course mth174

Okay, back to the problem at hand:I'm weighing the two possibilities of the chain, which seem that there would just be either no kinetic energy at all or just 1/2mv^2 for the falling chain.

You stated that the potential energy is dependent on the chain above the floor, so would it just be mg(h-y), but you had it wrote as ""its PE will be 1/2 m g * L/2, where m is its mass.""

I'm a little confused as to where you got the 1/2 and L/2.

If I can just correctly get the equation started everything else is straightforward afterward.

Sorry for the delay. Looking forward to finishing up the semester with flashing colors. "

You haven't included my previous comments; you need to include the entire thread of the discussion with every submission.
However I believe I specified the conditions in my previous notes. The 1/2 m g L/2 corresponds to the instant the last bit of the chain passes through the hole, as assumes furthermore that L < H.
However the expression should have been 1/2 m g L, not 1/2 m g L/2. Typed in an extra division by 2.
The mass of the falling portion of the chain at any instant depends on how much chain is falling at that instant. m g ( h - y) is equal to the mass of the entire chain, including any part that might still be on the table, multiplied by the distance of the end of the chain below the table. That quantity lacks a significant meaning and doesn't help.