Consider a rectangular block of wood moving with a velocity `v_(0)` in a gas at temperature T and mass density p. Assume the velocity is along x-axis and the are of cross-section of the block perpendicular to `v_(0)` is A. show that the drag force on the block is `4rAv_(0)sqrt((kT)/(m))` where,m is the mass of the gas molecule.

Explanation

Consider the digram

Let n = number of molecules per unit volume
`V_("rms") =` 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)` where m = mass of molecule.
Number of collision in time `Deltat=1/2(v +v_(o)) nDeltatA,` where A = area of croos- section of block and factor of `1//2` appears due to particles moving towards block.
`:.` Momentum transferred in time `Deltat =m(v+v_(0))^(2) nADeltat` from front surface.
Similarly, momentum transferred in time `Delta t = m(v -V_(0))^(2) nA Deltat` (from back suface)
`:. "Net force"" "("drag force") =mnA[v+v_(0))^(2)]" "["from front"]`
`=mn A(4vv_(0)) = (4mnAv) v_(0)`
`=(4 pAv) v_(0)`
Where we have assumed `p(mn)/(V) =(m)/(V)`
if v= velocity along x-axis
Then we can write `KE =1/2 mv^(2) =1/2 k_(B) T`
`rArr " "v=sqrt((k_(B)T)/(m)) " "["["underset( T="Temperature")underset(KE="Kinetic energy")underset(K_(B)="Boltzmann constnat")"]"]`
`:.` From eq .(i). Drag force `=(4pAv) v_(0) =4pAv . sqrt((K_(B)T)/(m))v_(0).`