Force $ma$ is the rate of change of momentum, or the derivative of momentum with respect to time $$\frac{d}{dt} mv = ma = F$$.

Kinetic energy is the integral of momentum with respect to velocity: $$\int mv \cdot dv = \frac{1}{2}mv^2$$

The fact that each of these are integrals/derivatives of the other probably hints at some deeper connection.

All of these quantities seem to describe how much "oomph" something has, but can anyone clarify in more specific terms why the relationships are like this?

  • $\begingroup$ But the are not derivatives of each other. $\endgroup$
    – paparazzo
    Jul 28, 2015 at 13:01

2 Answers 2


Observe that $v\dot v=\frac12\frac{\text d}{\text dt}v^2 = \frac12\frac{\text d}{\text dt}\mathbf v\cdot\mathbf v = \mathbf v\cdot\dot{\mathbf v}$ and therefore the integral of the OP becomes (assuming a constant mass $m$, e.g. a point particle) $$\Delta\left(\frac12mv^2\right)=\int_a^b mv\ \text dv = \int_a^b \frac{\text d}{\text dt}(m\mathbf v)\cdot\mathbf v\ \text dt = \int_a^b\mathbf F\cdot\mathbf v\ \text dt,$$ which translates into $$\Delta K = W,$$ i.e. the work of the net force acting on the point particle equals the change in kinetic energy. This result is known as the Work-Energy Theorem.


You are talking about two derivatives, one derivative to time and the other to velocity. If you would derive the kinetic energy two times to the velocity, you would end up with m, not F.


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