Answer:
Therefore, it would take approximately 32.54 seconds for a 5 kW motor to raise a 6,000 lbs load a vertical distance of 20 ft.
Explanation:
Calculate the work done to lift the load:
Work = Force x Distance
Work = 6,000 lbs x 20 ft
Work = 120,000 ft-lbs
Answer:
32.5 s
Explanation:
Power = work / time, and work = force × distance.
P = W / t
W = Fd
Substituting and solving for t:
P = Fd / t
t = Fd / P
Using SI units:
m = 6000 lbm × (1 kg / 2.205 lbm) = 2721 kg
F = mg = (2721 kg) (9.81 m/s²) = 26,694 N
d = 20 ft × (1 m / 3.28 ft) = 6.10 m
t = (26,694 N) (6.10 m) / (5000 W)
t = 32.5 s
Or, using English units:
P = 5 kW × (737.6 ft-lbf / kJ) = 3687.8 ft-lbf/s
t = (6000 lbf) (20 ft) / (3687.8 ft-lbf/s)
t = 32.5 s
