# Question #3a5d4

Jun 8, 2015

After correction from @stefan_zdre we got better agreement!!!
Thanks Stefan!

#### Explanation:

IF GPE is Gravitational Potential Energy and $d$ is density in $\frac{k g}{m} ^ 3$

You have from your formula that:
$G P {E}_{p y r a m i d} = d \cdot {h}^{2} \cdot {s}^{2} \cdot \frac{g}{12} =$
Where I think you refer to the expression fo r potential energy $U$ as:
$U = m g h$
Where:
$m a s s = m = d \cdot V$ density times volume ($V = {s}^{2} \frac{h}{3}$)
And $h =$ position of center of mass $= \frac{h}{4}$

With your data and $g = 10 \frac{m}{s} ^ 2$
I got:

$G P {E}_{p y r a m i d} = 2.57 \times {10}^{12} J$
$G P {E}_{B l o c k s} = 1.33 \times {10}^{12}$
$G P {E}_{T o t} = \textcolor{red}{3.9 \times {10}^{12}} J$

Now if one man generates $160 k J$ per day:
You have:
$160 \times {10}^{3} \cdot 2845 \cdot 23 \cdot 365 = \textcolor{red}{3.82 \times {10}^{12}} J$