Question

A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10s. If the airplane is putting out an average force of 5.8810x10^4 N during this time, what is the average friction force exerted on the airplane by the air?

Answer 1

Given :

A 2450 kg stunt airplane accelerates from 120 m/s to 162 m/s in 2.10 s.

If the airplane is putting out an average force of
`5.8810* 10^4 \ N`.

To Find :

The average friction force exerted on the airplane by the air.

Solution :

Acceleration is given by :

`a = (162-120)/(2.10)\ m/s^2\\\\a = 20 \ m/s^2`

Now, force equation is given by :

`F - F_(friction) = ma\\\\F_(friction) = F-ma\\\\F_(friction) = 58810 - (2450* 20 )\\\\F_(friction) = 9810\ N`

Therefore, frictional force exerted in the airplane by the air is 9810 N.