An advertisement shows a 1239 kg car slowly pulling at const

An advertisement shows a 1239 kg car slowly pulling at constant speed a large passenger airplane to demonstrate the power of its newly-designed 65.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 26.0 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 24.2% 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?

Solution

1 Hp =745.7 Watts

Output Power of car

P_output=n*P_input =0.242*(65.7*745.7)

P_output =11856.2 W

Work done by the car

W=Pt =11856.2*11.3 =133975 J

a)

Force that opposes the motion is

F=W/d =133975/26

F=5153 N

b)

Work performed by the car on the airplane is

W=133975 J (In forward direction)

c)

Work performed by the airplane on the car is

W=133975 J (In Reverse direction)