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Elevator Problem

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A 80.0 kg person stands on a scale in an elevator.
    (c) What does it read when the elevator is falling at 3.5 m/s?

    2. Relevant equations

    EF = MA

    3. The attempt at a solution

    I can't even attempt the solution because I don't know acceleration...
  2. jcsd
  3. Nov 2, 2008 #2


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    Hi Toxage! :smile:

    If the speed is constant, then the acceleration is … ? :wink:
  4. Nov 2, 2008 #3
    Then the acceleration is equal to 0.

    EF = MA

    EF = M(0)

    EF = 0

    Fn - Fg = 0

    Fn = Fg

    80(9.8) = 882 Newtons

    882/9.8 = 80kg

    Doesn't work.... Webassign says its the wrong answer...

    Did I do something wrong in my math?
  5. Nov 2, 2008 #4
    No your maths looks fine. What is the quoted answer? If your standing on scales and not accelerating the scales will tell you your weight. When you are accelerating use this formula:

    [tex] F_{net} = m(a + g) [/tex]

    But MAKE SURE YOU DEFINE A POSITIVE DIRECTION AND STICK TO IT. So if you define up as positive, g = -9.81. And then when accelerating up, a is positive and vice versa.
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