Frictional Force (and μ) & Acceleration (w/friction)

[ceceliaz]

1. The problem: 2 blocks are arranged at the ends of a
massless string (mass one is on a table and mass two is hanging straight down off the other side of the pulley that is connected to the edge of the table). The system starts from rest. When the 4.18 kg (the hanging mass) mass has fallen through 0.389 m, its down-ward speed is 1.29 m/s.
The acceleration of gravity is 9.8 m/s2 .
mass one (on table)=5.68 kg
mass two (hanging)=4.18 kg
μ=?
a=? (to the right towards the hanging mass)
What is the frictional force between the 5.68 kg mass and the table?
Answer in units of N.

2. I put together the SumFx and SumFy to make the equation to solve for frictional force:
μ(subk)= (m2)g-a(m1+m2) / (m1g)

3. However, I need acceleration. I tried using both (vf=vi+at) and (vf^2-vi^2/(2deltaX)), but it was wrong (probably because those formulas don't take into account friction.

QUESTION 1: How do you find friction or acceleration without each other?
QUESTION 2: Is "frictional force" the (μ) times (mass on the table) times (acceleration)?

 

[Doc Al]

3. However, I need acceleration. I tried using both (vf=vi+at) and (vf^2-vi^2/(2deltaX)), but it was wrong (probably because those formulas don't take into account friction.

Finding the acceleration is a kinematic exercise; friction is automatically included since the speeds are given. Redo your calculation--those kinematic equations should work just fine.

QUESTION 1: How do you find friction or acceleration without each other?

You don't. But you have all the information needed to solve for acceleration, and then for the friction.

QUESTION 2: Is "frictional force" the (μ) times (mass on the table) times (acceleration)?

No. But why are you solving for μ? Calling the friction force "F" and solve for it directly. (Friction equals μmg, but that's not needed unless you are asked to solve for μ.)