Answer: a = 2.8 m/s$^2$, T = 14 N

We can draw the free body diagrams for our problem as follows:

When the system is released, let the acceleration of the blocks be $a$.

From the Free Body Diagrams, for the bigger block $m_1 \rightarrow T = m_1a$ and for the lighter block $m_2 \rightarrow m_2g - T = m_2a$

Solving for $a$, we get $a = \large\frac{m_2g}{m_1+m_2}$

Substituting for $a$ and solving for $T$, we get $T = m_1a = \large\frac{m_1m_2g}{m1+m_2}$

Given that $m_1 = 5 kg$ and $m_2 = 2 kg$:

$\Rightarrow a = \large\frac{2 \times 9.8}{5 + 2}$$ =2.8\; m/s^2$

Solving for $T$, we get $T = m_1a = 5 \times 2.8 = 14\;N$

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