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Elevator problem - help

  1. Oct 13, 2004 #1
    elevator problem - urgent help!!!

    The cable of the 1800 kg elevator snaps when the elevator is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the elevator against guide rails so that a constant frictional force of 4.4 kN opposes the motion of the elevator.

    (image attached - i don't know how to make it embedded, sorry)

    (a) Find the speed of the elevator just before it hits the spring._____ m/s
    (b) Find the maximum distance x that the spring is compressed.______ m
    (c) Find the distance that the elevator will bounce back up the shaft._____m
    (d) Using conservation of energy, find the approximate total distance that the elevator will move before coming to rest._____m


    alrite let's see....

    a) i've gotten the first answer by dKinetic + dPotGrav (or Ug) + E thermal (from friction) = 0 from conservation of energy.... solve for V final from the kinetic part... 7.37ms-1

    b) i believe it should be -1/2mv^2 - mgS + 1/2kS^2 + F(of friction)*S = 0 but i'm not sure if I'm getting the signs wrong... should my answer be negative because its compressing or not?.... not getting the right answer... help please!!!!

    c) now i need my result from b) as x and is it 1/2kx^2 = mg(d up) + Friction(s +d up) solve for (d up).... Note: not sure if should just use answer from b) times the friction force or is it friction * (d up)... instead of what my equation says....?????

    d) to be honest i'm sorry but i have no clue to this one... and without b) and c)... i'm having trouble understanding it...

    i hope some1 can help me out???!!! thanx in advance!!!

    Attached Files:

  2. jcsd
  3. Oct 13, 2004 #2
    Hmm, Im stuck on c) and d) too. but i can help you on b) you have it right but use the velocity you got in a) and if you have a TI-86 or higher just use solver to find distance from the quadratic equation you get.

    I used: .5kS^2=.5mv^2-(force of friction)S+mgS
  4. Oct 13, 2004 #3

    i solved the quadratic, but is it the negative s because its in compression or not?...

    i got either 0.90 or -0.72... i'm thinking its the positive... but i only have 1 try left... which one did u use? just wanna make sure!!!
  5. Oct 13, 2004 #4
    It is .9

    but your equations for C and D dont work (I did the same thing as you did in C and the answer was wrong) so try it yourself and if it is still wrong were gonna have to brainstorm.
  6. Oct 13, 2004 #5
    actually, ur right.... friction only acts during d up... my original equation had friction acting at (s + d up)

    it should be:

    1/2kx^2 = mg(d up) + Friction(d up)
  7. Oct 13, 2004 #6
    other ?

    hey randomphyre, i was wondering... have u gotten around to ch8 #66 part (c)... any clues???? its the crates falling on the conveyor belt...

    energy supplied by motor???
  8. Oct 13, 2004 #7
    haha can get all the other parts except that. Its got me stumped too.
  9. Oct 13, 2004 #8
    did u get d) for the elevator???

    Ugrav (as a function of S + dup) = Uspring (as a function of S) + F(friction) L

    solve for L.... i thought it made sense... but i didn't get the right answer!... wut do u think??
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