In remark 1.2.6.2 (HTT), Lurie states that

Another possible approach to the problem of homotopy coherence is to restrict our attention to simplicial (or topological) categories C in which every homotopy coherent diagram is equivalent to a strictly commutative diagram. For example, this is always true when C arises from a simplicial model category (Proposition 4.2.4.4).

I am working with homotopy coherent diagrams from $\mathbb{Z}_{\ge 0}$ (thought as a poset) to $\text{Ch}_{\ge 0}(\mathbb{Z})$. Here $\text{Ch}_{\ge 0}(\mathbb{Z})$ stands for nonnegatively graded chain complexes of abelian groups, thought as an $\infty$ category. Since I want to compute the colimit of such a diagram, I am wondering the following:

Is $\text{Ch}_{\ge 0}(\mathbb{Z}) $ a simplicial model category?

In particular, in light of Corollary 4.2.4.7, it seems like for any diagram $N(\mathbb{Z}_{\ge 0}) \to N(\text{Ch}_{\ge 0}(\mathbb{Z})) $ I can find a simplicial functor $\mathbb{Z}_{\ge 0} \to \text{Ch}_{\ge 0}(\mathbb{Z})$ such that the $\infty$ -colimit of the former is equivalent to the homotopy colimit of the latter. Also, it seems like $\mathbb{Z}_{\ge 0}$ being a discrete category, such hocolimit should be computable from a discrete information. In other words, let $sk_0 : \text{sSet-Cat} \to \text{Cat}$ be the functor that takes the $0$ skeleton hom-wise and $LS: \text{Cat} \to \text{sSet-Cat}$ the functor that takes the trivial simplicial set home wise. Note that $LS$ is left adjoint to $sk_0$. Let $J$ be an ordinary category and $C$ a simplicial model category. Let $F: LS(J) \to C$ be a simplicial functor corresponding to $F': J \to sk_0(C)$. Is it true that the homotopy colimit of $F$ and of $F'$ are equivalent?

I hope I have not said too much stupid things :)