lec05-06-203-08Energy1mod

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

This preview shows pages 1–8. Sign up to view the full content.

5/6-1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
5/6-3 Collinear F and d Note kinetic friction force always opposes displacement always does - work Constant force F not // d example

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
5/6-3 5/6-3a
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
5/6-4 Power= work per unit time [ J/s= Watt] Δ Δ W P = t Average Power dW P = dt Instantaneous Power
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 26

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

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online