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Elevator free fall problem

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A bolt comes loose from underneath an elevator that is moving upward at a constant speed of 6.0 m/s. The bolt reaches the bottom of the elevator shaft in 3.0 s.

    a.) How high above the bottom of the shaft was the elevator when the bolt came loose?

    b.) What is the speed of the bolt when it hits the bottom of the shaft?

    2. Relevant equations
    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex]

    [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [/tex]


    3. The attempt at a solution

    I first used [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex] and plugged in V0 = 6 m/s , t = 3 sec and a = -9.8 m/s2 since the object is going down. I ended up getting a [tex]\Delta[/tex]y of -26.1 m. Is it ok for this value to be negative?

    Then I used [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [/tex] and plugged in my -26.1 m to get a Vf of 23.4 m/s


    I'm confused about the when to use a negative or positive velocity or [tex]\Delta[/tex]y.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 1, 2010 #2
    First you need to know what is [itex]x[/itex] in your equation. In this case it would be the distance from the bottom of the shaft to the bolt (at the certain time [itex]t[/itex]). So, you actually want to know [itex]x_0[/itex], because it's the initial height. After time interval [itex]t[/itex], [itex]x[/itex] is zero (bolt has reached the bottom).

    That's why it is important to consider such equations in vector form and define which vector points in which direction. In this case, it would be

    [tex]
    \vec{x}=\vec{x_0}+\vec{v_0}t+\frac{1}{2}\vec{g}t^2
    [/tex]

    [itex]x[/itex] point upwards to the bolt, [itex]x_0[/itex] also points upwards and denotes the initial height, [itex]v_0[/itex] also points upwards (because the elevator goes up) and [itex]g[/itex], of course, points downwards. So, it's best to take up as positive direction (we actually set the bottom of the shaft as the origin of coordinate system). Now, everything that points upwards is positive and everything else is negative.

    [tex]
    \vec{x}=(x_0+v_0 t-\frac{1}{2}g t^2)\hat{x}
    [/tex]

    [tex]\hat{x}[/tex] is unit vector pointing upwards (it's used to denote direction and mesurement unit - its length is 1).

    I hope all of this is not too much confusing, but these details are very important.
     
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