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 25.0 meters apart in 11.3 seconds. The passenger plane being towed is a Boeing 707, with a total weight of 62.5 tons (does the weight of the plane need to be divided by gravity to get mass or is it to just be converted into kg).

Calculate the force that opposes the motion. Assume that the efficiency of the car is such that 18.7% of the engine power is available to propel the car forward.

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 79.7*.187 = 14.9039HP (15 Hp) is used when dragging the plane that 25 m. 745.7 J/s = 1 HP; so 11185.5 J/s = 15 HP and work done in that 11.3 seconds is QE = 11.3*11185.5= 126396.15 Joules. And over that 25 m F = QE/S = 126396.15/25 = 5055.846 N is the force opposing the tow car. ANS.

And from earlier, QE = 126396.15 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. 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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