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Homework Help: Elevator free fall

  1. Dec 14, 2009 #1
    an elevator car weighing 50,000 N is 27m from the ground. the cable snaps and the elevator goes into free fall. Shock absorbers have been provided, the coil length is 1 m.

    Assume elevator is able to lock onto the shaft at the point of contact w/ the absorber so that it does not bounce back. When elevator comes into contact w/ absorbers there is friction force of 60,000N acting on it as it descends that 1 m distance.

    (A) Total energy of the elevator 27m above ground?
    MGH= 50,000N*27
    MGH = 1,350,000

    (B)At the bottom of the shaft,when elevator comes to rest what is the total energy (take into account work done by friction)

    MGH= 0 KE = 1/2 M V2- Friction

    1/2 5000 V2=1,350,000-60,000

    Total energy: 1290000 ?????

    (C)Use work energy principal to determine spring constant K.

    MGH= 50,000N* -1 = -50,000N
    SPE = 1/2Kx2 = 1/2 K

    -50,000+1/2 K = 1,290,000

    K= 2,680,000

    This doesn't seem correct.
  2. jcsd
  3. Dec 14, 2009 #2
    it looks very good (and a well posted question and attempt at solution!) up until part c:

    The energy stored in a spring is 1/2kX^2

    Equate this to the energy lost by "friction",
  4. Dec 14, 2009 #3
    Thank you Denverdoc.

    Is this what you mean?

    MGH= 50,000N* -1 = -50,000N
    SPE = 1/2Kx2 = 1/2 K (1)

    -50,000 + 1/2 K (1) = 60,000 N (??)

    1/2 K = 110,000

    K= 220,000
  5. Dec 14, 2009 #4
    not quite: mgh = 1/2 kx^2. The kneticenergy gets converted to the potemtial energy of the spring.
  6. Dec 15, 2009 #5
    Is this what you mean?

    50,000 = 1/2 K X ^2

    100,000 = K
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