1. Section 8.5: Power

2. Instantaneous Power: P ≡ (dE/dt) = (dW/dt)
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: Pavg = (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 = (103 Watt)  (3600 s)
= 3.6  106 W s = 3.6  106 J

3.  P = (dW/dt) = F(dr/dt) = Fv
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:
 Pavg = (W/Δt) = F(Δr/Δt) = Fvavg
vavg ≡ Average velocity of object

4. Ex. 8.10: Power Delivered by Elevator Motor
Elevator car, mass me = 1,600 kg, carries passengers of mass mp = 200 kg. Constant friction force
fk = 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/s2.
Figure 8.13: (Example 8.10) (a) The motor exerts an upward force T on the elevator car. The magnitude of this force is the tension T in the cable connecting the car and motor. The downward forces acting on the car are a friction force f and the gravitational force Fg = Mg. (b) The free-body diagram for the elevator car.

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: ∑Fx = 0
 F – FR – mg sinθ = 0
Gives required force F = 3,100 N
Power: P = Fv = (3,100)(22) = 6.8  104 W
b) When accelerates on level ground from vi = 25.0 m/s to vf = 30.6 m/s
in t = 10 s with a retarding force FR = 700 N
Solution: Now, θ = 0. ∑Fx = ma or F – FR= ma  F = FR+ ma (1)
Also, vf = vi + at  a = (v - v0)/t = (30.6 – 25)/(10) = 0.93 m/s2
Now, (1) gives required force F = 2,000 N
Maximum Power: P = Fvf = (2,000)(30.6) = 6.12  104 W