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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