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Tuesday, August 7, 2012

Final 10 - Falling Water


Given:

  • \mbox{Water is flowing down a wall of 1.5 meters and reaches a velocity of 4 meters/second at the bottom.}

Question:

  • \mbox{What is the work done by friction of a 1 gram water drop to retard its falling freely?}

Rationale:

  • V^2_{water-at-bottom} = 2 a_{retarded-from-normal-g} {height}
    • a_{retarded-from-normal-g} = \dfrac{V^2_{water-at-bottom}}{(2 height)}  
  • a_{friction} = g - a_{retarded-from-normal-g} 
  • W_{friction} = (m) (a_{friction}) {(height)}  

Calculate:

  • a_{retarded-from-normal-g} = \dfrac{(4)^2}{(2 * 1.5)} = 5.33333
  • a_{friction} = 10 - 5.33333 = 4.66667
  • W_{friction} =  (1 * 10^{-3})(4.66667)(1.5) = 7 * 10^{-3} = .007

References:

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