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 70.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 22.0 meters apart in 14.3 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. 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? Find the speed given the information posed in the problem, then use that information and the power rating to find the force of propulsion acting on the car. Note that there is no acceleration.

Explanation / Answer

m = 1239 kg
P = 70.7 hp = 52.72 * 10^3 W
S = 22.0 m
t = 14.3 s

As acceleration = 0
Net Force = 0, Therefore
Force that opposes the motion = Pulling Force

Power, p = 0.20 * 52.72 * 10^3 W
Work done = Power*time
W = (0.20 * 52.72 * 10^3)*14.3
W = 150779.2 J
Work performed by the airplane on the car during this time,W = 150779.2 J

Work done = Force * Distance
Force, = 150779.2/22.0
F = 6853.6 N
Force that opposes the motion,F = 6853.6 N

As it moves at constant speed, v = 22.0/14.3
v = 1.54 m/s

p.s - Force of propulsion acting on the car is same as force that opposes the motion, becoz there is no acceleration.

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