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

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

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/D

P = 19 kW and W = 6.8 X 10^7 J

19 = (6.8 X 10^7)/(t)

t = 3578.9 s

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