In the bernoulli experiment, the pressure on one side is greater than the other, thus the fluid flows faster through the thinner nozzle.In the image, is the pressure created shown correctly? Since liquids are incompressible, I thought that through the neck of the experiment, the volume wouldn't equal to the volume which is being moved over a distance with a constant velocity. So, in short, my question is : Is the pressure P2 shown on the figure the pressure which does negative work on the shaded area? Also, is this negative work done because the water encounters collisions and pressure from the walls of the neck part?
Q: In the image, is the pressure created shown correctly?
A: Yes, but P1=P2 if there is no flow (velocity).
Q: Is the pressure P2 shown on the figure the pressure which does negative work on the shaded area?
A: No, but as The A2 is smaller than the A1 so the net Force is also less.
Yet, this Force-thinking is also not enough, neither is the work.
If you want to create the complete picture you need to think the conservation of the energy. The energy must be same if losses are neglected;
pressure (+ elevation) = potential energy.
velocity = kinetic energy
and you have everything balanced, when you keep energy constant. This means that increased velocity (kinetic energy) must reduce potential energy (pressure).
If the fluid is incompressible then the fluid would speed up if going from left to right as the volume flux of the fluid cannot change.
$A_1v_1=A_2v_2$ where $A$ is the area and $v$ is the speed.
In this case the fluid is made to move faster so is increasing in kinetic energy.
It is the pressure difference which provides the net force on the fluid which does the work on the fluid to increase its kinetic energy.
The force to the left $P_1A_1$ does positive work on the fluid and the force on the right $P_2A_2$ does negative work on the fluid.
There is a fuller explanation in Chapter 28 Fluid Dynamics.