#### Answer

(a) The minimum mass that will stick and not slip is 1.83 kg
(b) $a = 1.32~m/s^2$

#### Work Step by Step

(a) To find the minimum mass, we can assume that the force of static friction is at a maximum. Then the component of the weight of block $m$ directed down the slope plus the force of static friction directed down the slope will be equal in magnitude to the weight of the 2.0-kg block.
$mg~sin(\theta)+F_f = (2.0~kg)~g$
$mg~sin(\theta)+mg~cos(\theta)~\mu_s = (2.0~kg)~g$
$m~sin(\theta)+m~cos(\theta)~\mu_s = 2.0~kg$
$m = \frac{2.0~kg}{sin(\theta)+cos(\theta)~\mu_s}$
$m = \frac{2.0~kg}{sin(20^{\circ})+cos(20^{\circ})(0.80)}$
$m = 1.83~kg$
The minimum mass that will stick and not slip is 1.83 kg
(b) Note that both blocks will have the same magnitude of acceleration. We can set up a force equation for the 2.0-kg block. Let $T$ be the tension in the rope. Let $M$ be the mass of this block. So;
$\sum F = Ma$
$Mg-T = Ma$
$T = M(g-a)$
We can set up a force equation for the 1.83-kg block. Let $m$ be the mass of this block.
$\sum F = ma$
$T - mg~sin(\theta) - F_f = ma$
$M(g-a) - mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$
$Mg - mg~sin(\theta) - mg~cos(\theta)~\mu_k = (M+m)a$
$a = \frac{Mg - mg~sin(\theta) - mg~cos(\theta)~\mu_k}{M+m}$
$a = \frac{(2.0~kg)(9.80~m/s^2) - (1.83~kg)(9.80~m/s^2)~sin(20^{\circ}) - (1.83~kg)(9.80~m/s^2)~cos(20^{\circ})(0.50)}{2.0~kg+1.83~kg}$
$a = 1.32~m/s^2$