The net vertical force on the caboose (C) is
∑ F[v, C] = F[n, C] - (1.4 kg) g = 0
where F[n, C] is the magnitude of the normal force exerted by the track on the caboose; so we have
F[n, C] = (2.3 kg) g = 13.72 N
which means the coefficient of kinetic friction µ between the caboose and the track is
F[f, C] = 0.46 N = µ (13.72 N) ⇒ µ = (0.46 N) / (13.72 N) ≈ 0.034
and presumably the coefficient is the same for the locomotive.
The net vertical force on the locomotive (L) is
∑ F[v, L] = F[n, L] - (2.3 kg) g = 0
where F[n, L] is the magnitude of the normal on the locomotive, so that
F[n, L] = (2.3 kg) g = 22.54 N
and so the locomotive is opposed by a frictional force with magnitude
F[f, L] = µ (22.54 N) ≈ 0.76 N
Taking the locomotive and caboose together, the net horizontal force on the system is
∑ F[h] ≈ - F[f, C] - F[f, L] + F[engine] = (2.3 kg + 1.4 kg) (3.1 m/s²)
Solve for F[engine] :
F[engine] ≈ (3.7 kg) (3.1 m/s²) + 0.46 N + 0.76 N ≈ 13 N