Two Blocks and a Pulley Friction Problem

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SUMMARY

The discussion focuses on solving a friction problem involving three blocks connected by a pulley system. The 9.0 kg block rests on a frictionless table, while the 12 kg block experiences kinetic friction with a coefficient of μk = 0.2. The solution involves calculating the net forces acting on each block, leading to the conclusion that the acceleration of the 5.0 kg block, when released, is 1.82 m/s². The tension in the rope connecting the blocks is determined to be 39.9 N.

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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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