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Power Jambalaya: Calculating Work, Power, Force, and Distance

Explore various physics problems and formulas involving power, work, force, and distance. Calculate power output for lifting, dragging, and moving objects.

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Power Jambalaya: Calculating Work, Power, Force, and Distance

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  1. Power all mixed up Things we can know P - Power (in W) W - Work (in J) F - Force (in N) s - distance (in m) t - time (in s) m - mass (in kg)  - coefficient of friction Formulas: F = mg (Lifting) F = mg (Dragging) P = W/t (Power) W = Fs (Work) Jambalaya! It’s a

  2. P - Power (in W) W - Work (in J) F - Force (in N) s - distance (in m) t - time (in s) m - mass (in kg)  - coefficient of friction Formulas: F = mg (Lifting) F = mg (Dragging) P = W/t (Power) W = Fs (Work) Sample 1 - A 840 W winch can lift a 2350 kg Land Rover up 8.2 m into a tree in what time?

  3. P - Power (in W) W - Work (in J) F - Force (in N) s - distance (in m) t - time (in s) m - mass (in kg)  - coefficient of friction Formulas: F = mg (Lifting) F = mg (Dragging) P = W/t (Power) W = Fs (Work) Sample 2 - Gumi Baere drags a 45.1 kg box that has a coefficient of friction between it and the floor of 0.34 a distance of 16 m in 11.7 seconds. What is her power output?

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

  5. What must be the power rating of a motor if it is to lift a 560 kg elevator up 3.2 m in 1.5 seconds? F = mg = (560 kg)(9.8 N/kg) = 5488 N W = Fd = (5488 N)(3.2 m) = 17561.6 J P = W/t = (17561.6 J)/(1.5 s) = 11707.7 W 11700 W

  6. A 560. W winch can lift a car 5.2 m in 37 seconds. What must be the mass of the car? 1 HP = 745.7 W P = W/t, W = Pt = (560 W)(37 s) = 20,720 J W = Fd, F = W/d = (20,720 J)/(5.2 m) = 3984.6 N F = mg, m = F/g = (3984.6 N)/(9.8 N/kg) = 406.6 kg 407 kg

  7. Red Elk leads a dog team that can put out 2.5 kW of power. They skid a 312 kg sled a distance of 340 m in 93 seconds. What must be the coefficient of friction? P = 2500 W P = W/t, W = Pt = (2500 W)(93 s) = 232500 J W = Fd, F = W/d = (232500 J)/(340 m) = 683.8 N F = mg, = F/(mg) = (683.8 N)/((312 kg)(9.8 N/kg)) = .22 .22

  8. A 150 HP tractor can drag a 350 kg load how far in a minute if the coefficient of friction between the load and the ground is 0.78. 1 HP = 745.7 W P = (150 HP)(745.7 W/HP) = 111855 W F = mg = (.78)(350 kg)(9.8 N/kg) = 2675.4 N P = W/t, W = Pt = (111855 W)(60 s) = 6711300 J W = Fd, d = W/F = (6711300 J)/(2675.4 N) = 2508 m 2500 m

  9. Power - All mixed up A 2340 kg car speeds up from 12 m/s to 25 m/s in 4.6 seconds on the level. What power?

  10. Power - All mixed up An elevator motor takes a 780 kg car and speeds it up from rest to 5.4 m/s lifting it 4.1 m in 1.51 s. Power = ?

  11. 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

  12. A 1520 kg car speeds up from 20. m/s to 30. m/s on the level. 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

  13. 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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