A coffee filter of mass 1.4g dropped from a height of 2m reaches the ground with a speed of 0.8m/s . How much kinetic energy${\mathbf{K}}_{\mathbf{air}}$did the air molecules gain from the falling coffee filter?

Kinetic energy is the energy possess by an object due to their motion. If we want something to accelerate, we must apply force to it. Using force necessitates exertion. Following the completion of the job, energy is transferred to the item, which then moves at a new steady speed. The amount of energy delivered is dependent on the mass and speed attained.

Calculate the kinetic energy of the coffee filter with the given mass and final speed.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{resitance}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2 \\ =\frac{1}{2}\left(1.4\times {10}^{-3}\mathrm{kg}\right){\left(0.8\frac{\mathrm{m}}{\mathrm{s}}\right)}^2 \\ =5.60\times {10}^{-4}\mathrm{J} \end{gathered}$$

No non-conservative forces act on the filter in this occasion, so energy is conserved this implies that $E_f=E_i$.

Calculate the kinetic energy for the hypothetical case in which there is no air resistance.

role="math" localid="1657790226012" $$\begin{gathered} {\mathrm{K}}_{\mathrm{no}\mathrm{resitance}}=\mathrm{mgh} \\ =\left(1.4\times {10}^{-3}\mathrm{kg}\right)\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(2\mathrm{m}\right) \\ =2.74\times {10}^{-2}\mathrm{J} \end{gathered}$$

Subtract ${\mathrm{K}}_{\mathrm{resitance}}$from role="math" localid="1657790298690" ${\mathrm{K}}_{\mathrm{no}\mathrm{resitance}}$to get energy that goes to air ${\mathrm{K}}_{\mathrm{air}}$.

role="math" localid="1657790369288" $$\begin{gathered} K_{air}={\mathrm{K}}_{\mathrm{no}\mathrm{resitance}}-{\mathrm{K}}_{\mathrm{resitance}} \\ =2.74\times {10}^{-2}\mathrm{J}-5.60\times {10}^{-4}\mathrm{J} \\ =2.68\times {10}^{-2}\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy obtained for air is $2.68\times {10}^{-2}\mathrm{J}$.