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 19.8% 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

Unless you know more about what the coefficients of friction and drag are, the way to calculate them is from the power it takes to oppose them.
I'm going to dolast two parts before force..


Work = Power * time.

79.7 horsepower = 59456.2 W
0.198* 59456.2 W * 15.3 seconds = 180.11661228 kJ

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.
176.514 kJ of work is performed by the car on the airplane.
The same amount of work is performed by the airplane on the car. =176.546 kJ

Force = work / distance.
Force = 176.514 kJ / 22 m = 8023.36 N

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