1 / 21

Power Contents: Definition of Power Whiteboards Power as change in energy/time Whiteboards

Power Contents: Definition of Power Whiteboards Power as change in energy/time Whiteboards. Power - The rate at which work is done. power = work/time = Fv F = Force to move something v = Velocity it is moving. A person does 145 J of work in 20.0 s. What is power output?

julietta
Download Presentation

Power Contents: Definition of Power Whiteboards Power as change in energy/time Whiteboards

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Power • Contents: • Definition of Power • Whiteboards • Power as change in energy/time • Whiteboards

  2. Power - The rate at which work is done • power = work/time = Fv • F = Force to move something • v = Velocity it is moving • A person does 145 J of work in 20.0 s. What is power output? • power = work/time = (145 J)/(20.0 s) = 7.25 J/s • = 7.25 Watts = 7.25 W • 1 horsepower = 745.7 Watts, 1 kW = 1000 Watts

  3. Power - The rate at which work is done • power = work/time = Fv • F = Force to move something • v = Velocity it is moving power = work/time = (Fs)/t = F(s/t) = Fv My 30. HP van could go 25 m/s top speed. What was the force resisting its motion? 1 horsepower = 745.7 Watts, 1 kW = 1000 Watts power = (30 HP)(745.7 Watt/HP) = 22,371 W power = Fv 22,371 W = F(25 m/s), F = 890 N

  4. Whiteboards: Power 1 | 2 | 3 | 4 | 5

  5. Joe Mama does 613 J of work in 2.13 seconds. What is his power output? P = W/t = (613 J)/(2.13 s) = 288 W 288 W

  6. Ima Wonder can put out 127 W of power. What time will it take her to do 671 J of work? P = W/t, t = W/P = (671 J)/(127 W) = 5.28 s 5.28 s

  7. What work does a 1.5 HP motor do in 1 minute? (3 hints) P = (1.5 HP)(745.7 W/HP) t = 60 s P = W/t, W = Pt = (1118.55 W)(60 s) = 67113 J 67,000 J

  8. Bob N. Frappels slides a box with 43 N of force at a constant speed of 5.3 m/s. What is his power output? P = Fv = (43 N)(5.3 m/s) = 227.9 W = 230 W 230 W

  9. Frieda People can put out 430. W of power. With what speed can she push a car if it takes 152 N to make it move at a constant velocity? P = Fv v = P/F = (430. W)/(152 N) = 2.83 m/s 2.83 m/s

  10. Power all mixed up P W t F s m μ • Power = W/t • Work = Fs • Tricky: • F = mg (lifting) • F = μmg (dragging) Example 1: If you do 45.0 J of work pushing a 5.20 kg box 6.70 m across the level floor, what is the coefficient of friction? (0.132)

  11. Power all mixed up P W t F s m μ • Power = W/t • Work = Fs • Tricky: • F = mg (lifting) • F = μmg (dragging) Example 2: A 450. W winch can lift a 1240 kg mass how far in a minute? (2.22 m)

  12. Whiteboards: Power Jambalaya 1 | 2 | 3

  13. How much work is it to drag a 62.0 kg sled 120. m across the level snow where the coefficient of friction is 0.105? • Power = W/t • Work = Fs • F = mg (lifting) • F = μmg (dragging) P W t F s m μ 7660 J

  14. You do 780. J of work lifting an 18.0 kg mass how high? • Power = W/t • Work = Fs • F = mg (lifting) • F = μmg (dragging) P W t F s m μ 4.42 m

  15. An electric motor must lift a 5610 kg elevator 18.3 m in 32.0 s. What must be its minimum power output? • Power = W/t • Work = Fs • F = mg (lifting) • F = μmg (dragging) P W t F s m μ 31,500 W

  16. A sled dog team can put out 702 W of sustained power. What mass can they drag 1320 m on the level in 492 s if the coefficient of friction is 0.150? • Power = W/t • Work = Fs • F = mg (lifting) • F = μmg (dragging) P W t F s m μ 178 kg

  17. An elevator motor needs to lift a 1,670 kg elevator at a constant speed a vertical distance of 18.3 m in 8.5 seconds. What is its minimum power rating? PE = mgh Power = work/time = (mgh)/t (1670 kg)(9.81 N/kg)(18.3 m) /(8.5 s) = 35,270.989… W 35 kW

  18. Iwanna Ferrari stretches a 150 N/m spring a distance of 1.35 m in .53 seconds. What is her average power output? Eelas = 1/2kx2 Power = work/time = (1/2kx2)/t 1/2(150 N/m)(1.35 m)2/(.53 s) = 257.9 W = 260 W 260 W

  19. A 1520 kg car speeds up from 20. m/s to 30. m/s. What time will it take with a 34 kW engine? Ek = 1/2mv2 Change in energy = work = 1/2mv2 - 1/2mu2 = 1/2(1520 kg)(30. m/s)2 - 1/2(1520 kg)(20. m/s)2 work = 380,000 J power = work/time, time = work/power = (380,000 J)/(34,000 J/s) = 11.17 s 11 s

  20. A 1452 kg car is going 11.0 m/s at the top of a 9.20 m tall hill. What is the power output of the engine if it is going 22.0 m/s at the bottom of the hill 13.0 seconds later? Total at top: = 1/2(1452 kg)(11.0 m/s)2 + (1452 kg)(9.8 N/kg)(9.20 m) Total at top: = 218758.32 J Total at bottom: = 1/2(1452 kg)(22.0 m/s)2 = 351384 J Work = Change = 351384 J - 218758.32 J = 132625.68 J Power = work/time = (132625.68 J)/(13.0 s) = 10201 W = 10.2 kW 10.2 kW

More Related