Posted by **Mary** on Saturday, March 10, 2012 at 8:35pm.

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.

- physics -
**drwls**, Sunday, March 11, 2012 at 7:38am
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

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