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Finding weight of a mass on an accelerating and braking elavator

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Zach, whose mass is 78 kg, is in an elevator descending at 11 m/s. The elevator takes 3.2 s to brake to a stop at the first floor.

    What is Zach's weight before the elevator starts braking?

    What is Zach's weight while the elevator is braking?


    2. Relevant equations

    F = mg

    F = ma ??

    3. The attempt at a solution


    What is Zach's weight before the elevator starts braking?
    I was thinking....

    F = mg

    So F = (78kg)*(11m/s) ????

    The second question i was completely clueless on....
    but an idea of mine was...
    F = ma

    a = (11m/s)/(3.2s) = 3.4375 m/(s^2)

    F = (78kg)(3.4475) ????
     
  2. jcsd
  3. Mar 18, 2009 #2

    LowlyPion

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    Your final idea gave you the right acceleration (deceleration) for the elevator, but not the force that some scales would measure.

    What about his original weight standing still or moving at constant velocity? It's not m*v as you've shown in part a)
     
  4. Mar 18, 2009 #3

    tiny-tim

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    uhh? :confused:
    if the braking deceleration was 0, would the weight be 0? :wink:
     
  5. Mar 18, 2009 #4

    djeitnstine

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    Try drawing a free body diagram to see the forces involved. That will make your life simpler.
     
  6. Mar 18, 2009 #5
    I figure for the first question that if the speed is constant that accleration is zero so it would simply just be mass*gravity.

    But for the second one since i need to factor in the deceleration i was thinking
    (9.8 + 3.4375)*(78kg)
    Is this correct?
     
  7. Mar 18, 2009 #6

    LowlyPion

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    That is correct.
     
  8. Mar 18, 2009 #7
    I dont get it. I type in the answer and it says i have a rounding error. lolll

    My final answer was: 1032.525

    it tells me "Express your answer using two significant figures."

    So 1032.52 or 1032.53 neither one works.
     
  9. Mar 18, 2009 #8

    tiny-tim

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    erm :redface: … that's six sig figs! :wink:
     
  10. Mar 18, 2009 #9
    Yeah they wanted the answer as 1000, :P

    Thanks for the help everyone.
     
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