Practice 5F: | 1 | 2 |
3 | 4 | 5 | Go up
Power - by Matt Henderson, 2003
1. A 1.0 X 10 ^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 X 10^3 retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to left the fully loaded elevator at a constant speed of 3.0 m/s?
Here's what you know, m = 1000 + 800 = 1800 kg and v = 3.0 m/s and Fr = 4.0 X 10^3and a = 9.8 m/s since gravity is what is pushing against it. Now we use the formula F = mDa + Fr
F = (1800)(9.8) + 4.0 X 10^3
F = 21640 Now using the formula P = Fv P = (21640)(3)
P = 64920 W = 65 kW
2. A car with a mass of 1.50 X 10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0 s. Assume that the force of resistance remains constant at 400.0 N during this time. what is the average power developed by the car's engine?
m = 1.50 X 10^3 and v = 18 ms/ and a = 12.0 s and Fr = 400 N
First use the formula aavg = Dv/Dt to find the time a = 18/12= 3/2 Then use the formula F = mDa + Fr again to find the F
F = (1.50 X 103)(3/2) + 400 F = 2650 N Then use the formula P = Fv to find the power - but use the average velocity (9 m/s - the average of 0 and 18) to find the power.
P = (2650 N)(9 m/s) =
P = 23850 W = 23.85 kW
(Table of contents)
3. A rain cloud contains 2.66 X 10^7 kg of water vapor. How long would it take for a 2.00 kW pump to raise the same amount of water to the cloud's altitude, 2.00 km?
m = 2.66 X 10^7 and P = 2.00 kW and h = 2.00 km = 2000 m
Since Work = PE and PE = mgh we can use the formula P = W/Dt 2000 W = ((2.66 X 10^7)(9.8)(2000)/t)
t = 260680000 = 2.61 X 108 s
(Table of contents)
4. How long does it take a 19 kW steam engine to do 6.8 X 10^7 J of work?
Use the formula P = W/Dt
P = 19 kW and W = 6.8 X 10^7 J
19 = (6.8 X 10^7)/(t)
t = 3578.9 s
(Table of contents)
5. A 1.50 X 10^3 kg car accelerates uniformly form rest to 10.0 m/s in 3.00s.
a. What is the work done on the car in this time interval?
We use the formula (1/2)(m)(v^2) = W which is the same as P = ((1/2)(m)(v^2))/Dt
m = 1.5 X 10^3 and v = 10 m/s
plug it in (1/2)(1.5 X 10^3)(10^2) = W = 7.5 X 10^4 J
b. What is the power delivered by the engine in this time interval?
We take the work (W) found from a. and plug it into the formula P = W/Dt
t = 3 s
P = 7.5 X 10^4 / 3 = 2.5 X 10^4 W