Connected objects with friction

[cfarrell]

Homework Statement

Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 2.33 kg mass has fallen through 0.315 m, its downward speed is 1.31 m/s.
The acceleration of gravity is 9.8 m/s^2.

What is the frictional force between the 3.81kg mass and the table?
Answer in units of N.

Homework Equations

ΣF=ma
F_f=μF_n
v²-u²=2ax

The Attempt at a Solution

First, I found the acceleration of the system by using the formula v²-u²=2ax.
1.31²-o²=2a*0.315
a=2.723 m/s^2

I then multiplied the mass of the block on the table by the acceleration because F=ma
F=3.81(2.723)
F=10.378I tried this answer, but it was wrong. I have no idea how to do this problem, but this is what I got to looking through my textbook and online. If anyone would explain how to get started solving it on the right track, I'd be extremely grateful.

Thanks!

 

[Herman Trivilino]

I then multiplied the mass of the block on the table by the acceleration because F=ma
F=3.81(2.723)
F=10.378

F=ma gives you the net force.

 

[cfarrell]

So if F=ma gives net force, is net force equal to applied force - friction force? How would I find the applied force on the block then, would I find the tension force in the rope between the blocks?

 

[Herman Trivilino]

Yes, and yes!

 

[haruspex]

would I find the tension force in the rope between the blocks?

Yes, but you cannot do that straight away because it is also related to the acceleration of the other mass. Consider the forces on the descending mass.