Answer
The acceleration of the 2.0-kg block is $2.29~m/s^2$
Work Step by Step
Note that the three blocks will all have the same magnitude of acceleration and the 3.0-kg will move downward while the 1.0-kg moves upward.
We can set up a force equation for the 1.0-kg block. Let $T_1$ be the tension in the rope attached to this block. Let $m_1$ be the mass of this block.
$\sum F = m_1~a$
$T_1- m_1~g= m_1~a$
$T_1= m_1~(g+a)$
We can set up a force equation for the 3.0-kg block. Let $T_2$ be the tension in the rope attached to this block. Let $m_3$ be the mass of this block.
$\sum F = m_3~a$
$m_3~g - T_2 = m_3~a$
$T_2 = m_3~(g-a)$
To find the acceleration of the 2.0-kg block, we can set up a force equation for the 2.0-kg block. Let $m_2$ be the mass of this block.
$\sum F = m_2~a$
$T_2 - T_1 - F_f = m_2~a$
$m_3~(g-a) - m_1~(g+a) - m_2~g~\mu_k = m_2~a$
$m_3~g - m_1~g - m_2~g~\mu_k = m_1~a+m_2~a+m_3~a$
$a = \frac{m_3~g - m_1~g - m_2~g~\mu_k}{ m_1+m_2+m_3}$
$a = \frac{(3.0~kg)(9.80~m/s^2) - (1.0~kg)(9.80~m/s^2) - (2.0~kg)(9.80~m/s^2)(0.30)}{1.0~kg+2.0~kg+3.0~kg}$
$a = 2.29~m/s^2$
The acceleration of the 2.0-kg block is $2.29~m/s^2$
