Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ . Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v0 is A. Show that the drag force on the block is 4ρAv0 √(KT/m), where m is the mass of the gas molecule.
Answers
Answer:
Explanation:
hope it helps
Answer:
Speed of wooden block =v
o
Cross-sectional area of block =A
Temperature of the gas =T
Let n be the number of molecules per unit volume.
rms speed of gas molecules =v
rms
Relative speed of the molecules with respect to front face of the block =v+v
0
In head on collision, momentum transferred to block per collision =2m(v+v
0
)
where m is the mass of molecule.
Number of collision in time Δt,=
2
1
(v+v
0
)nΔtA.
(Factor of
2
1
appears due to particles moving towards block)
Hence, momentum transferred in time Δt,
=m(v+v
0
)
2
nAΔt
It is from front surface only.
Similarly for back surface, momentum transferred in time =Δt=m(v−v
0
)
2
nAΔt
Net force
Δt
Δp
=m[(v+v
0
)
2
−(v−v
0
)
2
]nA
=mnA(4vv
0
)=(4mnAv)v
0
=(4ρAv)v
0
(∵mn=ρ) (i)
Also,
2
1
mv
2
=
2
1
kT (v along x-axis)
∴v=
m
kT
∴ Net force =4ρA
m
kT
v
0
[Using (i)
This is required drag force.