alt Hacker NewsWhat if you label standing still as 0 mph and start moving 10 mph, gaining x energy, then call that zero and start moving 10 mph from there? It's just as intuitive to say that you would gain x energy in that case, but you don't.
Replies
That's clever, and I can't imagine or explain it as easily. Something to do with a reference point moving away from you so when solving for bringing it back to zero it's different than just adding the two energies back together. You have to add up the energy of catching them all up to the initial starting reference. I think also because distance is one unit, so moving reference pointe is easier. Moving reference points on distance over time already gets my spidey senses going that it's not something you should do without some real understanding.
I suppose they are both "intuitive", but the example I gave was both intuitive and correct. Probably for anyone who has carried something or themselves up a hill, or climbed a set of stairs can relate to that from firsthand experience. I don't know what the kinetic energy corollary to that would be? "Stand still and I will throw a baseball at you going 15mph, and note how much it hurts. Now I will throw it at you going 30mph. See! It hurts 4x as much" :D
When you're already going 10 mph and you're about to add another 10 mph, you can only "call that zero" (i.e., go from 0 mph to 10 mph again) if your point of reference (i.e., the ground) also begins moving with you at that point. Since the ground is stationary, you're definitely about to increase from 10 mph to 20 mph relative to the ground, not from 0 mph to 10 mph, and that's harder to do. But if you're on a treadmill that was stationary for the first change, and then suddenly starts moving at 10 mph right before the second change without affecting your speed relative to the ground, then you can "call that zero" and you'll be able to add another 10 mph (ending up at 10 mph relative to the treadmill and 20 mph relative to the ground) with the same ease as the first go.