Two Blocks and a Pulley Friction Problem

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The problem involves a system of three blocks connected by a pulley, with specific weights and friction coefficients. The 9.0 kg block is on a smooth table, while the 12 kg block experiences friction with a coefficient of 0.2. The calculations for the forces acting on each block were outlined, leading to the determination of tension and acceleration. The final acceleration of the 5.0 kg block, when released, was calculated to be 1.82 m/s². The solution appears to be confirmed as correct by the poster.
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Homework Statement



System comprised blocks, a light frictionless pulley and connecting ropes (see diagram). The 9.0kg block is on a perfectly smooth horizontal table. The surfaces of the 12kg block are rough, with μk = .2 between the two blocks. If the 5.0 kg block accelerates downward when it is released, find its acceleration

Fairly certain my solution is correct, just would love a second opinion!

Diagram:
6k5Im7W.png

http://i.imgur.com/6k5Im7W.png

Homework Equations

The Attempt at a Solution



So for the 12kg Block my reasoining for its forces was:
Fnet = Ffriction - T1 = Ma = 0
Fnormal = (12kg)(9.8 m/s^2 _ = 117.6
ForceFriction = (.2)(11.6) = 23.52 N

Then for the 9 KG block:
Fnet = T2 - ForceFriction
T2 - 23.52N = Ma
T2 - 23.52N = (9kg)a
23.52-T2 = (9 kg) a

Then finally for the hanging 5.0 KG block:
T2 - Fg = ma
T2 - (5)(9.8) = (5)a
T2 - 49N = 5(a)

Then I simply solved the system of equations to find the value of A

Set two equations for the 9kg and the 5kg block in terms of a

(T2 - 49N)/5 kg = (23.52N - T2)/9kg
Solving to find that T2 = 39.9
Then simply plug that value back into T2 - 49N = 5a

a = 1.82 m/s^2
 
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Looks good.
 
gneill said:
Looks good.
Great. Thanks
 
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