Atmospheric pressure is very nearly 100 kPa. A sealed container of air at 1 atmospheric pressure has a door 1 m wide and 2 m high. This door is very hard to open during HIGH pressure days. If the atmospheric pressure on the outside of the container is just 1 percent greater than on the inside, then what added force in Newtons is required to open the container door?

F_net = 2 10³ N

Explanation:

The pressure expression is

P = F / A

F = P A

we use Newton's second law

F_ net = F₁ - F₂

Where F₁ and F₂ are the force otusidad and inside the room

F_net = P₁ A - P₂ A

F_net = (P₁ - P₂) A

Let's look for the outside pressure that is 1% higher than the inside pressure

P₁ = P₂ + 0.01 P₂ = 100 10³ (1 + 0.01)

P₁ = 101 10³ Pa

Let's look for the area

A = a h

A = 1 2

A = 2 m²

Let's calculate

F_net = (101 - 100) 10³ 2

F_net = 2 10³ N