Conservation of energy confusion Ideal Zorro (mass $M$, no height) swings down on a vine of length $L$ from a height
$H$ and grabs a kid of mass $m$ (zero height, standing on the ground) and together
they barely reach safety at a height $h$. Relate $H$ to the other parameters. Give $H$
in meters if $L = 40m, M = 100kg, h = 6m, m = 30kg$.
The given solution states that when Zorro reaches the ground, $(1/2) Mv^2 = Mgh$.
The solution does not account for the work from the force of tension in the vine. Why not? I thought the conservation of energy says that $(1/2) mv_f^2 - (1/2)mv_i ^2 = \int F(x) dx$, where the RHS F(x) would account for both gravity and force of tension in this problem.
Also, the solution does not account for the change in mass when the kid is picked up.  Shouldn't the change in mass be considered and how should it be accounted for?
 A: 
I thought the conservation of energy says that...

You're overthinking.  Conservation of energy says that if there's no losses, then energy stays the same.  Semantically, "conserve" means "stays the same".
So if you (or Zorro) change potential energy by $Mgh$, then that energy has to go somewhere.  If it doesn't go into heating up vines and tree branches and whatnot, then it has to go into kinetic energy.  Kinetic energy is equal to $\frac 1 2 M v^2$, so there's your relationship.
The whole point about the amount of attention that physicists (and engineers) put into these conservation laws is that they make things easy.  You don't have to calculate the solution to $\int F(x) \dot dx$ to get an answer.  If you know that the system is lossless, you don't even have to know $F(x)$ or $x(t)$ or anything else -- you just need to know that the energy has to go someplace, reason out where it has to go, and then you have your answer.
Sometime, when you have the time and the mathematical chops, you can go ahead and calculate $\vec F(\vec x)$ (yes, they're vector quantities) for Zorro swinging on a vine of length $h$.  Then you can calculate $\vec x(t)$, and you can find out what you already know* -- which is as long as the physical laws of the universe aren't changing over time**, energy is conserved.
* Thank you, again, Emily Noerther.
** Yes, yes, the universe is expanding.  But not by much.  You can ignore that bit.
A: The change in kinetic energy is the net work done by all the forces.  Gravity is a conservative force so we define the change in potential energy as the negative of the work done by gravity.  You can include the force of gravity in your $\int_{a}^{b}\vec F \cdot d\vec x$ where $\vec F$ is the total force including gravity, or you can treat the part of the work from the force of gravity as the negative of the change in potential energy, - $mg\Delta y$, where $y$ is the elevation above ground.  It is easier to treat the work from gravity using the potential energy as it depends only on the differences in elevation regardless of how complicated the path is between the initial and final elevations.
The force of tension does no work for a rigid vine since the angle between the tension force and the displacement is always 90 degrees. Also, you can treat this problem as a simple pendulum noting that the force of tension produces no torque about the top fixed point of the vine.
I think the given solution is wrong. At the ground before Zorro picks up the kid his speed $V$ is given by ${1 \over 2} MV^2 = MgH$.  When Zorro picks up the kid his speed changes from $V$ to $ V^* = V{M \over {m + M}}$ due to conservation of linear momentum.  Then ${1 \over 2} (M + m){V^*}^2 = (M + m)gh$.  You can solve for H.
