lec05-06-203-08Energy1mod

lec05-06-203-08Energy1mod - 5/6-1 Work done by force F...

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5/6-1
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θ F || F d θ d || F d W= F d || component (projection of) d along F OR Work done by force F acting over displacement d W= F d cos( θ ) F || component (projection of) F along d W= F || d !! J s m Kg m ) s m (Kg m (N) 2 2 2 = = = Units 5/6-2
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5/6-3 Collinear F and d Note kinetic friction force always opposes displacement always does - work Constant force F not // d example
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5/6-3 5/6-3a
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0 v - v a = t Work-Energy Theorem Object: m; constant force F; constant accelertation a=F/m 0 v = v +at 0 v + v x = [ ] t 2 W = F x = ma x 0 0 v - v v + v W = m ( ) {[ ] t} t 2 2 2 0 m W = (v - v ) 2 2 2 0 1 1 W = mv - mv 2 2 2 1 mv = K = Kinetic Energy 2 0 W = K - K W = K Δ Total work done on object= change in kinetic energy (work done by total force) ) = 2 2 m [K] = kg ( Joule = Nm s ) = 2 2 ft [K] = slug ( ft lb s or true for non-const. a also !!! 5/6-3b
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5/6-4 Power= work per unit time [ J/s= Watt] Δ Δ W P = t Average Power dW P = dt Instantaneous Power
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F 10 N d 3m 5 kg W = F || d = 10 (3) Nm = 30 J i = 0) v f 2 = 3.5 m/s W= Δ K What is v f ? W= F d = ½ mv
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lec05-06-203-08Energy1mod - 5/6-1 Work done by force F...

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