Answer:
The kinetic energy gained by the air molecules is 0.054313 J.
Step-by-step explanation:
Given that,
Mass of a coffee filter, m = 1.4 g
Height from which it is dropped, h = 4 m
Speed at ground, v = 0.9 m/s
Initially, the coffee filter has potential energy. It is given by :
![P=mgh\\\\P=1.4* 10^(-3)\ kg* 9.8\ m/s^2* 4\ m\\\\P=0.05488\ J](https://img.qammunity.org/2021/formulas/physics/college/e8jbwm6naqpiglv1fuz9hl1lckyq7bumgp.png)
Finally, it will have kinetic energy. It is given by :
![E=(1)/(2)mv^2\\\\E=(1)/(2)* 1.4* 10^(-3)* (0.9)^2\\\\E=0.000567\ J](https://img.qammunity.org/2021/formulas/physics/college/3l293qrwcktkgwi4kqvpd6pfrmrccib43b.png)
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
![E=0.000567-0.05488=0.054313\ J](https://img.qammunity.org/2021/formulas/physics/college/4622eamuzuds230bb5dvwlygtwnozy4y96.png)
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.054313 .