Answer
(a) $a = 1.55~m/s^2$
(b) $\mu_k = 0.53$
Work Step by Step
Let $a$ be the acceleration for both blocks. Let $m$ be the mass of each block. Let $F_t$ be the tension in the cord.
(a) We can set up a force equation for block B.
$ma = \sum F$
$ma = mg - F_t$
$F_t = mg - ma$
We can set up a force equation for block A. Note that we can use the equation for block A to replace $F_T$.
$ma = \sum F$
$ma = F_t - mg~sin(\theta) - mg~cos(\theta)\cdot \mu_k$
$ma = mg - ma- mg~sin(\theta) - mg~cos(\theta)\cdot \mu_k$
$2a = g - g~sin(\theta) - g~cos(\theta)\cdot \mu_k$
$a = \frac{g - g~sin(\theta) - g~cos(\theta)\cdot \mu_k}{2}$
$a = \frac{(9.80~m/s^2) - (9.80~m/s^2)~sin(34^{\circ}) - (9.80~m/s^2)~cos(34^{\circ})\cdot (0.15)}{2}$
$a = 1.55~m/s^2$
(b) If the system is not accelerating, then the forces acting on the system in opposite directions must have the same magnitude.
$mg~sin(\theta)+mg~cos(\theta)\cdot \mu_k = mg$
$cos(\theta)\cdot \mu_k = 1 - sin(\theta)$
$\mu_k = \frac{1 - sin(\theta)}{cos(\theta)} = \frac{1 - sin(34^{\circ})}{cos(34^{\circ})}$
$\mu_k = 0.53$
