What is the most acidic proton of 2-methylcyclopent-2-en-1-one?
There are two possible enolates that can be formed from your compound: the &#x3b1; deprotonation (left equilibrium arrow in the scheme below) and the &#x3b3; deprotonation. Both protons have roughly the same $\mathrm{p}K_\mathrm{a}$ values, so deciding which one is more acidic is non-trivial.



However, experimental procedures give us a hint. Usually, the &#x3b1; deprotonation is achieved with strong bases at low temperatures — kinetic conditions. Typically conditions such as LDA in THF at $-78~\mathrm{^\circ C}$. This leads to the conclusion, that the &#x3b1; deprotonation is kinetically favoured. This is more evident in open chain compounds and especially if the &#x3b1; carbon is a methyl group.

The &#x3b3; deprotonation is generally achieved under more thermodynamic conditions: Weaker bases, warmer temperatures. I didn’t do it myself, so I cannot state conditions off the top of my head, but I would guess conditions such as $\ce{$t$BuOK}$ and $0~\mathrm{^\circ C}$. Therefore, this system must be the thermodynamically favoured one and the &#x3b3; proton must be slightly more acidic. (But only a tad, or the kinetic product would not be so easily accessable.)

I cannot provide a proper explanation, but it would seem to me like the non-branched &#x3c0; system achieved by &#x3b3; deprotonation is more stable than the branched one following &#x3b1; deprotonation.