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 $5.8810\times 10^4 \ N$.

To Find :

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

Solution :

Acceleration is given by :

$a = \dfrac{162-120}{2.10}\ m/s^2\\\\a = 20 \ m/s^2$

Now, force equation is given by :

$F - F_{friction} = ma\\\\F_{friction} = F-ma\\\\F_{friction} = 58810 - (2450\times 20 )\\\\F_{friction} = 9810\ N$

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