Work of Bouyant Force

  1. I am pretty sure i know how to answer my hw problem if someone can tell me how to calculate the potential energy due to the bouyant force? Im assuming its Fr but what is r and where would you get it from?

    In the problem i am given the depth from the surface at which an object is at.

    THANKS!
     
  2. jcsd
  3. Assuming a 1D problem and a calculus based class, I'd use F = - (d/dx) V

    Where F is the buoyant force and V is the potential energy associated with the buoyant force. (d/dx) is a spatial derivatve. Solve this equation for V by integration.
     
  4. HallsofIvy

    HallsofIvy 40,675
    Staff Emeritus
    Science Advisor

    You need to give more information. There is NO potential energy due to a force alone. Potential energy is the work a force could (or would be done against the force) in moving an object. No motion- no work. You have to be given the distance in the problem. Exactly what does the problem say?
     
  5. Well he said he was given the depth from the surface that the object was located and it should float that distance to the surface.

    If your class isn't calculus based, figuring out the units of 'r' should get you started.
     
  6. ok i think i got it... the problem is basically all variables and says you drop a sphere of mass m , volume v and density p into water. The sphere drops to a distance d below the surface of the water. What it the total potential energy due to gravity and the bouyant force. It then says to assume p is lower than water. I think this would be the sphere rising to the top? so the distance the bouyant force pushes up is d so the energy is pgvd + the gravitational potential energy and thats -mgd. the total would be...

    pgvd - mgd right?
     
  7. Careful with your densities, the p given is different than p(water). Remember the buoyant force in water is equal to the weight of *water* displaced. Also, remember that only a *change* in potential energy is meaningful, what you calculated is the change in potential energy as the sphere rises to the surface.

    To answer your question, is p or p(water) a better fit for your final equation?
     
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