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Energy of a Falling Elevator

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data
    In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator.

    1) What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring?

    2) When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?

    2. Relevant equations
    K1+U1 = K2+U2-Work
    K=.5*m*v^2
    U=mgh; .5*k*x^2
    W=FD

    3. The attempt at a solution
    Using the formulas I got above, I came up with:
    k=8500 N/m??
    .5*2000*4^2+2000*9.8*2=.5*2000*v^2+.5*8500*1^2 - 17000.
    Doing this gave me v = 8.24 m/s^2 which I don't think is right...

    I can't begin part 2 until I understand part 1, so no attempt at the moment.
     
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2

    mgb_phys

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    First you need to work out the spring constant.
    The energy stored in a spring is E = 1/2 k x2 as you said.
    The energy in the falling elavator is E = 1/2 m v2
    Energy in the brake is E = f x

    Assuming the question means the elavator would stop with the spring compressed by 2m then all the kinetic energy of the falling elevator (except that lost to the brake) goes into the spring - you can get k.
     
  4. Oct 7, 2008 #3
    So,
    .5*m*v^2 = .5*k*x^2+f*x
    .5*2000*16 = .5*k*4+17000*2

    If thats correct, then k = -9000 N/m.

    Plugging that back into my original of:
    .5*2000*42+2000*9.8*2=.5*2000*v2+.5*(-9000)*12 - 17000(1).

    I get v=8.76 m/s^2 which is still incorrect.
     
  5. Oct 7, 2008 #4

    mgb_phys

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    You can't really have a negative spring constant.
    According to the numbers given the brake will stop the elveator in less than the two meters - are you sure you copied it correctly?
     
  6. Oct 8, 2008 #5
    Yes, I copied it word for word.
    Even if I use k = 9000 N/m, I still come out with the wrong velocity of 8.23.

    If you know what you are doing, and sure you are right, that might be possible that the elevator stops before it reaches 2m, but the problem should still be valid if it is not stopped at 1m.
     
  7. Oct 8, 2008 #6

    mgb_phys

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    I discussed it with someone else, we forgot the loss of potential energy as the lift falls the last 2m!
    So the energy lost to the spring compression is KE + mgh, where m=2000kg g=9.8m/s^2 h=2m
     
  8. Oct 9, 2008 #7
    Ok, so using:
    Initial Kinetic Energy + Potential Energy due to Gravity = Energy of the Spring + Energy of the Frictional Brake

    .5*2000*42+2000*9.8*2 = .5*k*22+17000*2
    I get k = 10600 N/m

    Then plugging it back into:
    Initial Kinetic Energy + Potential Energy due to Gravity = Energy of the Spring + Final Kinetic Energy + Energy of the Frictional Brake
    .5*2000*42+2000*9.8*2=.5*2000*v2+.5*10600*12 + 17000*1

    I get the velocity = 5.74 m/s2

    Let me know if these values are correct before I proceed with the rest of the problem.
     
  9. Oct 9, 2008 #8

    mgb_phys

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    Looks reasonable
     
  10. Oct 9, 2008 #9
    Hmmm, still incorrect...
     
  11. Oct 9, 2008 #10

    mgb_phys

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    Before contact:
    KE = 1/2 mv^2 = 0.5 * 2000 * 4^2 = 16Kj
    PE = mgh = 2000 * 9.8 * 2 = 39.2Kj

    Energy lost to brake = f d = 17000 * 2 = 34kj
    Energy lost to spring = 1/2 k x^2 = 0.5 * k *2^2 = 2k
    2k + 34Kj = 16Kj + 39.2Kj, k = 10600 N/m

    Speed after 1m, initial ke + change in pe = brake loss + spring compression + final ke
    (0.5 * 2000 * 4*4) + (2000*9.8*1) = (17000*1) + (0.5*10600*1*1) + (0.5*2000*v^2)
    16Kj + 19.6kj = 17kj + 5.3Kj + v^2
    v = sqrt(16+19.6kj - 17+5.3Kj) = 3.6 m/s

    Note - you used 2m loss to PE, the elevator stops after 1m.
     
  12. Oct 9, 2008 #11
    Ah, that is the right answer. Thanks for pointing out my dumb mistake.

    And now for part b, the acceleration:
    I don't know how to approach this problem at all. Maybe,
    Sum of the Forces = ma; where the sum of the forces would be 1/2kx^2-frictional force?

    I don't know, Im stumped.
     
  13. Oct 9, 2008 #12

    mgb_phys

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    Add up all the forces acting on the elevator (don't forget gravity - and watch the signs)
    Then ,as you said, f=ma
     
  14. Oct 9, 2008 #13
    In the up direction of my FBD:
    Force of friction = 17000

    In the downward direction of my FBD:
    Weight = m*g = 19600
    Spring force = k*x = 10600

    So, up - down:
    17000-19600-10600=ma; a= -6.6 m/s^2 which is not right. This seems logical, but where did I go wrong?
     
  15. Oct 9, 2008 #14

    mgb_phys

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    Which direction does the spring force act?
     
  16. Oct 9, 2008 #15
    Ugh I'm so dumb and it makes me so mad. I should have considered your initial warning more. Okay, that makes perfect sense now. Thank you soooooo much for your help, it was much appreciated towards not only getting the correct answer, but fully understanding the problem at hand.
     
  17. Oct 9, 2008 #16

    mgb_phys

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    One step at a time and draw lots of diagrams!
     
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