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 84.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 26.0 meters apart in 13.3 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 23.1% 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

The net force f = MA = P - F = 0; where F is the force opposing the pull P.

As there is no acceleration, we can say is that P = F; so that A = 0 is the acceleration (none).

If we assume all the power available is actually used, then 84.7*.231 = 19.5657 HP is used when dragging the plane that is 26 m.

1 HP = 745.7 J/s ; 19.5657HP = 14590 J/s and work done in that 13.3 seconds is QE = 13.3*14590 = 194048.9 Joules.

And over that 26 m; F = QE/S = 194048.9/26 = 7463.42 N is the force opposing the tow car. ANS.

And from earlier, QE = 194048.9 Joules is the work done by the tow car. ANS.

The airplane's drag/friction works to offset the energy input by the tow car. So it's equal to minus QE = -194048.9. ANS.

Note...both QE's are done on or by the entire system, not just the car or the plane, but both, the total system. This is obvious if you just do a thought experiment and put the brakes on for the airplane. Both the car and the plane will slow down, both are acted on by what the plane does.

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