A dastardly villain has tied an innocent victim to a train track. Stop him with your eye laser beams

The laser eyes shine some quantity of photons per unit time, $N_{\gamma}$ , with some energy per photon, $E_{\gamma}$ , so that the power, $P_\textrm{eyes}$ is given by $N_{\gamma}E_{\gamma}$ .

Each photon that hits the train delivers twice its original momentum to the train (since it reverses directions).

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The momentum of a single photon is given by $E_{\gamma}/c$ and so the rate of momentum change for the train (the force provided by the incident photons) is given by $\displaystyle\dot{p}_\textrm{train} = 2N_{\gamma}E_{\gamma}/c$ .

For the train to come to a standstill before annihilating the victim, the incident photons must shed all the initial energy of the train.

The amount of work done by a force $F$ through a distance $l$ is given by

$\displaystyle W=\int\limits_{l_0}^{l_f} \vec{F} \cdot \vec{dl} = \int\limits_{l_0}^{l_f}\mbox{ }\dot{p}_\textrm{train}dl = \frac{2}{c}\left(l_f-l_0\right)N_{\gamma}E_{\gamma}$ which for the train must equal $\frac12 M v_0^2$ , yielding $\boxed{\displaystyle P_\textrm{eyes} = N_{\gamma}E_{\gamma} = \frac{cMv_0^2}{4L}}$

Force Due to a light beam is given by $F =\frac {P}{c}$ where P is power of source and c is speed of light.

Since this train is perfectly reflecting, $F =\frac {2P}{c}$

now, calculate force using kinematics and subs values to get the answer. P = 3,000,000,000,000 W !!