Atwood with Sliding mass and real pulley

[SadDan]

Homework Statement

Block 1 with mass m1=4.04 kg rests on a very low friction horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to the hanging block 2 of mass m2=2.02 kg, as shown.

The pulley is a uniform disk of radius 11.85 cm and mass 1.980 kg. Calculate the speed of block 2 after it is released from rest and falls a distance of 1.84 m.

What is the angular speed of the pulley at the instant when block 2 has fallen a distance of 1.84 m ?

Homework Equations

Wtot=change in Energy
KE=1/2 mv^2
W=integral of force*displacement
N2L for rotation and translation

The Attempt at a Solution

The tension of 2 and 1 on the pulley should be different but the accelerations of the blocks would be the same?
a=m2*g/(mp/2 +m1+m2)
T2=m2(g-a)=13.27
T1=m1*a=13.0829
WEm2=integral from 0 to 1.84 (T2xdx) = 22.47 J
WT1m1=integral from 0 to 1.84 (T1xdx) = 22.1467 J
work energy theorem:
KE=W
1/2mv^2=WEm2+WT1m1
v=sqrt(2(WEm2+WT1m1)/mp)=3.33m/s

This is not the correct answer, I'm not sure what I am doing wrong would the energy side of the equation be only kinetic?

 

[Mastermind01]

I think the diagram would be helpful. Is the pulley frictionless? Also why would the accelerations be same?

 

[SadDan]

m1 and m2 are atached by a string on a pulley so wouldn't their accelerations be the same?

 

[SadDan]

And all the information we were given is on the question so I'm assuming its not frictionless

 

[Mastermind01]

m1 and m2 are atached by a string on a pulley so wouldn't their accelerations be the same?

Correct. I can't really understand what the equations for work you've written, up you can just use conservation of energy to do the problem.