Class 26Sol

Class 26Sol - mgh i – F k s F k θ s Solve for s F k s = 1 2 mv i 2 mgh i s = 1 2 mv i 2 mgh i F k s = 1 2(5(4 2(5(9.8(2(10 = 13.8 m v i = 4 m/s

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Oregon State University Physics 201 Class Notes Class 25 (Nov 23), Fall 2009 h s m Analysis of the (external) work being done: W ext = F k s cos θ , where θ = 180°: The fundamental Work-Energy equation: E mech.f = E mech.i + W ext K T.f + . K R.f + . U G.f = . K T. i + K R.i + U G.i + W ext Detailed expansion of the Work-Energy equation for this situation: ( / 2 ) mv f 2 + 0 + mgh f = ( 1 / 2 ) mv i 2 + 0 + mgh i + W ext 0 = ( / 2 ) mv i 2 +
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Unformatted text preview: mgh i – F k s F k θ s Solve for s: F k s = ( 1 / 2 ) mv i 2 + mgh i s = [( 1 / 2 ) mv i 2 + mgh i ]/ F k s = [( 1 / 2 )(5)(4) 2 + (5)(9.8)(2)]/(10) = 13.8 m v i = 4 m/s v f = 0 m/s h i = 2 m h f = 0 m m = 5 kg F k = 10 N Let h = 0 here Simplify here by substituting values that are 0....
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This note was uploaded on 01/13/2010 for the course PH 201 taught by Professor Staff during the Fall '08 term at Oregon State.

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