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Fluid Mechanics and Energy

  1. Jun 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Spanish galleon is about to be boarded by bloodthirsty pirates in the
    shallows of a Caribbean island. To save a box of treasure on board, the
    captain orders his crew to secretly toss the box overboard, planning to come
    back for it later. The rectangular box is waterproof and measures 40.0 cm by
    25.0 cm by 30.0 cm. It is made of wood and has mostly gold pieces inside,
    resulting in an average box density three times that of seawater.

    Sinking below the surface, the box moves at constant vertical velocity of 1.15
    m/s for 12.0 m before hitting the bottom. (a) Draw the free-body diagram for
    the box, (b) determine the magnitudes of the forces on the box, and (c)
    calculate the work done by each force and the net work done on the box. (d)
    Calculate the change in the box’s gravitational potential energy. (e) What is
    the change in the box’s total energy ?


    2. Relevant equations
    V= lwh
    P= M/V
    Buoyant force= pgV for floating
    completely submerged buoyant force= (P of fluid/P of object)weight of object
    W= Fd
    PE= mgh
    E= PE + KE
    Drag force= .5pCAv^2

    3. The attempt at a solution
    i don't need help with part a.

    part b) i got the buoyant force to be 908.46 when it is floating. for submerged, it is 294 N
    Drag force= 138.88 (not sure, because i couldn't find the coefficient in my textbook)
    weight= 908.46

    for part c) i just multiplied y= -12 to each of the forces to find the work

    for part d) i just did PE= mg(deltah) and got -10901.5 J

    for part e i know how to do it (E= PE + KE) but i'm not getting the correct answer. I think i'm doing something wrong in either part b or part c.
    I used the buoyant force when it submerged.

    the correct answer for part e is -10901.5 J
     
    Last edited: Jun 29, 2012
  2. jcsd
  3. Jun 29, 2012 #2
    There are 3 forces acting on the box: the weight (downward), the upward buoyant force, and drag force (upward). You already calculated the buoyant force (294N) and the weight (908 N... this actually should probably have been 882N =3 x 294N ). The key to solving part b are the words "the box moves at constant vertical velocity of 1.15 m/s." This means that the box is not accelerating, and, thus, that it is in force equilibrium. You now have enough information to calculate the drag force, without needing to estimate the drag coefficient.
     
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