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

Learn about power as the rate of doing work, equations for power calculation, examples of power output, and how to determine force and velocity in physics problems.

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

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  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

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