An advertisement shows a 1239 kg car slowly pulling at constant speed a large pa
ID: 1287352 • Letter: A
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 23.0 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 18.7% 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
Given data
Mass of the car m = 1239 kg
Power of the engine P = 84.7 hp
= 63186.2 W
Distance traveled by air plane x = 23.0 m
Time taken to travel is t = 11.3 s
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The speed of the car is,
v = x/t
= 23 m / 11.3 s
= 2.035 m/s
As the power 18.7 % is used to moving the car.
P' = 0.187 (63186.2 W)
= 11815.8 W
The amount of power is used to pull the plane is
P' = Fv
So,the force acting on the plane is
F = P'/v
=11815.8 W / 2.035 m/s
= 5806.3 N
=5.81x103 N
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The work is performed by the car on the airplane during this time
W = Fx
= (5806.3 N) (23 m)
=133544.887 J
= 1.34*105 J
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The work is performed by the airplane on the car during this time
W = - 133544.887 J
= - 1.34*105 J
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