211–220 of 231 posts
Velocity is itself an expression of energy, and energy does scale linearly. Why does it take quardically more energy for each extra unit velocity? Well, because the closer you get to the speed of light the harder it is to get closer still. Why that is simply requires a deeper appreciation of the system than I expect is possible to muster in a comment.
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…
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 Squener…
You can convert any energy form into thermal energy. When you boil water with an immersion heater, what happens if you use two heaters? Right, they add up. But not if you think in Squenergy. Or you can measure the effect (in heat) that a ball impact has. In Squenergy two balls don't add up add up, so it's less natural to think that way.
A stationary but hot object has kinetic energy due the the motion of the individual atoms that make it up, even though its overall momentum is 0. I.e. ∑ⱼ mⱼ v⃗ⱼ = 0⃗ where the mⱼ are the masses of the parts of the object and the v⃗ⱼ are the velocities of those parts. If the object initially has 0 velocity, its kinetic energy is: T = ½∑ⱼ mⱼ v⃗ⱼ² Now we give the object a kick (or just switch reference frames) to change…
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…
> 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
I can hit the window many times lightly, with zero damage at the end.
Or I could roll of of those taps into one big energetic swing and break it. There are thresholds below which no damage happens.
Kinetic energy is the work needed to accelerate something from rest to speed v (velocity). Below we'll use the regular derivative representation to represent an tiny change in a value using notation such as dE (tiny change in energy).
Work can be described as dE (tiny change of energy) = F (force) * dx (tiny change in position). -> dE = F * dx
Force though changes momentum and algebraically can be described as:
F (force) = dp/dt (tiny change in momentum over tiny change in time)
The distance travelled over a infinitesimally small time frame is:
dx (tiny change in position) = v (velocity) * dt (tiny change in time) -> dx = v * dt
So now we have:
dE = F * dx (see our definition above) = dp/dt * v * dt (due to F = dp/dt and x = vt) = v dp (the dt terms cancel each other out and we are left with velocity and rate of change in momentum) -> dE = v * dp
So we are therefore left with:
dE (tiny change in energy) = v (velocity) * dp (tiny change in momentum) -> dE = v * dp
We can translate this as stating: the energy cost of adding a tiny bit of momentum depends on how fast the object is already moving. Now we also know that algebraically:
p (momentum) = m (mass) * v (velocity) -> p = mv
Using derivatives we also have:
dp (tiny change in momentum) = m (mass) * dv (tiny change in velocity) -> dp = m * dv
Going back to our original equation, we now have:
dE (tiny change in energy) = v * dp = v * m * dv (since dp = m * dv)
To get the total energy that it takes to move a mass m from 0 to velocity v, we need to add up all the tiny energy costs from 0 to v, and for this we use the integral to get:
E = ∫(from 0 to v) add up m (mass) * v (velocity) * dv (rate of change in velocity) = 1/2mvv (basic integration) = 1/2m*v^2
So from the above we can see that the kinetic energy rises quadratically with velocty. Now the 'why' this is encoded in the universe: read a bit more about symmetry and Emmy Noether. The laws of physics do not care where you are, when you are, or whether you are moving at a constant speed in a straight line. This is called Galilean symmetry in ordinary classical mechanics. Because the laws of motion have to stay consistent under changes of reference frame, energy cannot just be proportional to velocity. A linear energy law would break the symmetry.
Earlier quoted context omitted.
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.
"Work" is the weird thing in physics, I'd say is about as opposite to intuitive as you can get when introducing a concept. It's only intuitive when considering lifting an object - say, a bag of groceries. Heavier the bag, higher the lift -> more work. But then you carry that heavy bag a couple kilometers, arrive at home exhausted, only to be told by the physics teacher that you did exactly 0 work. Or in fact negative…
Very intuitive analogy is that running an engine in neutral burns fuel, but doesn't do work to move the car forward.
Earlier quoted context omitted.
It takes a stupid cow, but when they can climb mountains as they do it is not inconcievable that one can roll down. (I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)
However, they can go up stairs, but not down them.
A stationary but hot object has kinetic energy due the the motion of the individual atoms that make it up, even though its overall momentum is 0. I.e. ∑ⱼ mⱼ v⃗ⱼ = 0⃗ where the mⱼ are the masses of the parts of the object and the v⃗ⱼ are the velocities of those parts. If the object initially has 0 velocity, its kinetic energy is: T = ½∑ⱼ mⱼ v⃗ⱼ² Now we give the object a kick (or just switch reference frames) to change…
Looks like in your 2nd equation you've already assumed kinetic energy is quadratic with speed T = ½∑ⱼ mⱼ v⃗ⱼ²