How Do Pulley Systems Affect Block Acceleration?

[jack10]

Homework Statement

two blocks are arranged as shown in this diagram and held at rest
-the table top is frictionless

http://i55.tinypic.com/py894.jpg

what would the accelerations be for each block when they are released?

Homework Equations

Fnet=ma

The Attempt at a Solution

edit: nvm, i solved it.

5.0kg block = s
2.5kg block = t

Fnet(s)= Fg(s)-2(Ft) = 5kg A(s)
= 49N - 2 (2.5kg x 2 x A(s)) = 5kg A(s)
49N/5.0kg - (10kg x A(s))/5kg = A(s)
9.8m/s^2 -2A(s) = A(s)
A(s) = 3.3m/s^2
A(t) = A(s) x 2 = 3.3m/s^2 x 2 = 6.6m/s^2

 

[anathema]

Therefore, the acceleration of the 5.0kg block is 3.3m/s^2 and the acceleration of the 2.5kg block is 6.6m/s^2. This is because the 2.5kg block experiences twice the force of the 5.0kg block due to the weight of the 5.0kg block. Since the table is frictionless, there is no opposing force to the force of gravity, resulting in a net force and acceleration for both blocks.