = -kx
    (0.15)(-9.81) = -k(0.07)
    k = 21.02 N/m

    To find the distance the frame moves downward:
    mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2 = mgh+(0.5)(m)(v)2 + (0.5)(k)(x)2
    (0.35)(9.81)(h-0.07) + (0.5)(0.35)(1.386)2 + (0.5)(21.02)(0.070)2 = (0) + (0) + (0.5)(42)(h)2
    (3.4335h - 0.240345) + (0.336) + (0.051449) = 21h2
    0 = 21h2 - 3.4335h - 0.1475
    h = 0.1988m or -0.0353m
    h - 0.07 = 0.1288m or -0.1053m

    So the answer (which is positive) must be 0.1288m

    I have put this as the answer, but it says it's still wrong. Any ideas where I went wrong this time?
     
  9. Nov 8, 2011 #8
    Ooops, I just realized that I should have added 0.07 m to the positive value of h, so it would be 0.1988 + 0.07 = 0.2688 m. It also says that this is wrong ...
     
  10. Nov 8, 2011 #9
    The 0.15kg frame has a weight of 1.47N and this causes an extension of 0.07m.
    This gives a spring constant of 21N/m
    A 0.2kg lump of putty dropped from 0.30m will have KE (= PE) of 0.589J
    When the putty lands on the platform the spring will extend and this energy is stored in the spring.
    The energy stored in a spring is 0.5F x ext or,
    since k(spring constant) = F/e
    F = k x ext then energy stored = 0.5 x k x e^2
    The spring will stop extending when the 0.589J of KE have been stored (absorbed)
    so 0.589 = 0.5 x 21 x e^2
    This gives e (extension) of 0.237m
    This is the maximum extension and the spring will oscillate.
     
  11. Nov 8, 2011 #10
    EXTRA
    when the oscillations stop the spring will essentially have a frame of mass 0.15kg with a 0.2kg lump of putty. The total mass is 0.35kg, weight = 0.35 x 9.81 = 3.43N and therefore an extension of 0.16m.
    The falling lump of putty caused an extension of 0.237m over and above the original extension of 0.07m caused by the frame alone.
    This means that the oscillations would be greater than the original extension and the whole thing would be bouncing around a bit.
     
  12. Nov 8, 2011 #11

    PhanthomJay

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    The problem is looking for the max stretch of the spring from the initial position of the frame. It's original position is 0.070 m beloe the unstretched lenght of the spring. Thus, you are looking for the positive value of (h - 0.07), once you correct the k value in your last term. Also, the problem has 3 significant figures in the given values, so round off your answer to 3 sig. figs.
     
  13. Nov 8, 2011 #12
    Wow, thank you so much PhanthomJay! You are a life saver!

    So I got h=0.365 or -0.0384. Take the positive (0.365) and minus 0.07 and I get 0.295m. This is now correct?
     
  14. Nov 8, 2011 #13

    PhanthomJay

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    It looks good to me! Are you entering your answer one last time ...3 strikes 'yer out?
     
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