A 12 second simulation may be too long of a simulation window for Event Simulation. It's much better at handling much faster simulations. Please see the quickstart guide posted in this forum thread for more detail.
https://forums.autodesk.com/t5/design-validate-document/quick-start-for-event-simulation/td-p/668065...
Can you share your design with us to have a closer look?
If you look at the Solver Data

... and the Solver Output

You can see the External Work and what the value is at the time that the solver outputs to the results.
Time Procedure Total Kinetic Internal Viscous External Total Elapsed Remaining
Inc Step Time Time Energy Energy Energy Work Energy Wall Time Wall Time
0 2.50e-08 0.00e+00 0.00e+00 0.00e+00 0.00e+00 0.00e+00 0.00e+00 0.00e+00 0 00:00:00 n/a
50 2.50e-08 1.25e-06 1.25e-06 2.66e-02 2.58e-02 7.15e-06 5.24e-02 1.47e-05 0 00:00:00 0 00:10:45
100 2.50e-08 2.50e-06 2.50e-06 5.53e-02 5.02e-02 3.29e-06 1.06e-01 1.08e-05 0 00:00:01 0 00:10:47
....
....
....
>>>>>> Writing FEMAP output for time = 0.000900022 <<<<<<<
36300 2.49e-08 9.01e-04 9.01e-04 6.28e+00 2.89e+01 1.40e-04 3.52e+01 7.38e-04 0 00:10:06 0 00:01:06
36350 2.49e-08 9.03e-04 9.03e-04 6.29e+00 2.89e+01 1.44e-04 3.52e+01 7.22e-04 0 00:10:07 0 00:01:05
....
....
40200 2.49e-08 9.98e-04 9.98e-04 6.41e+00 2.89e+01 1.04e-04 3.53e+01 4.84e-04 0 00:11:11 0 00:00:01
40250 2.49e-08 1.00e-03 1.00e-03 6.49e+00 2.89e+01 1.07e-04 3.53e+01 4.61e-04 0 00:11:12 0 00:00:00
>>>>>> Writing FEMAP output for time = 0.00100002 <<<<<<<
40269 2.49e-08 1.00e-03 1.00e-03 6.45e+00 2.89e+01 1.05e-04 3.53e+01 4.87e-04 0 00:11:12 0 00:00:00
In this case, there should be a conservation of energy for each step.
Total Energy = Kinetic Energy + Internal Energy + Viscous Energy - External Work
If you keep the value of Kinetic Energy very low (meaning there's not a lot of inertial forces), the solution will be quasi-static (I don't think I did a great job of this in the above example).
I don't believe there is a way to get the Reaction Force of a constraint at each time step at this time. I will have to ask around and circle back.
However, you may be able to use the External Work value above to understand the forces acting on your bottle.
I hope this helps.
Nathan Chandler
Principal Specialist