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Terminal Velocity and Bouyancy - F = kv

  1. Sep 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock of mass 0.400 kg is released from the surface and sinks in the ocean. As the rock descends it is acted upon by three forces: gravity, buoyancy, and drag. The buoyancy is an upward force equal to half its weight. Drag from the water can be modeled by F = kv, where k = 0.650 kg/s.

    (a) Determine the terminal speed of the sinking rock.

    (b) Determine its depth, speed, and acceleration 1.50 seconds after it is released.

    (c) At what depth will it be at 99.0% of its terminal speed?


    2. Relevant equations

    Net force (vertical), F = ma, F = kv

    3. The attempt at a solution

    The solution to a) was easy enough.

    Fnet = mg + FB + Fdrag

    Fnet = -3.92 N + 1.96 N + Fdrag = 0 N (at terminal speed)

    Fdrag = 1.96 N = 0.650 kg/s *vt

    vt = 3.02m/s.

    b) For calculating the depth, speed, and acceleration, I think I need to integrate the function of the net force with respect to time to get the equations for velocity and position. But I have not had much luck doing so because I am rusty on my integration.

    The equation for the acceleration is

    a(t) = -4.90 N + 1.63s-1 *v(t)

    so I'm thinking

    dv/dt = -4.90 N + 1.63s-1 *dx/dt

    but then I get something like

    dv = -4.90 N dt + 1.63s-1 *dx

    and I don't know what to do with dx?

    c) I'm thinking the answer for this part will be easy enough once I have the answer to b...
     
    Last edited: Sep 28, 2009
  2. jcsd
  3. Sep 28, 2009 #2
  4. Sep 29, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Instead of dx/dt, try leaving that as v in this equation.
     
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