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Block falls on a spring

  1. Nov 17, 2007 #1
    A 9.3 kg block is dropped onto a spring of spring constant 1086 N/m from a height of 800 cm. When the block is momentarily at rest, the spring has been compressed by 50 cm. Find the speed of the block when the compression of the spring is 10 cm. The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.


    My attempt:

    Ki + Ui = Kf + Us
    Ki = o so...
    Kf = Ui - Us
    1/2 mv^2 = mgh - 1/2kx^2

    v = (sqrt)mgh - 1/2kx^2/2m
    v = (sqrt)(9.3)(9.81)(8) - 1/2(1086)(.1)^2/2(9.3)
    v= 6.240838447m/s which is wrong.
     
  2. jcsd
  3. Nov 17, 2007 #2

    dynamicsolo

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    This is a favorite homework or exam problem because it catches just about everybody out. Don't forget that the block continues to descend as the spring is compressed, so the change in gravitational potential energy is a bit larger than what you get from falling just 8.00 m.

    Be careful also about how you divided out that 1/2 when you were solving for v. There is at least one algebra error...
     
  4. Nov 17, 2007 #3
    uhh... so should it be 8.5m instead of just 8? and with the 1/2 fix it should be

    v = (sqrt) 2(9.3)(9.81)(8.5) - (1086)(.1)^2/(9.3) ???(on my last try so i cant afford to screw it up again :(
     
    Last edited: Nov 17, 2007
  5. Nov 17, 2007 #4

    dynamicsolo

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    I take it you're on one of those computer-entry problem systems. Could you check the problem statement again, please? As I've been working through the numbers, the spring constant seems inconsistent with the other information in the problem. (This has become important now that you tell me you're down to one more try.)
     
  6. Nov 17, 2007 #5
    m = 9.3 kg
    Dropped from 800 cm = 8 m
    gravity = 9.81 m/s^2
    Spring Constant = 1086 N/m
    When the block is momentarily at rest, the spring has been compressed by 50 cm = .5 m

    and the question asks for the speed of the block when the compression of the spring is 10 cm = .1 m

    those are all the values given in the problem
     
  7. Nov 17, 2007 #6

    dynamicsolo

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    All right, here's the thing. The energy you get by dropping the block by 8 meters should compress the spring a lot more than 0.5 m. I can find no way to make the first part of the problem consistent with the second part by adjusting any of the numbers in a way that could be caused by a typographical error. You may need to talk to the course instructor about this one. (It wouldn't be the first time there was a mistake in a computer system problem...)

    In fact, the problem is really overdetermined as it stands. You don't need more than two of the following: the height from which the block is dropped, the spring constant, or the maximum compression distance of the spring. One of the three values for these in the problem would seem to be in error.

    Compare the potential energy released by the block falling 8 + 0.5 meters with the spring potential energy when it is compressed 0.5 meters. There is a substantial discrepancy -- or are we supposed to consider that there is a large transformation of mechanical energy?

    If you must solve this problem during the weekend, I'd take a chance that the maximum compression distance and spring constant are correct and use that information to find the speed of the block when the spring was compressed by 0.1 m.

    I'm sorry I can't be more helpful than this, but it looks like there is a mistake in the statement of the problem. (That wouldn't be the only time that's happened this month...)
     
  8. Nov 17, 2007 #7
    Hey, thx for your help. I managed to get the correct answers too lol. Here it is for someone who might have a similar problem.

    Kf + Us = Ki + Ui
    1/2mv^2 + 1/2kx^2 = mgh (Ki = 0 so it is dropped)

    [move everything over so you get v alone]

    v = [sqrt] (2gh - (kx^2/m))
    v = [sqrt] (2*9.81*8 - (1086*.1^2/9.3))
    v = 12.56 m/s
     
  9. Nov 17, 2007 #8

    dynamicsolo

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    I get your answer using 8.1 m, the complete descent of the block to this point. (If you use only 8.0 m, you get v = 12.48 m/sec, which is probably still within the accepted precision for the answer.)

    If you use this to find the compression at which the velocity will be zero, you get about 1.25 m (which is about what I was finding earlier), with the block descending a total of 9.25 m. So the statement about where the block is momentarily at rest should be discarded from the problem as it is, because the 50 cm. figure is wrong (and unneeded)...
     
  10. Feb 20, 2008 #9
    I used this equation for my exact same problem and got it wrong twice so far. I only have two tries left.

    A 2.45 kg block is dropped onto a spring with a force constant of 4223 N/m from a height of 4.82 m. When the block is momentarily at rest, the spring is compressed by 24.0 cm. Find the speed of the block when the compression of the spring is 13.3 cm.

    I even reworked the problem myself and it makes total sense to me. I ended up with 9.71m/s as my answer yet it says it's incorrect.

    Am I missing something here?
     
  11. Feb 20, 2008 #10
    ok figured it out. You do have to add the amount of spring compression to the height.
     
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