An advertisement shows a 1239 kg car slowly pulling at constant speed a large pa
ID: 1966168 • Letter: A
Question
An advertisement shows a 1239 kg car slowly pulling at constant speed a large passenger airplane to demonstrate the power of its newly-designed 92.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 24.0 meters apart in 15.3 seconds. The passenger plane being towed is a Boeing 707, with a total weight of 62.5 tons.
Calculate the force that opposes the motion. Assume that the efficiency of the car is such that 16.5% of the engine power is available to propel the car forward.
How much work is performed by the car on the airplane during this time?
How much work is performed by the airplane on the car during this time?
Explanation / Answer
Mass of the car m = 1239 kg Power of the engine P = 92.7 hp = 69126.37 Watt Distance traveled by plane x = 24 m Time taken to travel t = 15.3 s ----------------------------------------------------------------------------------------- The speed of the car v = x/t = 24 m / 15.3 s = 1.56 m/s Since the power 16.5% is used in forwarding the car then P' = 0.165 (69126.37 W) = 11405.85 W Now this power is used to pull the plane then P' = Fv Therefore the force acting on the plane is F = P'/v = 11405.85 W / 1.56 m/s = 7311.44 N = 7.311*103 N ---------------------------------------------------------------------------------------- Work done by car W = Fx = (7311.44 N) (24 m) = 175474 .56 J = 1.75*105 J --------------------------------------------------------------------------------------- Work done by plane on car is W = - 175474 .56 J = - 1.75*105 J = 7.311*103 N ---------------------------------------------------------------------------------------- Work done by car W = Fx = (7311.44 N) (24 m) = 175474 .56 J = 1.75*105 J --------------------------------------------------------------------------------------- Work done by plane on car is W = - 175474 .56 J = - 1.75*105 JRelated Questions
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