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Power

Power. Chapter 5.6 WOD is underlined. 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). Question.

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Power

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  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 m2/s3 • 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

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