Why does inflation end when the potential becomes too steep? Inflation ends when the potential becomes too steep.
Is it because kinetic energy increase when the potential becomes steeper?
 A: The inflaton's equation of motion is $\square\phi=-V'(\phi)$ in a $+---$ metric (if we work in the $\xi=0$ gauge). It often helps to compare such field-theoretic PDEs to their counterparts in classical mechanics, viz. $\ddot{x}=-V'(x)$ for motion in a gravitational field. Just as a ball rolling on a hill can only get to so high a potential, and in particular will only move so far horizontally if the hill is steep up to that cutoff height, inflation also has a natural cutoff. Note that these arguments neglect quantum tunnelling, but that doesn't make much difference with high probability.
A: A homogeneous and isotropic fluid will give rise to inflation if its equation of state satisfies $w = p/\rho < -1/3$, which follows from the equation, $$\frac{\ddot{a}}{a}= -\frac{4\pi G}{3}(\rho + 3p).$$  For a uniform scalar field, we have $\rho = \dot{\phi}^2/2 + V$ and $p = \dot{\phi}^2/2 - V$, and so the condition on $w$ in terms of the kinetic and potential energies is $$2(\dot{\phi}^2/2) < V.$$  Thus, as the field rolls down the potential and gains kinetic energy, eventually this condition will fail and the scalar field will no longer support inflation. 
