One way of looking at the question is to say that all induction is over some well-ordered well-founded set, and strong induction is useful when the order type of that well-ordered well-founded set is not the same as that of the natural numbers. Rather than elaborate on this general remark, let me simply give two examples of slightly more sophisticated induction that would appear in many undergraduate courses.
The first is the statement that every positive integer has a prime factorization. Here if we are trying to factorize n it is convenient to assume that all smaller numbers can be factorized, since if n is not prime then we will not use the factorization of n-1 but rather the factorizations of two factors of n about which we know nothing. In a sense we could say that the well-ordered well-founded set that underlies this proof is N with the relation "is a proper factor of". (The simplest proof that this is a well-ordering well founded is to use the fact that if a is a proper factor of b then a is less than b, though once we know the fundamental theorem of arithmetic then this seems slightly unnatural. But we don't want to assume the fundamental theorem of arithmetic in order to prove the easy part of the fundamental theorem of arithmetic.)
The second is the statement that every tree with n vertices has n-1 edges. Here the proof is that every tree must have at least one vertex of degree 1 (or you could keep following a path until you hit a point you've hit before -- to make this completely rigorous will itself involve induction but let's forget that one) and that if you remove that vertex and its edge then you must still have a tree, which by induction has n-2 edges. Here we could either do straightforward induction on the number of vertices or we could do induction on the more complicated well-ordered well founded set of all graphs under strict containment. The advantage of the first is that it is straightforward induction, but the disadvantage is that we have to create a slightly artificial inductive hypothesis -- that all trees with n vertices have n-1 edges. Graph theory beginners often trip up here and try to prove results like this by arguing that if you take a tree and add a new vertex to it, joining it to one of the existing vertices, then by induction the old tree has n-2 edges so the new one must have n-1 edges. In other words, they assume the result for n-1 and prove it for n, but unfortunately the result they assume for n-1 is not the right one and strictly speaking all they prove is that for each n there is a tree with n vertices and n-1 edges.
The second example isn't exactly strong induction, but I think it fits into the general discussion and helps to give some idea of what forms of induction are appropriate when.
