physics

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(a) a spherically shaped buoy is made from wood, which has a specific gravity of 0.6. find the acceleration of the buoy when released from rest at the bottom of a freshwater lake. (b) if the buoy starts out 10 m below the surface of the water, determine the height above the water that it will rise after it shoots out of the water.

  • physics -

    Buoyancy = 1000 kg/m^3 * 9.81 * vol
    weight = 600 kg/m^3 * 9.81 * vol

    net force up = 400 * 9.81 * vol

    F = m a

    400 * 9.81 * vol = 600 * vol * a

    a = (2/3) * 9.81
    --------------------------------
    work done in rise to surface
    = F * 10 = 4000*9.81*vol
    = Kinetic energy on exit from water
    = (1/2) m v^2
    = .5 * 600 * vol * v^2
    so
    300 v^2 = 4000* 9.81
    v^2 = 130.8
    v = 11.4 m/s initial speed up

    Ke = Pe at stop = m g h
    but oh. I did not have to do that
    work done during rise = m g h directly
    4000*9.81*vol = 600*vol*9.81 * h

    h = (40/6) = 6 3/3 m

    note g did not matter in part b

    NOte also that this problem ignores the force required to accelerate the water around the sphere. This "added mass" is actually comparable to the mass of water displaced by the sphere. I bet the text writer did not even know that (hydrodymanics, not usually covered in school or college physics but very important if you are designing fishing gear)

  • physics -

    why is work done during the rise= mgh directly?

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