Answer:
The kinetic energy gained by the air molecules is 0.054313 J.
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 :
[tex]P=mgh\\\\P=1.4\times 10^{-3}\ kg\times 9.8\ m/s^2\times 4\ m\\\\P=0.05488\ J[/tex]
Finally, it will have kinetic energy. It is given by :
[tex]E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 1.4\times 10^{-3}\times (0.9)^2\\\\E=0.000567\ J[/tex]
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
[tex]E=0.000567-0.05488=0.054313\ J[/tex]
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.054313 .
