If the velocity is the same (rate of change) then I think work is the same (Physics was a long time ago).
That’s from a basic physics perspective.
Biologically I would assume 2 steps requires more calories as we don’t go slower, and I suspect the mechanics of motion don’t work as simply as a basic physics experiment.
Also been a minute since I took physics 1, but IIRC, Work w = Force f × Displacement s in its simplest form.
We can plug in Force f = Mass m × Acceleration a such that w=(ma)s to see that the primary components of work in this scenario are mass, acceleration, and displacement, so velocity doesn’t directly affect work.
I guess you could do some further derivation to get it in as a term via a=d/dtv(t), but you still end up having to calculate acceleration anyways.
If the velocity is the same (rate of change) then I think work is the same (Physics was a long time ago).
That’s from a basic physics perspective.
Biologically I would assume 2 steps requires more calories as we don’t go slower, and I suspect the mechanics of motion don’t work as simply as a basic physics experiment.
Also been a minute since I took physics 1, but IIRC, Work w = Force f × Displacement s in its simplest form.
We can plug in Force f = Mass m × Acceleration a such that w=(ma)s to see that the primary components of work in this scenario are mass, acceleration, and displacement, so velocity doesn’t directly affect work.
I guess you could do some further derivation to get it in as a term via a=d/dtv(t), but you still end up having to calculate acceleration anyways.
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