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Integrals of Force

  1. Oct 16, 2008 #1
    The problem is: A leaky 10-kg bucket is lifted from the ground to a height of
    12 m at a constant speed with a rope that weighs 0.8 kg/m. Initially the
    bucket contains 36 kg of water, but the water leaks at a constant rate and
    finishes draining just as the bucket reaches the 12 m level. How much work
    is done?

    Work = Force*Distance
    Force = Mass*Acceleration
    Mass = Volume*Density
    So, Force = Volume*Density*Acceleration

    don't know how relevent some of those are...

    For work that does not have a constant force
    Work = the definate intregral of [tex]\int[/tex] f(x)dx,

    f(x) being the force.

    Acceleration = 9.8 m/s^2

    Density of Water = 1000 kg/m^3

    I put the bounds of my integral as being [0,12]. x is the depth of the bucket
    so x = 0 is the top and x = 12 is the bottom

    36/12 = 3 kg/m water lost

    Force = ((36-3x) + 10 + 0.8x)*9.8

    Take the Integral of Force from 0 to 12

    Answer = 3857.28 J

    Does this look right?
  2. jcsd
  3. Oct 16, 2008 #2

    D H

    Staff: Mentor

    Everything but the rope itself looks good.
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