Potential Energy determination for the One Floor Free-Fall Drop of the topmost M(16), (1/4) inch thick steel Block onto the top of the M(94) Steel Block, Floor 1-94 plus the 6 super thick steel Baseme

This is the second important Energy Calculation for WTC-1, the North Tower, for M(16) Free Falling one floor or 3.8 meters onto the top of M(94), directly below.

The only Force allowed to act on M(16) through 3.8 meters of Free-Fall, is Gravity….
Use basic High School Newtonian Physics to calculate your answer in Energy units of Joules (J).
You might chose to use this Equation: Potential Energy = M(16)∙g∙h(3.8m)
We know the Gravitational Free Fall acceleration constant for earth is (g) = (9.8 m/s^2)

We also can know that h(1 flr) = 3.8 meters, since h(1) = (1/110)(h(110)) = (0.0091)(415 m)

The mass of the alleged free-falling 16 floor block, M(16), needs now to be accurately calculated based on the correct relative steel thicknesses of the steel walls contained in M(16) as
compared to M(94), on a % Mass or % Weight ratio basis.

It turns out, after viewing my Steel Thicknesses Power Point Graphics at:
http://911scholars.ning.com/photo/zaks-wtc-tt-crosssection-of?context=user, and http://911scholars.ning.com/photo/graphical-mathematical?context=user, and
http://911scholars.ning.com/photo/tall-crosscut-wall-of-wtc-tt-1?context=user,
that M(94) is essentially 55 times more massive as M(16) due to the greater
thicknesses of the steel in M(94), at 4”, 5”, and 6” reported in the lower floors
and the 6 additional basement floors. It follows then that:

M(94) = 98.2% of the total Tower’s Steel.
M(110), the total Tower’s weight is 5 x 10^5 Tons = 500,000 Tons = 1 x 10^9 Lb = 4.6 x 10^8 Kg
M(94) = (98.2%)(4.6 x 10^8 Kg) = 452 x 10^6 Kg = M(94)
M(16) = (1.8%)(4.6 x 10^8 Kg) = 8.3 x 10^6 Kg = M(16)

We now have all of the quantities, in red, required to calculate the Energy of a 3.8 meter, 1 floor, M(16), Free Fall and the necessary Physics Equation, in red, to calculate the final Energy result.
PE(16)FF(3.8m) = (M(16))∙g∙(h(1 floor))
PE(16)FF(3.8m) = (8.3 x 10^6 Kg)(9.8 m/s^2)(3.8 m) = 3.1 x 10^8 Joules

The Energy released by one Little Boy Atomic Bomb (LBAB) = 6.3 x 10^13 Joules = 15 KiloTons TNT.
Therefore the PE of M(16) free Falling 3.8m, divided into the PE(LBAB) gives the following:
[PE(LBAB)/PE(of M(16)FF(3.8m))] = [(6.3 x 10^13 Joules)/(3.1 x 10^8 Joules)] = 203,000:1
Meaning PE(16)FF(3.8m) is (1/203,000) the energy of 1 LBAB, which is 0.074 Ton of TNT = 74 Kg of TNT
= 164 Pounds of TNT explosive (is very, very small amt. compared to 1LBAB = 33,000,000 pounds TNT)