Free body diagram and Newton's second law

Two bodies of mass $m_1=50 \text{kg}$ and $m_2 = 10 \text{kg}$ are connected with a light rope through a pulley (no friction between the rope and the pulley). $m_2$ moves on surface with angle $37^{\circ}$ with a friction coefficient $\mu_k=0.5$. Force $\vec F$ acts on $m_1$. The system was moving with a constant velocity for some time until the force stopped and then the system started to accelerate. What's the acceleration of the system?

I assumed that the system accelerates to the left, therefore the equation for the $m_1$ would be: $\begin{cases} N_1=m_1 g\\ T-N_1 \mu=m_1a \end{cases}$

Therefore $T=m_1(a+g \mu)$

For $m_2$ the equation would be: $\begin{cases} N_2=m_2 g \cos 37^{\circ}\\ m_2 g \sin 37^{\circ} - T - N_2 \mu=m_2a \end{cases}$

Therefore $m_2 g \sin 37^{\circ} -T- m_2 g \cos 37^{\circ} \mu = m_2a$

The solution would be:

$100 \cdot 0.2024 - 50a -250=10a \\ a \approx -3.829 \frac{\text{m}}{\text{s}^2}$

Which is wrong according to the answers. The answer should be $\approx -5.83\frac{\text{m}}{\text{s}^2}$ which you can get pretty close if instead of $10a$ in the last step it was $-10a$. So I'm pretty sure that my mistake here is with the signs (of the acceleration), but I can't figure out why I'm wrong.

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Could you take the numbers out and replace them with variables (or at least numbers with units)? It's really hard to follow what you're doing when all I see are numbers with no way of determining their significance. –  Dan Mar 30 '13 at 23:01
For one thing, it is not explicitly mentioned that $\mu$ is the friction co-efficient for $m_1$ too. Another thing, $F$ must got to do something with the problem, they wouldn't mention it in the diagram else... Is there some info that you've not mentioned? Also, the answer is negative, the system doesn't accelerate to left, so it does to right. Does that ring a bell? –  Cheeku Mar 31 '13 at 0:32
@Dan - I mentioned the numbers just to show the fact that there's a problem with a sign. I do not ask to solve the question itself, I just ask whether there is a problem with a signs. –  brmch8 Mar 31 '13 at 4:53
@Cheeku - There is another question in the problem and it is to determine the force, but I solved it already. –  brmch8 Mar 31 '13 at 4:55

Answer: The problem says that the system was moving right with constant velocity. So, all the friction forces will work towards left.

Hence, the equations will become $$T+N_1 \mu = m_1 a$$

and

$$m_2 g \sin 37°- T + m_2 g \cos 37° = m_2a$$

Solving these will give you the right answer!

Elementary, my dear Watson!

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Thank you very much, sir! –  brmch8 Mar 31 '13 at 5:04