Question: Consider the process of a woman lifting a barbell discussed in Section. Analyse the energy changes in this process,choosing the woman alone as the system. What quantities can be calculated with this choice of system?

The free energy principle defines how non-equilibrium steady-states are maintained in living and non-living systems by restricting their states to a minimal number. It establishes that systems minimise a free energy function of their internal states, implying hidden states in the environment.

First, write the Energy Principle, considering the woman as the system. Note that no heat is transferred, so the term is zero.

On the other hand, the barbell exerts work on the woman (), which happens to be negative because the force the barbell exerts on the her is directed downwards (because of Newton's third law), and the displacement is directed upwards. Similarly, the Earth does negative work on the woman (her center of mass displaces an unknown distance).

Note that in this occasioncontains the change in internal energy of the woman, as well as the change in kinetic energy of her arm when lifting the barbell.

Then continue to write Newton's second law for the barbell, in which it is solved for the force .

$$\begin{gathered} \sum {\text{F}}_{\text{y}}=F-mg \\ =ma \\ F=m(g+a) \end{gathered}$$

Use basic motion formulas to obtain the acceleration of the barbell in terms of its final speed and the displacement

$$\begin{gathered} v=at \\ {v}^2=a^2{t}^2 \end{gathered}$$

Divide both sides by

$\frac{v^2}{h}=\frac{a^2{t}^2}{h}$

Substitute $h=\frac{a{t}^2}{2}$into the obtained equation.

$$\begin{gathered} \frac{v^2}{h}=2a \\ a=\frac{v^2}{2h} \end{gathered}$$

Substituteinto the last equation of the previous step to obtain n terms of the given variables.

$F=m\left(g+\frac{v^2}{2h}\right)$

Substitute the obtained value of into the last equation of the second step in order to obtain to calculate the work that Earth does on the woman and the displacement of her centre of mass.

$$\begin{gathered} \Delta E_{\text{woman}}=-mgh-\frac{1}{2}m{v}^2-m_{\text{woman}}gd \\ {\overbrace{\Delta E_{woman}+m_{\text{woman}}gd}}^{\Delta E_w}=-mgh-\frac{1}{2}m{v}^2 \\ \Delta E_w=-\left(mgh+\frac{1}{2}m{v}^2\right) \end{gathered}$$

Group the terms$\Delta E_{\text{woman}}$ and $m_{\text{woman}}gd$into one term .