A coffee filter of mass 1.8kgdropped from a height of 4mreaches 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? Start from the Energy principle, and choose as the system the coffee filter, the Earth and the air.

Mass of coffee filter$\mathrm{m}=1.8\mathrm{g}$ , height$\mathrm{h}=4\mathrm{m}$ , speed$\mathrm{v}=0.8\mathrm{m}/\mathrm{s}$ .

The potential energy is given by:

$\mathbf{U}\mathbf{=}\mathbf{mgh}$

Here denotes mass. Theis acceleration due to gravity that is given by$\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$,h stands for height.

The kinetic energy is given by:

$\mathbf{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here,is mass of the object andis the velocity of the object.

The mass of coffee filter is m=1.8g.

Convert the unit of mass into kg.

$$\begin{gathered} \mathrm{m}=1.8\mathrm{g}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right) \\ =1.8\times {10}^{-3}\mathrm{kg} \end{gathered}$$

Thus, out of all, the observers in situation may feel that Newton’s first law has been violated.

Substitute , $\mathrm{m}=1.8\times {10}^{-3}\mathrm{kg},\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{and}\mathrm{h}=4\mathrm{m}$into the formula of potential energy.

role="math" localid="1657795625494" $$\begin{gathered} \mathrm{U}=\left(1.8\times {10}^{-3}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right) \\ =70.56\times {10}^{-3}\mathrm{J} \end{gathered}$$

Therefore, the potential energy is $70.56\times {10}^{-3}\mathrm{J}$.

Substitute$\mathrm{m}=1.8\times {10}^{-3}\mathrm{kg}\mathrm{and}\mathrm{v}=0.8\mathrm{m}/\mathrm{s}$ andinto the formula of kinetic energy.

$$\begin{gathered} \mathrm{k}=\frac{1}{2}\left(1.8\times {10}^{-3}\mathrm{kg}\right){\left(0.8\mathrm{m}/\mathrm{s}\right)}^2 \\ =0.576\times {10}^{-3}\mathrm{J} \end{gathered}$$

Therefore, the kinetic energy is $0.576\times {10}^{-3}\mathrm{J}$.

The loss of energy is given by $$\begin{gathered} \mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{PE}\mathbf{-}\mathbf{KE} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$.

Subtract the obtained kinetic energy from the obtained potential energy to obtain the amount of energy loss.

$$\begin{gathered} \mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{PE}\mathbf{-}\mathbf{KE} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{70}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{576}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{69}\mathbf{.}\mathbf{984}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{J} \end{gathered}$$

The loss of energy is said to be transferred in the air molecules.

Therefore, loss of kinetic energy is$69.984\times {10}^{-3}\mathrm{J}$ and energy is transferred in air molecules.