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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 -

    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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