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Density of Block Submerged into Swimming Pool

  1. Apr 22, 2012 #1
    A spring is mounted vertically at the bottom of an empty swimming pool. A block of mass M = 857 kg is attached to the spring, and it compresses the spring from its equilibrium position by distance xo. Now the swimming pool is filled with fresh water, and the block is submerged. You find the the spring is now extended distance x = 1.73xo from its equilibrium position. Find the density of the block.

    I have zero clue how to start...can someone please help?
     
  2. jcsd
  3. Apr 22, 2012 #2

    ehild

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    Perhaps starting with a free body diagram of forces, including the buoyant force?

    ehild
     
  4. Apr 23, 2012 #3
    ok..

    mg down = 857 * 9.81 = 8407.17

    buoyant force (when block is submerged) = density of fluid x volume submerged in fluid x gravity
     
  5. Apr 23, 2012 #4
    a little more help please...
     
  6. Apr 23, 2012 #5

    ehild

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    The block is in equilibrium. What does it mean for the forces acting on it?

    ehild
     
  7. Apr 23, 2012 #6
    Sum of the forces acting on block = 0. So: Mg = density of fluid x volume submerged x gravity...right?
     
  8. Apr 23, 2012 #7

    ehild

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    And what about the spring?

    ehild
     
  9. Apr 23, 2012 #8
    This is where I'm confused...for spring, Mg = 1/2kx^2...but how does this factor in?
     
  10. Apr 23, 2012 #9

    ehild

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    That is the energy. What is the force a spring exerts when it is stretched or compressed by x? Think of Hook's Law.

    ehild
     
    Last edited: Apr 23, 2012
  11. Apr 23, 2012 #10
    It exerts a force equal to -kx
     
  12. Apr 23, 2012 #11

    ehild

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    See picture. What is the direction of this force when the tank is filled with water and the spring is extended?
    What is the direction when there is no water in the tank?

    ehild
     

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