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 79.7 hp (horsepower) engine. During the pull, the car passed two landmarks spaced 22.0 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 22.0% of the engine power is available to propel the car forward.

? N

How much work is performed by the car on the airplane during this time?

? J

How much work is performed by the airplane on the car during this time?

? J

Explanation / Answer

Power = 79.7* 745.7 W = 59432.3 W

Available (22%) = 0.22*59432.3 = 13075 W

velocity of pull = d/t = 22/15.3 = 1.4379 m/s

Force = Power/velo = 13075/1.4379 = 9093 N

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

Work (car on plane) = Force * dis = 9093*22 = 200046 J = 200 kJ

Work (Plane on car) = -200 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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