I’d agree with the air part.
However, with a heavy wheelbarrow on grass and soft ground? At slow speeds it would not surprise me if that was the dominate consideration.
Another way to look at it:
Work = Force x Distance (Newtonian definition of work)
Change in speed = Force x Time (x a constant to account for the mass involved)
So if you apply some given force for 1 second, you will increase the speed by 1 MPH regardless the starting speed.
BUT if the starting speed was higher, then you covered more distance during that second, so you did more work.
The above neglects friction and a number of other factors.
Yes, but you still have to pick some particular reference frame and stick to it (or if you’re changing reference frames, take that change into account). The energy of a system and its components will be different in reference frame A from what it is in reference frame B, but in any given reference frame, it’s conserved and everything works out. What’s confusing about the rocket example is that you’re trying to use a reference frame attached to the rocket, but it’s accelerating, so it isn’t always in the same reference frame.
Ah, that makes more sense. I still have no idea what’s going on, but at least now I know why.
It’s not doing twice as much work. It is only doing twice as much work specifically to its winch brake. It’s doing much less work to the rest of its environment, by virtue of leaving exhaust gasses that are traveling backwards more slowly. There are a few different buckets to keep track of here. When you keep track of them all, both rockets do the same amount of work. They have to. They are converting the same amount of chemical potential energy into heat and motion in the same end products.
We’re saying the same thing, but you’re right, I should have been more specific: the faster rocket is doing twice as much work to its vehicle as the slower rocket is to its (slower) vehicle.