Find the acceleration of 2 objects given the coefficient of kinetic friction?

Question

Find the acceleration reached by each of the two objects attached by pulleys. One, 7.00 kg, is on an inclined plane of 37 degrees and the other, 12 kg, is hanging freely off the side. The coefficient of kinetic friction between the 7.00 kg object and the plane is 0.300.

I have tried this two different ways.
First:
m2gsinΘ -ukm2gcosΘ - T = m2a
T - m1g = m1a
I found T, and set the two equations equal to eachother and found
a = (m2gsinΘ - ukm2gcosΘ - m1g) / (m1 + m2).
After plugging everything in I found a = -4.882 m/s2 but this is the wrong answer.

Second try:
For 12kg object
m1g - T = m1a
(12)(9.8) - T = 12a
T = 117.6 -12a
For 7kg object
the frictional force
f = μ N = μ mg cos 37 =16.4359 N
T = f + m2g + m2a
T = 16.436 + (7)(9.8) + 7a = 117.6 - 12a
19a = 117.6 -85.036
a = 1.714
This answer is still wrong. Where did I go wrong?

Answer 1

http://en.wikipedia.org/wiki/Coefficient_of_friction