GoKawiil$\begingroup$
The previous answers all restate the problem as "Work is force dot/times distance". But this is not really satisfying, because you could then ask "Why is work force dot distance?" and the mystery is the same.
The only way to answer questions like this is to rely on symmetry principles, since these are more fundamental than the laws of motion. Using Galilean invariance, the symmetry that says that the laws of physics look the same to you on a moving train, you can explain why energy must be proportional to the mass times the velocity squared.
First, you need to define kinetic energy. I will define it as follows: the kinetic energy $E(m,v)$ of a ball of clay of mass $m$ moving with velocity $v$ is the amount of calories of heat that it makes when it smacks into a wall. This definition does not make reference to any mechanical quantity, and it can be determined using thermometers. I will show that, assuming Galilean invariance, $E(v)$ must be the square of the velocity.
$E(m,v)$, if it is invariant, must be proportional to the mass, because you can smack two clay balls side by side and get twice the heating, so
$$ E(m,v) = m E(v)$$
Further, if you smack two identical clay balls of mass $m$ moving with velocity $v$ head-on into each other, both balls stop, by symmetry. The result is that each acts as a wall for the other, and you must get an amount of heating equal to $2m E(v)$.
But now look at this in a train which is moving along with one of the balls before the collision. In this frame of reference, the first ball starts out stopped, the second ball hits it at $2v$, and the two-ball stuck system ends up moving with velocity $v$.
The kinetic energy of the second ball is $mE(2v)$ at the start, and after the collision, you have $2mE(v)$ kinetic energy stored in the combined ball. But the heating generated by the collision is the same as in the earlier case. So there are now two $2mE(v)$ terms to consider: one representing the heat generated by the collision, which we saw earlier was $2mE(v)$, and the other representing the energy stored in the moving, double-mass ball, which is also $2mE(v)$. Due to conservation of energy, those two terms need to add up to the kinetic energy of the second ball before the collision:
$$ mE(2v) = 2mE(v) + 2mE(v)$$
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