Total mechanical energy¶
Lets assume an object is falling at point \(y_i\) it has speed \(y_i\) at \(y_f\) its speed is \(v_f\). The total work done is:
\[
W_{\text{total}} = K_f - K_i
\]
And since the object is falling
\[\begin{split}
W_{\text{total}} = W_{\text{grav}} \\
K_f - K_i = U_i - U_f \\
K_i + U_i = K_f + U_f \\
\frac{1}{2} mv_i^2 + mgy_i = \frac{1}{2} mv_f^2 + mgy_f
\end{split}\]
Thus the total mechanical energy is:
\[\begin{split}
E = K + U \\
E_i = K_i + U_i \space \text{ energy at } y_i \\
E_f = K_f + U_f \space \text{ energy at } y_f
\end{split}\]
By energy conservation we have \(E_i = E_f\) thus E is constant at any \(y_i\) if there is no other energy \(\sum \vec{F}_{\text{other}} = 0\)
The intuition is that if we throw a object up, the kinetic energy is always converted to potential energy and vice versa.