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Bouyancy Problem

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    You stand at the edge of a pool of water with a rock of mass M in your hand.
    You throw the rock up in the air. The rock leaves your hand at a height h0 and rises to a
    maximum height h1 before falling into the water. After entering the water, the rock
    eventually submerges and settles to the bottom at a depth of h2 and comes to a gentle stop with its velocity approaching zero at the bottom. Consider the density of the rock to be
    equal to the density of water (it is a porous rock) and neglect air friction. Assume the
    thermal energy of the rock remains unchanged in the problem. Assume any horizontal
    motion or horizontal velocity is zero and only consider the vertical motion of the rock.

    b. [10 pts] Consider just the rock as the system and think about the part of the path
    between part 4 and part 5. Write down the energy equation between point 4 and point 5
    listing all possible terms. Explain why any terms are zero and which terms cancel after
    simplifying this equation. Model the interaction of the rock with the water by both a
    buoyancy force as some additional work done by the water drag force, Wdrag.
    c)As the rock loses energy in the water, it adds 5 J of energy to the thermal energy
    of the water. If the heat capacity of the water is 4.2 J/g/K, the density of water is
    1 g cm-3, and the pool of water is 10 m by 10 m by 10 m in size, by how much does the
    temperature of the water change?

    2. Relevant equations

    KE=1/2 mv^2
    U= -mgh
    Fbouyancy=density of water/density of object*mg
    DeltaEthermal= mCdeltaT


    3. The attempt at a solution

    KE4+U4=U5+KE5
    KE5(rock is at rest)=0

    .5mv^2-mgh=-mgh2
    .5mv^2-mg(0)=-mgh2
    .5mv^2-0=-mgh2
    v^2=-2mgh2/m
    v=sqrt(-2gh2)

    and then for deltatE=Wsurrounings +Q (no thermal energy released thus Q=0)
    deltatE = Fbouyancy - Wdrag
    deltatE = d of water/d of object *mg - Wdrag
    deltatE = 1(d of water=d of object) *mg - Wdrag
    deltatE = mg-Wdrag

    I am not sure if I am doing this correctly and what else I should include

    c) Change in E thermal - mCdeltaT
    Delta T=Change in Thermal Energy/ mCv
    Delta T = 1.19e-9 K

    I just want to make sure i got this part right

    Thank you very much
     
  2. jcsd
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