# Calculating watt output (not consumption) for lasers

Force is mass times acceleration. Energy is force imposed on an object times the length that object traveled whilst under the influence of that force. Power is energy divided by the time it took to move that object. Because photons are massless, wouldn't the power output of any laser be zero?

Note: I know this can't be true because obviously you can see lasers. This is more of a "why" question rather than a "yes or no" question.

The definitions you give for force, energy and power are actually for special cases and not generally applicable, for instance to photons, as you correctly notice. A more general definition of force is: \begin{align}\vec{F}&=\frac{{\rm d}\vec{p}}{{\rm d}t}\\&=\vec{v}\frac{{\rm d}m}{{\rm d}t}+m\frac{{\rm d}\vec{v}}{{\rm d}t}\\&=\vec{v}\frac{{\rm d}m}{{\rm d}t}+m\vec{a}\end{align} where $\vec{p}=m\vec{v}$ is the momentum. If the rate of change of the mass is $0$, then this reduces to the familiar $\vec{F}=m\vec{a}$.
The definition you give for energy is what's usually called 'work'. The more useful quantity in this context is the analogue of the kinetic energy, which you probably know as $E=mv^2$. The more general formula is: $$E=\sqrt{p^2c^2 + m^2c^4}$$ Again $p$ is the momentum, and $c$ is the speed of light. Notice that for a particle with no velocity ($p=0$) this reduces to $E=mc^2$, which I'm sure you've seen before. It turns out that for photons it makes sense to define the momentum not as $p=mv$ but as $p=\frac{h\nu}{c}$ (this can be/has been checked experimentally), where $h$ is Planck's constant and $\nu$ is the photon frequency (related to the wavelength as $\nu=\frac{c}{\lambda}$). It then follows that the energy of a photon or other massless particle is: $$E=h\nu$$ Finally, power is defined in general as: $$P = \frac{{\rm d}E}{{\rm d}t}$$ or in words, as an energy rate. For a laser output you could estimate this by counting the number $N$ of photons which come out in a time interval $\Delta t$. A laser emits a single frequency, so all the photons have energy $h\nu$. Thus: $$P=\frac{Nh\nu}{\Delta t}$$