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 65.7-hp (horsepower) engine. During the pull, the car passed two landmarks spaced 22.0 meters apart in 13.3 seconds. The passenger plane being towed is a Boeing 707, with a total weight of 62.5 tons. % of the Calculate the force that opposes the motion Assume that the efficiency of the car is such that 14 engine power is available to propel the car forward Number 4498 1 0 3 How much work is performed by the car on the airplane during this time? Number 98801.5 How much work is performed by the airplane on the car during this time? Number -98801 .5

Answer 1

Given,

m = 1239 kg ; P = 65.7 hp = 48992.5 W ; d = 22 m ; t = 13.3 s ; w = 62.5 tons

14.3 % of Power is

P' = 7006 W

We know that,

Power = Workdone/time

workdone = Power x time

W = 7006 x 13.3 = 93179.8 J

W = 9.32 x 10^4 J

F = W/d = 9.32 x 10^4 J/22 = 4.24 x 10^3 N

Hence, F =  4.24 x 10^3 N

Work by car is:

Wc = 9.32 x 10^4 J

The work done by the resistive force is:

Wa = - 9.32 x 10^4 J (in apposite direction so negative)