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Scale in an elevator physics problem

  1. Sep 14, 2007 #1
    A 62-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when the elevator is ascending at 11 m/s but its speed is decreasing by 5 m/s in each second?

    I'm not really sure where to begin.
     
  2. jcsd
  3. Sep 14, 2007 #2

    learningphysics

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    Hint, you need the force exerted by the girl on the scale... that's the weight the scale reads...

    You also know that the force the girl exerts on the scale = the force the scale exerts on the girl.
     
  4. Sep 14, 2007 #3
    Well I know that normally the scale would read 608.5 N if the elevator was not moving, but I'm not sure what to do here.
     
  5. Sep 14, 2007 #4
    First, find what the deceleration of the elevator is.
     
  6. Sep 14, 2007 #5

    learningphysics

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    draw the freebody diagram of the girl... what are the forces acting on the girl... you know that [tex]\Sigma\vec{F} = ma[/tex]. follow l46kok's suggestion and find the acceleration of the girl.
     
  7. Sep 15, 2007 #6
    Would the deceleration be -3.45 m/s squared?
     
  8. Sep 15, 2007 #7

    learningphysics

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    No, the deceleration is the same given in the question...
     
  9. Sep 15, 2007 #8
    only the velocity is given?
     
  10. Sep 16, 2007 #9

    learningphysics

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    They give 5m/s per second... so the acceleration is -5m/s^2
     
  11. Sep 19, 2007 #10
    iidesjo!

    okay,

    Fg=m*g (Fg is gravitational force)
    Girl weighs (m) 62 [kg].
    Gravitational constant (g) is 9.81 [m/s^2]
    Fg= 62*9.81= 608.22 [N]
    that is the static value.

    With F=m*a you can calculate the difference in force compared to static (v=c)
    a= -5 [m/s^2]
    F= 62*-5= -310 [N]

    so the scale will indicate 608.22-310= 298.22 [N]
    to calculate back to mass divide through the graviotational constant.
    298.22/9.81= 30.4 [Kg]
    you're welcome!
    :D
     
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