Question

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?

Short answer

The amount of kinetic energy the air molecules gained from the falling coffee filter is${\mathrm{K}}_{\mathrm{air}}=2.68\times {10}^{-2}\mathrm{J}$ .

Step-by-step solution

Definition of Kinetic Energy

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.

Calculation of kinetic energy of coffee filter

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

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

Calculation of kinetic energy for no air resistance

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

Final calculation for kinetic energy of air

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 $2.68\times {10}^{-2}\mathrm{J}$.