When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

Solution 54E Step 1 of 3: In the given problem, we need to Calculate the power used to make the plane climb against gravity. Given data, Speed, v=2.5 m/s Mass, m=700 kg Power of engine, P=75 kW As the gravity is acting vertically downward, we need to consider the vertical motion. Step 2 of 3: The rate at which work is being done against gravity is P=Fv Since F= mg P=mgv 2 Substituting m=700 kg, v=2.5 m/s and g= 9.8m/s 2 P=(700 kg)(9.8m/s )(2.5 m/s) P=17.15 kW 17.15 kW part of the engine power that is being used to make the airplane climb.