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Calculate the initial velocity of the elevator

  1. Sep 25, 2007 #1
    A descending elevator of mass 290kg is uniformly decelerated to rest over a distance of 4m by a cable in which the tension is 3631N

    the acceleration due to gravity is 9.8m/s^2
    calculate the initial velocity of the elevator at the beginning of the 4m descent

    answer in units of m/s

    i am confused on this one due to the introduction of velocity
     
  2. jcsd
  3. Sep 25, 2007 #2
    using what we did in the last problem where T = MA + MG

    3691 = 290A + 290(9.8)
    A = 2.720

    so now we have the acceleration
     
  4. Sep 25, 2007 #3

    learningphysics

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    First find the acceleration using forces... then you can find the initial velocity using kinematics.

    There's also a method to do it using work/energy (have you covered this yet?)
     
  5. Sep 25, 2007 #4

    learningphysics

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    good. now use kinematics to get the initial velocity. careful about signs.
     
  6. Sep 25, 2007 #5
    we havent covered work/energy

    so i found the acceleration to be 2.720

    so i have Vf= 0, Vi =?, A = 2.720, D = 4m

    so now i can solve for Vi correct????
     
  7. Sep 25, 2007 #6
    do i give the 2.720a negative sign

    and then use the formula

    Vf^2 = Vi^2 + 2AD
     
  8. Sep 25, 2007 #7

    learningphysics

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    yes, use this formula. acceleration is positivie. d is negative.
     
  9. Sep 25, 2007 #8
    ok so i get Vi = 4.66535 m/s ???
     
  10. Sep 25, 2007 #9

    learningphysics

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    looks good.
     
  11. Sep 25, 2007 #10
    ok 2 more left if u got the time ur my saviour
     
  12. Sep 25, 2007 #11
    A person weighing .8kN rides in an elevator that has a downward acceleration of 1.9m/s^2
    the acceleration of gravity is 9.8m/s^2
    what is the magnitude of the force of the elevator floor on the person??
    answer in units of kN

    what is kN?

    do i do this one like the last problem?
     
  13. Sep 25, 2007 #12

    learningphysics

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    it's very similar to the last problem... just that instead of tension, you have the normal force that the elevator exerts upward on the person. again be careful about signs.

    1000N = 1kN.
     
  14. Sep 25, 2007 #13
    so i can solve it like this

    X = Normal Force

    X = MA + MG

    X= (.8)(1.9) + (.8)(9.8)

    X = 9.36kN
     
  15. Sep 25, 2007 #14
    or did i screw up the signs again
     
  16. Sep 25, 2007 #15

    learningphysics

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    0.8kN is the weight, not the mass...

    also a = -1.9m/s^2
     
  17. Sep 25, 2007 #16
    ahh so i have to convert that weight to mass.... and then solve using the equation i had used
     
  18. Sep 25, 2007 #17
    so the new equation looks like this

    X = (.081kN)(-1.9m/s^2) + (.081kN)(-9.8)
     
  19. Sep 25, 2007 #18
    or was the -9.8 supposed to be positive

    where X would equal .6444
     
  20. Sep 25, 2007 #19

    learningphysics

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    no...

    I recommend working the whole problem in N and then converting to kN at the end...

    The weight is 0.8kN = 800N. what is the mass in kg?

    Also, you already know the weight (ie mg) is 0.8kN = 800N

    X = mass*(-1.9) + 800N

    if you are confused about signs and directions, start at the freebody diagram.
     
  21. Sep 25, 2007 #20
    ahhh ok my conversions were off

    so mass is 81.63 N

    so then X = -155.1020408N + 800N

    X = 644.89795 N

    X = .64489795 kN
     
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