It's easiest to visualize in terms of conversion from potential energy.
We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder. And we also know when they fall, by the time they reach the ground and all the potential energy has been converted to kinetic energy, the previously higher ball will have twice the kinetic energy too.
But a twice higher ball won't have even close to twice the speed at impact. So let's look at why not.
The force of gravity is a constant force that causes constant acceleration in free fall regardless of speed. (Ignoring air resistance, inverse sq considerations, etc.)
Suppose it takes 1 second for the ball on the 10ft ladder to hit the ground with kinetic energy of 10 and a speed of 100. Again, gravity as a constant acceleration force is speed increase per time... not speed per distance. In the ladder example, it took 1 full second for gravity to accelerate the object to speed 100.
Now think about the 20ft ladder: the ball is dropped. How much kinetic energy and speed does the ball have after it has fallen 10 feet (but still has 10 left to go)? Well it has the same exact amount as the other ball did after falling 10 feet for a duration of 1 second: kinetic energy of 10 and speed of 100.
Now the crux: thinking about when the final 10 feet of the fall look like. We know for sure the ball still has 10 ft of potential energy to covert into kinetic, and that that will happen as it falls. But what of the impact speed? Since the current velocity of the ball as it enters the last 10 feet is already 100, we know it will spend less time transiting this distance than it did the first half where it started at off at speed 0. Since gravity imparts speed in free fall as a function of time - consequently less speed will be imparted over the second 10 foot interval. That concept is enough to prove the relationship isn't linear.
If you do the actual calculation or tests, you will see one ball needs to be dropped from 4x the hight of another to hit the ground at 2x the speed, but yet with still 4x the kinetic energy.
> We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder.
What makes this intuitive? The foundation of the asker’s question is that it seems intuitive that kinetic energy would increase linearly with speed, but that turns out to be wrong.
The effort to move a piece of furniture from 1st to 2nd floor is the same as the effort to move it from the 2nd to the 3rd. We have good intuition for this by our experience, which derives a linear relationship. The effort to move a piece of furniture up two floors is double the effort of moving it up one floor (ie you have to put the same effort twice, assuming enough rest).
I would not say we have the same intuition for kinetics. Increasing walking/running from 0 to 5 km/h doesn’t feel the same as than moving from 5 to 10, which does not feel the same as moving from 10 to 15. I don’t think we have an experience of linear relationship between running speed and effort, or other types of speed/energy types of relationships.
Can someone help me understand the following?
Getting up from a seat als walking a couple steps feels that same at home and in a flying airplane (or does it?). But the base speed is 0 in the former and several hundred mph in the latter case
When you get up from a seat and walk a few steps you are already doing that on something that is hurtling down space. We don’t notice that our planet moves a lot, because we can’t really see the movement in our reference frame. If you were on a plane without any windows, no turbulence and no sound cues from the engine, you wouldn’t know when getting up from your plane seat that you are in a moving object either.
Acceleration is a real force that we can feel. But once moving at a constant speed, physics dictates that it’s all the same. That’s also why you can throw a tennis ball up on a plane and not have it fly backwards immediately smacking into the person behind you.
In the reference frame of you and the aircraft, you are not moving at all and neither is the plane. In the reference frame of the ground you and the plane are moving.
I guess Galileo came up with it first:
https://en.wikipedia.org/wiki/Galilean_invariance
you are still moving against reference frame (floor) that is at speed 0.
and also pushing that reference frame down when moving up
Scale up the numbers in you example: The effort to move a piece of furniture from 10,000th to 20,000th floor is NOT the same as the effort to move it from the 20,000th to the 3rd. The reduced gravity will help you.
If you're talking about intuitions, you have no firsthand intuitions about lifting effort decreasing with distance to the Earth. We can intuit about constant gravity, and the math of constant gravity works fine for this description.
And while the real situation at scale is more complicated, the math is going to come out to the same answer, albeit with extra terms muddying everything up.
If someone says that something true can be illustrated intuitively with a thought experiment, "sure, but what if we take that to a scale where our intuitions fail" is a sort of odd place to take the discussion unless you're genuinely curious how the math is going to shake out.
On earth, it just about is... you haven't scaled up enough. Low earth orbit doesn't have much less gravity, it's just that there's no air resistance so you can move fast enough sideways so that you don't run into the earth. Hence orbit and not just floating.
But more to the point the kinetic energy here is being turned into gravitational potential energy. If you move to a place with a weaker gradient in gravitational potential of course the same amount of kinetic energy moves you farther up.
What intuitive understanding do you have of moving furniture up 10,000 floors? None.
That's a good question, and I suppose the mgh formula isn't a suitable answer, so my answer would be something like: if you lift an object to some height, and then you repeat that action (lifting it from there to twice the height), you've done twice the work, and doing twice the work requires twice the caloric intake.
> if you lift an object to some height, and then you repeat that action (lifting it from there to twice the height), you've done twice the work, and doing twice the work requires twice the caloric intake.
You’re introducing two new intuitions, and it’s not intuitively obvious how they are related to each other. Why would work correlate 100% with caloric intake, and caloric intake 100% with kinetic energy?
Certainly, ‘work’ is highly counterintuitive. If I move a concrete block over loose sand on a beach, I’m doing zero work, in the physics definition, so moving it over a kilometer should be as easy as moving it for a millimeter.
Even ignoring the difference between caloric intake and caloric expenditure, it also isn’t intuitive to me that caloric expenditure is independent of the speed at which one lifts an object.
In the end, the answer is “because the math works out that way, and kinetic energy is a useful concept”
Friction. Work isn't just about height.
Holding that block stationary at arms length then. 0 work.
Okay but that depends on the intuitions the question is trying to justify, which makes it circular. We also know, for example, that the body uses more than twice as much energy to do twice as much work (because of fatigue on the muscles or whatever the right term is here). In fact it takes positive energy just told a weight at a fixed height, doing zero mechanical work! So you’re actually appealing to even weaker intuition than the one the question is trying to ground!
All intuitions are wrong, but some are usefull. You have to do the experiment, discover the formula, and then adapt your intuitions accordingly.
> In fact it takes positive energy just told a weight at a fixed height, doing zero mechanical work!
Stacking a weight on top of a table holds it at a fixed height and requires zero mechanical work.
The failure in intuition here relates to physiology and the mechanism by which muscles work, not physics. Myosin and actin are constantly cycling through bonding and release during muscle contraction, as this is how the shortening action actually occurs. In fact, muscle contraction is particularly unintuitive because people frequently consider ATP the "energy currency", yet the ATP-consuming steps are actually the release/relaxation and preparation for binding, not the pulling action. This is also why the phenomena of rigor mortis upon death occurs.
I think if you define energy as force X distance then integration alone will give you the squared term.
How I got banned from some reddit channel. Flip this around ask if a ball were fired out of a gun up into the air what height would it reach? A ball twice as fast goes up 4 times as high. If energy is force times distance it had 4 times the energy.
At some point you just have to shut up and calculate.
Because things like energy are relative. So if you label the ground 0, and go up 10 feet, you get x energy. Going up another exact same x from your 10 foot ladder spot you could now call 0 again, would mean you gain x energy again. Since they're both the same height, and you gained the same energy, you could infer double the height has double energy.
What 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.
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.
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
Not really. Potential energy in a gravitational well obviously has absolute coordinates.
Because physical movement is intuitively transitive. Going from A to B then B to C is the same as going from A to C.
The journey from Y to Z might feel more tiring than the journey from A to B, but only if you do them all in one day :)
So why isn't increasing your velocity from A to B then B to C the same as A to C? Isn't that intuitively transitive too?
it is if your reference point is A in both cases.
> Going from A to B then B to C is the same as going from A to C.
Not really, no. Not all forces are conservative.
Feels like what OP meant to say is, “you could rightly assume that a ball…” instead. Seems like a fair starting point if you’re just doubling things because if the height difference. I really liked cubic’s explanation overall.
That indeed would have been better. Much too late for that edit now. But the subsequent debate over the intuitive claim is fun.
Because if the one falling 20ft lands on a seesaw, the other side of it will toss two balls each of the same mass 10ft up.
Then 20ft should not be used in the explanation. They should just have one ball going at 2x speed hit the seesaw and have 4 of those balls go up at 1x speed.
That ends up begging the question, because the next step is "how high do you have to drop it from so that it's travelling twice as fast?" and you're immediately going round in circles.
Nothing of the sort. The seesaw can be in space far from gravitational influences. Potential energy is extraneous in this explanation.
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Brilliant. For those wanting more numbers [0], the ball on the 10ft ladder hits the ground at (I'll stick with imperial units) 17.296 MPH, the ball on the 20ft ladder hits the ground at 24.46 MPH or 41.42% faster, and the ball on the 40ft ladder hits the ground at 34.59 MPH or 100% faster.
[0] https://www.omnicalculator.com/physics/free-fall
I agree that this feels intuitive, that potential energy should increase linearly with height.
But in the end, it's all up to the units/quantities we choose to measure, no? If we, say, decided to measure "Squenergy" in Sqoules, with 1Sq² = 1J, then suddenly, squenergy does increase linearly with speed! The formula for kinetic Squenergy becomes sqrt(m/2)v.
Of course this complicates other stuff, like potential Squenergy becoming sqrt(MgH), it not being additive, etc.
Nice. Nitpick: in the middle paragraph you put "speed 10" instead of 100.
Fixed. Thanks.
> We know intuitively that a ball atop a 20ft ladder has twice the potential energy of a ball atop a 10ft ladder.
...no ? dropping something 10 times from 1ft is nowhere near energetic/damaging as once from 10tf
lifting something 10 times 1 foot is exactly the same as lifting it 10 feet :)
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