Answer:
The force acting on the train due to the air resistance is given by the formula:
F = (1/2) * p * A * v^2 * Cd
where p is the density of air, A is the cross-sectional area of the train, v is the speed of the train, and Cd is the drag coefficient.
We can assume that the drag coefficient is constant for a given shape of the train, and we can take it to be 0.5 for a train moving through air.
The pressure difference between the front and back of the train is given by:
P = (1/2) * p * v^2
The pressure difference between the sides of the train and the tunnel walls is given by:
2 * P = 2 * (1/2) * p * v^2 = p * v^2
The force acting on the train due to this pressure difference is given by:
F' = 2 * P * S = 2 * p * v^2 * S
The net force acting on the train is given by:
Fnet = F - F' = (1/2) * p * A * v^2 * Cd - 2 * p * v^2 * S
We can simplify this expression by factoring out p * v^2:
Fnet = p * v^2 * [(1/2) * A * Cd - 2 * S]
We are given that So = 4St, so we can substitute A = St and S = St/4:
Fnet = p * v^2 * [(1/2) * St * Cd - 1/2 * St * Cd] = 0
Therefore, the net force acting on the train is zero. This means that the force due to air resistance is exactly balanced by the force due to the pressure difference between the sides of the train and the tunnel walls.
We are given that Po - P = 2 * Pu, so we can substitute P = (1/2) * Po - Pu:
F' = 2 * P * S = 2 * [(1/2) * Po - Pu] * S = (Po - Pu) * 4 * St
We can equate this expression for F' to the expression for air resistance force:
F' = p * v^2 * [(1/2) * St * Cd - 2 * St/4] = p * v^2 * [(1/2) * St * Cd - St/2]
Equating these two expressions, we get:
(Po - Pu) * 4 * St = p * v^2 * [(1/2) * St * Cd - St/2]
Simplifying and solving for Pu, we get:
Pu = Po - (p * v^2 * Cd) / 8
We are given that Cd = 2N, so we can substitute this value:
Pu = Po - (p * v^2 * 2N) / 8
Pu = Po - (p * v^2 * N) / 4
Therefore, the pressure inside the train is Po - (p * v^2 * N) / 4.