Bungee jumper and conservation of energy A bungee jumper of mass $75 kg$ is standing on a platform $53 m$ above a river. The length of the upstretched bungee cord is 11m. The spring constant is $65.6 N/m$. Calculate the jumper's speed at $19 m$ below the platform on his first fall.
I'm able to solve it using conservation of energy $E_g= E_k+E_e$ to get correct answer.
But why isn't $E_g$ used for conservational energy at end. Why is it not $E_g=E_k+E_e+E_g$, because he still has gravity potential energy at end, doesn't he?
 A: Well, first you have to tell, what $E_g$ is and how do you measure it. You are probably using formula for change of potential energy:
$$
E_g=mg h
$$
Where $m$ is mass, $g$ is gravity acceleration and $h$ is height. This particular formula is not completely right, to be completely formal we should rewrite it as:
$$
\Delta E_g=mg\Delta h
$$
$\Delta$ represents the difference, eg. $\Delta E_g=E_{g_{END}}-E_{g_{BEGINING}}$ (final state - initial state). So your equation is technically correct:
$$
E_{g_{BEGINING}}=E_k+E_e+E_{g_{END}}.
$$
Now we should talk, how we define $E_g$. In classical mechanic you will normaly define it in Kepler potential as $E_g=-G\frac{mM}{R}$, which is useful for calculating celestial mechanics, but not simple kinematics...
For your case you can use formula $\Delta E_g=mg\Delta h$, which gives only energy difference as mentioned. But you don't know nothing about initial or final $E_g$. So you generaly say that one of this two energiyes iz 0, since you are always calulating with energy differences. You generally say, that 0 is the lowest position possible, to avoid calculating with negative $E$. So if we take in count $E_{g_{END}}=0$, your formula yields:
$$
E_{g_{BEGINING}}=E_k+E_e
$$
NOTE: $g$ is function of distance from center of Earth $R$. This is already taken in count in formula for gravitational potential energy in Kepler potential. But for exeriments near Earth surface, g is changing so slowly, that we assume, thaat it is constant, and therfore calcuate only with energy differences.
A: You are always allowed to pick the $0$ point of potential energy. For $U_g=mgh$, this means you can choose where $h=0$ is. Typically in physics problems like these you choose this point to be a starting or ending point of the object so that $U_g=0$ for some point of interest.
In your first case $U_g=K+U_e$, $h=0$ is set at $19\,\mathrm m$ below the platform. In what you propose $h=0$ could be somewhere else, for example $18\,\rm m$ below the platform. Then you would have two $U_g$ terms to worry about:
$$mg(18\,\mathrm m)=K+U_e+mg(-1\,\mathrm m)$$
Notice how this reduces to the other case though when you move the gravitational potential energy term on the right side of the equation over to the left side of the equation.
This points to the more general principle that changes in potential energy determine the physics, not absolute values of potential energy.
