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 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 15.4% 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

1hp=745.7 watts
Work = Power * time.
15.4% * 79.7 *745.7watts * 11.3seconds = 103.42kJ

If we assume the same coefficient of kinetic friction and air resistance (which is probably NOT true), then the work is split between car and airplane according to their masses. About 98% of the work goes into pulling the plane
103.42 kJ of work is performed by the car on the airplane.
The same amount of work is performed by the airplane on the car.

98%103.42=101.35

Force = work / distance.
Force = 101.35kJ / 22 m = 4606 N

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