1. Power Chapter 5.6 WOD is underlined

2. Question A 2400 kg car can slow from 10 m/s to rest in 3.2 meters.If the car has traditional friction brakes, what force do they apply while decelerating? (Show how to get vf^2 = vi^2 + 2av from w = Delta KE)

3. Question If this same car is traveling at 15 m/s, what will the stopping distance be? At 20 m/s? 25 m/s? 30 m/s?

4. Show that Vel vs distance is a Quadratic function. Can also solver eqn. Relate to a student Speeding at 95 vs 65. Speed increase is 50%. Distance increase is 100%

5. Power Power is work / time P = W / t Power is Capital P Do not confuse this with lowercase p (p = rho = momentum)

6. Power Add units to WOD. Power is work / time P = W / t Units are Joules/seconds or “Watts” 1 Watt = 1 kg m 2 /s 3 1 W = 1 J / s The concept of POWER focuses on TIME. English units is horsepower. Based upon actual power 1 horse can generate.

7. Power Add units to WOD. Power is work / time P = W / t Do Power companies sell us power?

8. Power Add units to WOD. Power is work / time P = W / t Do Power companies sell us power? No, Power is kWatts. They sell us kW-Hours. P * time = (Work / time) * time = Work = Energy

9. Power Question: If someone uses 1000 watts of power for one hour, how much energy do they use? (shown next slide)

10. Power Question: If someone uses 1000 watts of power for one hour, how much energy do they use? (Work out here) P = 1000 watts = 1000 J /s Problem is 1000 J /s * 1 h = 1000 J/s * 3600 s/hr * 1 h

11. Power Question: If someone uses 1000 watts of power for one hour, how much energy do they use? 3.6 * 10 6 Joules

12. Different Units: American Unit is “Horse Power” or “Power supplied by one horse” 1hp = 550 foot*lbs/second 1 hp = 746 Watts Also, energy can be measured in KiloWatt Hours 1 kwh = 3.6 * 10 6 Joules Kwh = power * time = energy Your electric bill probably uses kwh instead of Joules

13. If you run your hair dryer for 20 minutes at 1500 watts at 12 cents a kWh. How much did you spend? Cost = kW * hr * $ /kWh * 30 days = 1.5 * 20/60 *.12 * 30 = $1.80 How about switching 15 light bulbs from 100 watt to 13 watt? Watt savings is 15*(100-13) =1305W= 1.3 kW Cost = kW * hr * $ /kWh * 30 days = 1.3 * 8 hr/day *.12 * 30 = $37.44

14. How about switching off your computer and monitor? How much do you save? Those use about 250w/hr each. Assume you turn it off 20 hours a day instead of leaving it on all the time. Know for test: Cost = kW * hr * $ /kWh Like test question, calculate for 1 day. Cost = kW * hr * $ /kWh =.500 * 20 hr/day *.12 = $1.2/day Savings on utility bill by turning it off for the 20 hours a day that the computer sits idle = $1.2/day * 30 = $36 a month

15. Problem An elevator must lift 1000 kg mass a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip?

16. Problem An elevator must lift 1000 kg mass a distance of 100 m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip? Power = work / time = mgh/t = mg (h/t) = mg*v = 1000 kg (9.8m/s2) (4m/s) = 39200 watts