#### Answer

$\mu_k = 0.450$

#### Work Step by Step

Since the block moves with constant speed, the force directed down the ramp must be equal in magnitude to the sum of the friction forces directed up the ramp.
$mg~sin(\theta) = F_{N_1}~\mu_k + F_{N_2}~\mu_k$
$3w~sin(\theta) = 4w~cos(\theta)~\mu_k + w~cos(\theta)~\mu_k$
$3~sin(\theta) = 5~cos(\theta)~\mu_k$
$\mu_k = 0.6~tan(\theta)$
$\mu_k = 0.6~tan(36.9^{\circ})$
$\mu_k = 0.450$