Consider a rectangular block of wood moving with a velocity $v_0$ in a gas at temperature $T$ and mass density $\rho$. Assume the velocity is along $x$-axis and the area of cross-section of the block perpendicular to $v_0$ is $A$. Show that the drag force on the block is $4 r A v_0 \sqrt{\frac{k T}{m}}$, where, $m$ is the mass of the gas molecule.
Consider the diagram
Let $n=$ number of molecules per unit volume
$v_{\mathrm{rms}}=\mathrm{rms}$ speed of the gas molecules
When block is moving with speed $v_o$, relative speed of molecules w.r.t. front face $=v+v_o$
Coming head on, momentum transferred to block per collision $=2 m\left(v+v_o\right)$, where, $m=$ mass of molecule.
Number of collission in time $\Delta t=\frac{1}{2}\left(v+v_o\right) n \Delta t A$, where, $A=$ area of cross-section of block and factor of $1 / 2$ appears due to particles moving towards block.
$\therefore$ Momentum transferred in time $\Delta t=m\left(v+v_0\right)^2 n A \Delta t$ from front surface.
Similarly, momentum transferred in time $\Delta t=m\left(v-v_0\right)^2 n A \Delta t$ (from back surface)
$\therefore$ Net force
$$\begin{aligned} (\text { drag force }) & =m n A\left[\left(v+v_0\right)^2-\left(v-v_0\right)^2\right] \quad \text{[from front]}\\ & =m n A\left(4 \vee v_0\right)=(4 m n A v) v_0 \\ & =(4 \rho A v) v_0 \quad \text{... (i)}\\ \text { where we have assumed} \quad \rho & =\frac{m n}{V}=\frac{M}{V} \end{aligned}$$
If $v=$ velocity along $x$-axis
Then, we can write
$$\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2} k_B T
$$\Rightarrow \quad v=\sqrt{\frac{k_B T}{m}} \quad \left[\begin{array}{rl} K_B & =\text { Boltzmann constant } \\ K E & =\text { Kinetic energy } \\ T & =\text { Temperature } \end{array}\right]$$
$\therefore$ From Eq. (i), $\quad$ Drag force $=(4 \rho A v) v_0=4 \rho A \sqrt{\frac{k_B T}{m}} v_0$.