A car misses a turn and sinks into a shallow lake to a depth of 5.8 m. If the area of the car door is 0.54 m2, what is the force exerted on the outside of the door by the water?
Note: Use pressure of air = 101 kPa

You can estimate it by Pressure = density of water * gravitational constant * depth of the door * area of door
We are taking a few liberties since depth of the door is not constant but since the problem does not contain the actual dimensions of the door, I guess you just have to go with this.

This post was edited by Azrad on Sep 25 2013 02:37pm

The wording is a little confusing to me, the part that says "what is the force exerted on the outside of the door by the water" makes me wonder if we are supposed to add 1 atm of pressure to the result...

force*

yea, it's truly vague
I guess you would have to estimate the height being less than max depth , as center of door.
also add pressure of air, for abs pressure at depth

Yes you would need to add atmospheric pressure.
Pabsolute = Prelative + Patmosphere

P = rho*g*h = rho(kg/m^3)*(9.81 m/s^2)*(5.8m) + Patm -> Units are in N/m^2 or Pascals
F = PA

Assumptions: rho is constant and does not vary with height/depth (this is an okay assumption for water which is incompressible)