Fix a reference frame and suppose you are on a frictionless surface standing next to a heavy box equal to your own mass, and you and the box always start at rest relative to one another. In every example, you will push the box leftward, adding 1 m/s leftward velocity to the box, and adding 1 m/s rightward velocity to yourself.
Let’s suppose we didn’t know what “kinetic energy” is, but let’s suppose such a concept exists, and that whatever it is, an object of your mass has 0 units of it when at rest, and it is a continuous monotonic function of the absolute value of that object’s velocity. Let’s also take as an assumption that when you perform such a push like the above, you are always adding precisely 1 unit of this thing called “kinetic energy” to you and the box combined.
Okay so suppose the box and you are at rest and you perform this push, and start moving at 1m/s left and right, respectively. You and the box started with 0 units of kinetic energy, and you added 1 unit total. Since you and the box have the same absolute value of velocity, your energies are equal, so you each must have gotten 1⁄2 of a unit. Great, therefore we derive 1 m/s is 1⁄2 unit of kinetic energy.
Now suppose you and the box start out at a velocity of 1m/s rightward, so you have 1⁄2 unit of energy each, for a total of 1 unit. You perform the same push, bringing the total kinetic energy to 2 units. The box ends up at 0 m/s, so it has 0 units of energy now. You end up going 2m/s rightward, with all the energy. Great, therefore we derive 2 m/s is 2 units of kinetic energy.
Now suppose you and the box start out at a velocity of 2m/s rightward, so you have 2 units of energy each, for a total of 4 units. You perform the same push, bringing the total kinetic energy to 5 units. The box ends up at 1 m/s, so it has 1⁄2 unit of energy now, since we derived earlier that 1m/s is 1⁄2 unit of energy. You end up going 3m/s rightward. So you must have the other 4.5 units of energy. Therefore we derive 3 m/s is 4.5 units of kinetic energy.
We can continue this indefinitely, without running into any inconsistencies or contradictions. This “kinetic energy” thing so far seems to be a self-consistent concept given these assumptions! In general, we derive that an object of our mass moving at velocity v is has a kinetic energy of 1⁄2 v^2 units.
And I hope this makes it clearer why kinetic energy has to behave quadratically. A quadratic function f is precisely the kind of function such the quantity f(x+c) + f(x-c) − 2f(x) is constant with respect to x. It’s the only function that satisfies the property that a fixed “amount of push” of the propellant you are carrying away from you always adds the same total energy into the combined system of you + propellant.
And it also gives some intuition for why you end up with more energy when you fire the propellant while moving faster. When you pushed the box while initially at 0 m/s, your kinetic energy went from 0 units to 0.5 units (+0.5), but when you pushed the box while initially at 1 m/s, your kinetic energy went from 0.5 units to 2 units (+1.5), and when you pushed the box while initially at 2 m/s, your kinetic energy went from 2 units to 4.5 units (+2.5) and in all cases you only added 1 unit of energy yourself. Where does the extra energy come from? From slowing down the box’s rightward motion, and/or from not speeding up the box to go leftward from rest.
Here’s my intuition-driving example/derivation.
Fix a reference frame and suppose you are on a frictionless surface standing next to a heavy box equal to your own mass, and you and the box always start at rest relative to one another. In every example, you will push the box leftward, adding 1 m/s leftward velocity to the box, and adding 1 m/s rightward velocity to yourself.
Let’s suppose we didn’t know what “kinetic energy” is, but let’s suppose such a concept exists, and that whatever it is, an object of your mass has 0 units of it when at rest, and it is a continuous monotonic function of the absolute value of that object’s velocity. Let’s also take as an assumption that when you perform such a push like the above, you are always adding precisely 1 unit of this thing called “kinetic energy” to you and the box combined.
Okay so suppose the box and you are at rest and you perform this push, and start moving at 1m/s left and right, respectively. You and the box started with 0 units of kinetic energy, and you added 1 unit total. Since you and the box have the same absolute value of velocity, your energies are equal, so you each must have gotten 1⁄2 of a unit. Great, therefore we derive 1 m/s is 1⁄2 unit of kinetic energy.
Now suppose you and the box start out at a velocity of 1m/s rightward, so you have 1⁄2 unit of energy each, for a total of 1 unit. You perform the same push, bringing the total kinetic energy to 2 units. The box ends up at 0 m/s, so it has 0 units of energy now. You end up going 2m/s rightward, with all the energy. Great, therefore we derive 2 m/s is 2 units of kinetic energy.
Now suppose you and the box start out at a velocity of 2m/s rightward, so you have 2 units of energy each, for a total of 4 units. You perform the same push, bringing the total kinetic energy to 5 units. The box ends up at 1 m/s, so it has 1⁄2 unit of energy now, since we derived earlier that 1m/s is 1⁄2 unit of energy. You end up going 3m/s rightward. So you must have the other 4.5 units of energy. Therefore we derive 3 m/s is 4.5 units of kinetic energy.
We can continue this indefinitely, without running into any inconsistencies or contradictions. This “kinetic energy” thing so far seems to be a self-consistent concept given these assumptions! In general, we derive that an object of our mass moving at velocity v is has a kinetic energy of 1⁄2 v^2 units.
And I hope this makes it clearer why kinetic energy has to behave quadratically. A quadratic function f is precisely the kind of function such the quantity f(x+c) + f(x-c) − 2f(x) is constant with respect to x. It’s the only function that satisfies the property that a fixed “amount of push” of the propellant you are carrying away from you always adds the same total energy into the combined system of you + propellant.
And it also gives some intuition for why you end up with more energy when you fire the propellant while moving faster. When you pushed the box while initially at 0 m/s, your kinetic energy went from 0 units to 0.5 units (+0.5), but when you pushed the box while initially at 1 m/s, your kinetic energy went from 0.5 units to 2 units (+1.5), and when you pushed the box while initially at 2 m/s, your kinetic energy went from 2 units to 4.5 units (+2.5) and in all cases you only added 1 unit of energy yourself. Where does the extra energy come from? From slowing down the box’s rightward motion, and/or from not speeding up the box to go leftward from rest.