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Velocity and Loss in Energy onedimensional

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data
    If a 15kg object is released from a height of 11.58m and the objects loss in energy is 302 J, determine its velocity as it enters the container (object is being dropped straight down into a container)
    Assume g = 9.8m/s^2
    m(object) = 15kg = w(object) = 147N
    s = 11.58m
    ΔE = -320 J


    2. Relevant equations
    Had no idea. But the ones i know that i thought could apply:
    ΔKE = KE(final) - KE(initial)
    KE = 1/2 * mv^2
    GPE = mgh
    v^2 = u^2 + 2as


    3. The attempt at a solution
    Okay, I had no idea how to do this, but i tried to work out as many things as i can then mash em together to get an answer :p

    GPE = mgh
    =15*9.8*11.58
    =1702.26

    Then i thought; well if the loss in energy is 320 J then i could take that from the original GPE
    1702.26 - 320 = 1382.26 J

    and now i could work out the new height by subbing it back in
    1382.26 = 15*9.8*h
    h=9.403m

    And here is where it gets iffy... they say the velocity when it enters the container, but the container isn't at 9.403m its at 0...right? thats what i got from the question anyway. Therefore how did they only lose 320N? So I assumed that I could say its KINETIC energy is the 1382.26 J, and the height of the container is the difference because they say when it "enters".
    Therefore..
    KE = 1/2 * m * v^2
    1382.26 = 1/2 * 15 * v^2
    2764.52 = 15 * v^2
    v^2 = 184.30
    v = 13.58 m/s

    We don't have the answer to this question yet, but *a lot* of other people got 13.66m/s as the answer... so since i had no idea how to do this question and no basis to support whether im right or not, could someone help me out and either point out where i went wrong or if I'm right? thank you!
     
    Last edited: Mar 22, 2012
  2. jcsd
  3. Mar 22, 2012 #2
    I think they mean by lose of energy that it's due to friction, because when dropping an object from a certain hight the object does't lose energy but the energy is converted from potential to kinetic, and so when the question states that there there was loss in energy i think its due to friction(i might be wrong i don't know). If this is the case then i think you should do this:
    Energy (at h=11.58 )-Energy lost=1702.26-320=1382.26
    Ek=1/2*mv^2
    v=13.57m/s
    in this case you take the barrol at hight 0.
     
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