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

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 58.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 25.0 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 17.6% 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

A).

Power = 58.7* 745.7 W = 43772.59 W

Available (17.6%) = 0.176*43772.59 = 7704 W

velocity of pull = d/t = 25/15.3 = 1.63 m/s

Force = Power/velo = 7704/1.63 = 4726 N

Since, the system moves with constant speed. the same force oposes the motion

B).

Work (car on plane) = Force * dis = 4726*25 = 118150 J = 118 kJ

C).

Work (Plane on car) = -118 kJ same amount but opposite. [since work done is change in KE and no KE

change occur, no change in velocity, so total work must be zero.]

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