I presume you mean when it reaches the end of the slack! For your energy equation to hold good, the truck must freewheel (frictionlessly, of course), so that it is using only its known kinetic energy at the end of the slack. If, as in the real world, the driver continues to apply throttle, then the energy in the rope will be supplemented by the additional pull of the engine. (Which would of course be a very significant increaase in the force and energy!).
Given that, your e=mv^2 equation obviously yields the same result as my equations in my post above.
Indeed. It does simplify the mathemetics a bit to assume the driver only uses momentum to provide the snatch. This is the way I begin all snatch recoveries, as it doesn't shock load the driveline. Only if necessary will I still apply gas after the snatch occurs. I'd rather bounce off the rope a few times than over-drag the other vehicle.
The other element here, is if you are on soil with a low shear strength (such as sand), you can dig yourself into trouble if you continue to apply power.
... But proper recovery technique is not the question at hand, eh?...
energy = force x displacement.
Why do you use e=1/2 * f * s, and then double it? It gets the same result, but I don't follow the logic of the "average" force part. We are only interested in the terminal force and terminal energy, surely?
Ahh! You're right. Why did I halve it at first...? I must have confused the energy of a spring equation (1/2 * k * s^2) with the energy of force equation (f * s). Crap... anyways, moving on.
If we use the f * s equation, it implies that the force is
constant over the distance 's'. That is a problem, because in our scenario it is not.
In our scenario, the strap/chain/etc most closely represents a linear spring. At displacement = 0, force = 0, and the force line increases at some unknown, approximately linear, rate from there.
The full energy equation is an integral of the force equation over the displacement:
.
All we are looking for is the max force though. If you compare the force vs. distance graph of a constant force and a linear force over the same displacement range from the origin, both with the same total areas (or energy)... you will find the 'max force' of the linear force graph = twice that of the constant force graph. Thus the final * 2.
So if you don't mind, I will amend my response below:
(hopefully my explanation makes sense, I am having trouble putting my thoughts into words here)
The two vehicles are attached with the strap, loose at first. The recovering vehicle proceeds forward with a bit of gas, reaching only 5 mph when reaching the end of the strap. At this point the vehicle has gathered kinetic energy equal to 1/2*mass*velocity^2.
KE = 0.5 * 5000 lb * (5 mi/hr)^2 = 5.7 kJ
We will assume for this instance that the stuck vehicle will remain stuck and will not budge (worst case)... so all of the recovering vehicle's energy transfers into the strap and is turned into elastic potential energy. This stored energy will be equal to the kinetic energy that the truck had. This stored energy relates to the force exerted on each end by the following: energy = average force * distance. The distance is how far the strap stretches. The average force is assuming the rope exerts constant force, which ours does not. Because it's force exerted most closely resembles a linear relationship to the stretch, the average force should be multiplied by 2 to get the maximum exerted force (which is all we are interested in here)... assuming the system reaches equilibrium without failure.
In instance 1, the truck used a static strap, which we will assume stretches only 4 inches before reaching equilibrium.
5.7 kJ / (4 in) * 2 = 25,072 lbf (enough to snap a strap or possibly rack your frame)
For instance 2, we will use a dynamic strap, which will stretch about 6 feet.
5.7 kJ / (6 ft) * 2 = 2,089 lbf (well within the safe range of most straps)
For the last instance, what if we used a chain, which has extremely minimal stretch. So we will say 0.5"...
5.7 kJ / (0.5 in) * 2 = 200,576 lbf (you will certainly break something!!)
So, I hope this gives you a real world, numerical understanding of why dynamic straps should ALWAYS be used in dynamic vehicle-to-vehicle recoveries.