If we keep accelerating in K a particle of mass m > 0, from a standing position by exerting a constant force F ≠ 0, it will have the same magnitude of acceleration, a' = F / m, in each of its momentarily comoving inertial frames along the way. (In classical mechanics, the magnitude of acceleration would be the same F / m in every inertial frame, not only in the momentarily comoving ones.)
Let's divide the motion of the particle in K into infinitely many sections as follows: the first section starts at v = 0; then, iteratively, the next section will always start as soon as the speed has reached v + (c – v) / 2. So the first section is [0; c / 2), the second is [c / 2; c · 3 / 4), and so on. If Δtv denotes the duration of the section in K that started at speed v, and av the magnitude of the particle's acceleration at the beginning of that section relative to K, then:
Δtv > ((c – v) / 2) / av = ((c – v) / 2) / (a' · (1 – v2 / c2)3/2) = c3 / (2 · a' · (c – v)1/2 · (c + v)3/2) > c / (25/2 · a')
The last expression is a positive constant independent of v, which means that the particle can never reach the speed of light this way, as there are infinitely many sections.
Now let's consider the case when the force is not constant but its magnitude is increasing with v, namely let F = m / (1 – v2 / c2)3/2, i.e. a function of the speed already reached. Then in K, a = 1 m/s2 all the time, so the particle's speed would eventually reach c after having been accelerated during the time interval of [0; c) seconds. Within this (half-open) interval, F is always of finite value (although it's converging to infinity). Therefore, this way of accelerating is feasible in theory; there is no such law in mechanics that would limit the value of F.
Note: energy is out of scope in this essay.