Loads in decelerating a falling object I remove trees in confined spaces setting up rigging systems, I am keen to look at the science behind it in more depth so far the formulas that I have found don't take into effect the ongoing effect of gravity once the object or log begins decelerating.
So far,

If a $300\:kg$ log falls $1\:m$ and then is slowed down (evenly) over $2\:m$ to a stop how much force is required (or exerted on the elements within the system)?

$$KE = \frac12mv^2 = Fd$$
where $d$ is stopping distance, or 
$$F=\frac{\frac12mv^2}{d}$$
$$F=\frac{\frac12\cdot300\cdot(1\cdot9.806)^2}{2}=7212\:N$$
is the stopping force required to balance the force of the log. as this force is on both sides of the pulley it is doubled (I know widening the angle at the pulley can reduce this).
I'm after confirmation that this is the correct formula for this scenario. 
Scratch that; I know this formula is wrong for the scenario as if the log's velocity is zero there is still force on the system, but not according to the formula???
 A: There can still be forces involved even if nothing is moving.  If all the forces balance, then the only thing that will say is that there should be no net acceleration (which is a change in velocity, either through change in speed or change in direction… remember these are vectors).  In your system, gravity is constant, but the acceleration of the tree is due to the combination of forces, which are not in balance (otherwise the tree would never stop moving or never start).
The trick is in the part of the statement that says the tree is "slowed down (evenly) over 2m to a stop," which tells you two things:  (1) the problem implies a constant acceleration and (2) the displacement over which the acceleration occurs.  This sounds like a problem handled by the kinematic equations of motion.  The problem is further simplified by being essentially one-dimensional, but I will show the equations in vector form below.  The first equation to consider is:
$$
\mathbf{x}(t) = \mathbf{x}_{o} + \mathbf{v}_{o}t + \frac{1}{2}\mathbf{a} \ t^{2}
$$
where $\mathbf{x}_{o}$ is some initial displacement [meters] (i.e., 2 meters here), $\mathbf{v}_{o}$ is some initial velocity [meters per second] (determined from the first part of the problem where the tree is in free fall for 1 meter), $\mathbf{a}$ is a constant acceleration [meters per second per second or m/s$^{2}$], and $t$ is the amount of elapsed time [seconds].  Since you are given distances and not times for durations of fall, then a slightly different equation is more appropriate, which is given by:
$$
\mathbf{v}_{f}^{2} = \mathbf{v}_{o}^{2} + 2 \mathbf{a} \left( \mathbf{x}_{f} - \mathbf{x}_{o} \right)
$$
where $\mathbf{v}_{j}$ is the initial (j = o) and final (j = f) velocities and $\mathbf{x}_{j}$ is the initial (j = o) and final (j = f) displacements.  You can choose many of these parameters to make your life easier.  For instance, in the first part, you could choose $\mathbf{x}_{o}$ = 0 and $\mathbf{x}_{f}$ = 1 meter, $\mathbf{a}$ = acceleration of gravity, $\mathbf{g}$ (or ~9.8 m/s$^{2}$), and $\mathbf{v}_{o}$ = 0.  Then you can solve for $\mathbf{v}_{f}$ for the first part and use it in the second part of the problem, with a similar methodology (though in the second part you solve for $\mathbf{a}$ which no longer equals $\mathbf{g}$).
Side Note:  Having been a former logger myself (I worked manual labor to pay for college), is there any way to start controlling a falling tree before letting it fall 1 meter?  Those things can twist a lot even in that short distance and cause all sorts of headaches, not to mention the extra burden on the equipment (it takes more force because you added momentum to the tree by letting it fall initially).
A: To obtain the fall velocity we have:
$$mgh=\frac{1}{2}mv^2$$
From this you can extract a value for the velocity, as a function of the altitude.
Subsequently, we need to slow the tree down. If we have a distance to stop, we know
$$\Delta s= \Delta v \cdot \Delta t$$
If we know the impulse, and the time available to stop, we can use:
$$m\Delta v = F \Delta t$$.
In short, use the first equation to find $v$, find the second equation to find $\Delta t$, and use the last equation to find $F$.
Please note that this a kinematics story, and this is the force necassary to slow the tree down. If the velocity is zero there will still be the gravitational force which needs to be counteracted.
A: It may help to decompose the force required into two parts. 
First, notice that if you provide mg, or 300 x 9.8 newtons in opposition, the log will not accelerate. If stationary it will remain so, and if moving at some velocity it will continue to move at that velocity.
Now look at the 1 meter fall. This will provide a velocity as you have described, although your force calculation is incorrect (v is 9.8 m/sec if the log falls for 1 second, not it if falls for 1 meter) in terms of counteracting the momentum and energy produced, but only in the absence of any extra forces such as gravity. Fortunately, you dealt with that in the first step. Actually, your force calculation doesn't need to deal with velocity at all. $$mgh = Fd$$ (force times distance equals force times distance) and $$F=\frac{mgh}{d} = \frac{300 \times 9.8 \times 1}{2} = 1470 N$$
So the force required will be$$F_{total}= F + mg = 1470 + (300\times9.8) = 4410 N$$
This, of course, is a wildly optimistic figure, since it essentially assumes that you can produce just the right force so that deceleration is constant over the 2 meters of controlled fall. Things like the elasticity of the rope and the difficulty of guaging just how much force is being applied will cause peak forces to be considerably higher.
