Energy as an Integral
We've learned that Work is \(\int F dx\). By substituting \(F = m \frac{dv}{dt}\) and using the Chain Rule, we can show that:
\[W = \int m \frac{dv}{dt} dx = \int mv dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\]
This is the Work-Energy Theorem: the work done on a particle equals the change in its Kinetic Energy (\(T\)).
Worked Examples
Example 1: Braking Distance
If a car with mass \(m\) and speed \(v\) hits the brakes (constant force \(F\)), how far does it travel? The work done by the brakes (\(F \cdot d\)) must equal the total kinetic energy (\(\frac{1}{2}mv^2\)). Solving for \(d\) gives \(d = \frac{mv^2}{2F}\). Notice that distance increases with the square of the speed.
The Bridge to Quantum Mechanics
Kinetic energy in Quantum Mechanics is an operator: \(\hat{T} = \frac{\hat{p}^2}{2m} = -\frac{\hbar^2}{2m}\nabla^2\). Just as classical work "accumulates" kinetic energy, a quantum system's energy is the result of the "curvature" (\(\nabla^2\)) of its wavefunction. When a particle is squeezed into a smaller space, its wavefunction must curve more sharply, which increases its kinetic energy. This "Kinetic Pressure" is what keeps atoms from being crushed into the nucleus.