1. Section 8.5: Power

2. Power ≡ Rate at which work W is done or rate at which energy E is transformed: Instantaneous Power: P ≡ (dE/dt) = (dW/dt) For work W done in time Δt: Average Power: P avg = (W/Δt) SI power units: P = (Energy)/(time) = (Work)/(Time) Unit = Joule/Second = Watt (W). 1 W = 1J/s British units: Horsepower (hp). 1hp = 746 W “Kilowatt-Hours” (from your power bill). Energy! 1 KWh = (10 3 Watt)  (3600 s) = 3.6  10 6 W s = 3.6  10 6 J

3. Convenient to write power in terms of force & velocity v. For force F & displacement Δr making angle θ with F. We know: W = F  Δr = FΔr cosθ Instantaneous Power:  P = (dW/dt) = F  (dr/dt) = F  v Average Power:  P avg = (W/Δt) = F  (Δr/Δt) = F  v avg v avg ≡ Average velocity of object

4. Ex. 8.10: Power Delivered by Elevator Motor Elevator car, mass m e = 1,600 kg, carries passengers of mass m p = 200 kg. Constant friction force f k = 4,000 N acts against motion. (A) Find power needed for a motor to lift car + passengers at a constant velocity v = 3 m/s. (B) Find power needed for motor to lift car at instantaneous velocity v at upward acceleration a = 1.0 m/s 2.

5. Example: Power Needs of a Car Calculate the power required of a car of mass, m = 1,400 kg: a) When climbs a hill with θ = 10 ° at constant velocity v = 22 m/s. Solution: ∑F x = 0  F – F R – mg sinθ = 0 Gives required force F = 3,100 N Power: P = Fv = (3,100)(22) = 6.8  10 4 W b) When accelerates on level ground from v i = 25.0 m/s to v f = 30.6 m/s in t = 10 s with a retarding force F R = 700 N Solution: Now, θ = 0. ∑F x = ma or F – F R = ma  F = F R + ma (1) Also, v f = v i + at  a = (v - v 0 )/t = (30.6 – 25)/(10) = 0.93 m/s 2 Now, (1) gives required force F = 2,000 N Maximum Power: P = Fv f = (2,000)(30.6) = 6.12  10 4 W