- #1

joemost12

- 20

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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:

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

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