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Brick Falling Off a Rising Pallet

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A crane is in the process of lifting a pallet of bricks to the upper floor of a building. Suddenly a brick slides and falls off the rising pallet. You clock the time it takes for the brick to hit the ground at 2.5 seconds.

    The crane has height marking and you see that the brick fell off the pallet from a height of 22 meters. How fast is the brick going when it hits the ground?

    2. Relevant equations

    y = yo + Vot + (1/2)at2

    V = Vo + at

    3. The attempt at a solution

    So we are given that:

    t = 2.5 s
    y = 22 m
    yo = 0
    a = g = 9.80 m/s2
    Vo = ?

    Find: V = ?

    I understand that I need to use the kinematic equations to solve this. What I am confused by is that the words 'lifting' and 'rising' are emphasized in the problem statement.

    Does this imply that Vo is NOT equal to 0 ?

    Is so, I am confused as to how to find it?

    Any advice is appreciated :smile:
     
  2. jcsd
  3. Sep 23, 2009 #2

    rl.bhat

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    Brick slides off from the rising pallet with some velocity. Before that brick and pallet have the same velocity. So when the brick slips form the rising pallet, what will be its initial velocity and in which direction?
     
  4. Sep 23, 2009 #3
    It will be up; but I do not know its value. Or is it just Vo?

    Can I solve the 2 equations in 2 unknowns now? Is Vo the same for both of the equations?

    Thank you!
     
  5. Sep 23, 2009 #4

    rl.bhat

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    Velocity vo is the same as the velocity of the pallet.
    Use the first relevant equation.
    In that y = 0, yo = 22 m. t is given. Find vo. Take care of the directions of the various terms.
     
  6. Sep 24, 2009 #5
    Can I use y0= 0 and y = 22 if I choose positive as going downwards?

    So,

    y = Vot + at2

    Vo = (y - at2) / t

    Vo = (22 - 9.80*2.52) / 2.5

    Vo = - 15.7 m/s

    So Vo is negative because it is going upwards and I chose downwards as positive.

    Now I use this in the second equation. Again choosing down as positive

    V = Vo + at

    V = (-15.7) + 9.8 * 2.5

    V = 8.8 m/s going downwards.

    Does this look correct?
     
  7. Sep 24, 2009 #6

    rl.bhat

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    y = Vot + at2
    It should be
    y = Vot + 1/2*at2
     
  8. Sep 24, 2009 #7
    Oops.* So it should be"


    y = Vot + (1/2)at2

    Vo = (y - (1/2)at2) / t

    Vo = (22 - 0.5*9.80*2.52) / 2.5

    Vo = - 3.45 m/s

    So Vo is negative because it is going upwards and I chose downwards as positive.

    Now I use this in the second equation.* Again choosing down as positive

    V = Vo + at

    V = (-3.45) + 9.8 * 2.5

    V = 21.1 m/s going downwards.

    Is this correct?* It seems like a better answer in that the first one (with the error) was too slow.
     
  9. Sep 24, 2009 #8

    rl.bhat

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    Yes. It is correct.
     
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