## University Physics with Modern Physics (14th Edition)

$\mu_k = 0.450$
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. $mg~sin(\theta) = F_{N_1}~\mu_k + F_{N_2}~\mu_k$ $3w~sin(\theta) = 4w~cos(\theta)~\mu_k + w~cos(\theta)~\mu_k$ $3~sin(\theta) = 5~cos(\theta)~\mu_k$ $\mu_k = 0.6~tan(\theta)$ $\mu_k = 0.6~tan(36.9^{\circ})$ $\mu_k = 0.450$