Question:
Consider a rectangular block of wood moving with a velocity vo in a gas at temperature T and mass density ρ. Assume the velocity is along the x-axis and the area of cross-section of the block perpendicular to vo is A. Show that the drag force on the block is
$4 \rho A v_{0} \sqrt{\frac{k T}{m}}$, where $m$ is the mass of the gas molecule.
Solution:
Let ρm is the no.of molecules per unit volume
Change in momentum by a molecule on front side = 2m (v + v0)
Change in momentum by a molecule on backside = 2m (v – v0)
No.of molecules striking front side = 1/2 [A(v+v0)∆t] ρm
No.of molecules striking back side = 1/2 [A(v-v0)∆t] ρm
Solving the above equation by considering the KE of the gas molecule, we get the dragging force as 4m A ρmv0√kgT/m