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An advertisement shows a 1239 kg car slowly pulling at constant speed a large pa

ID: 1777748 • 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 63.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 26.0 meters apart in 11.3 seconds. The passenger plane being towed is a Boeing 707, with a total weight of 62.5 tons.

An advertisement shows a 1239 kg car slowly pulling at constant speed a large passenger airplane to demonstrate the power of its newly-designed 63.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 26.0 meters apart in 11.3 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 20.9% of the engine power is available to propel the car forward. Number How much work is performed by the car on the airplane during this time? Number How much work is performed by the airplane on the car during this time? Number

Explanation / Answer

GIven Data

Mass of the car m = 1239 kg

Power of the engine P = 63.7 hp = 47501.08 Watt

Distance traveled by plane x = 26 m

Time taken to travel t = 11.3 s

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The speed of the car

             v = x/t = 26 m / 11.3 s

                        = 2.3 m/s

Since the power 20.9% is used in forwarding the car then

               P' = 0.209 (47501.08 W)

                    = 9927.73 W

Now this power is used to pull the plane then

              P' = Fv

Therefore the force acting on the plane is

              F = P'/v = 9927.73 W / 2.3 m/s

                            = 4316.4 N

                           = 4.316*103 N

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Work done by car

            W = Fx = (4316.4 N) (26 m)

                         = 112226.5 J

                         = 1.12*105 J

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Work done by plane on car is

             W = -  112226.56 J

                 = - 1.12*105 J

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