Posted by Mary on .
A small plane mass 600 kg lifts off from the runway at a speed of 145 Km/Kr. 2 Minutes later, the plane has reached an altitude of 300 meters and its airspeed has dropped to 90 Km/Hr. Assume the power output of the engine was constant during this time. Ignore air resistance and compute the HP (horsepower) produced by the engine.
Divide total energy gain by 120 seconds for the energy of power delivered by the engine.
P.E. gain = M g H = 1.746*10^6 J
K.E. loss = (M/2)(V1^2 - V2^2)
V1 = 40.28 m/s
V2 = 25.00 m/s
K.E. loss = 2.992*10^5 J
Net energy gain = 1.465*10^6 J
Power during interval = 1.465*10^6/120 = 12210 W = 164 horsepower
This is an underestimate because there is also air friction drag that the engine must overcome, and because the propellers are not 100% efficient