Does (2-fluoroethyl)benzene undergo elimination via E1cb or E2?
I think that it is easier for the molecule $\ce{Ph-CH2-CH2-F}$ to undergo E1cb mechanism than E2 mechanism. It can be also possible that the molecule undergoes elimination through a mixture of E1cb and E2, but it can be said  that majority portion of the molecules will prefer the E1cb rather than any other pathway. 

First of all, an E2 mechanism goes via simultaneous cleavage of the $\ce{C_{\beta}-H}$ and $\ce{C_{\alpha}-LG}$ bonds (LG being the leaving group). So, the rate of the reaction depends on the ease of breaking these two bonds. In this case the leaving group is fluoride ($\ce{F-}$), which has a very strong bond with the $\alpha$-carbon, making it a bad leaving group. So, due to this reason, the rate of E2 should lower for this molecule. 


Now, if we check the plausibility of E1cb reaction pathway, we will find that it is most suitable in this case. 

Firstly, the proton abstraction by the base gives a highly resonance stabilised benzylic carbanion intermediate, and this carbanion formation is the slow step. As this intermediate's energy will decrease, the activation energy will decrease resulting in an increase in the rate constant as, $k = A\exp(-E_\mathrm a/RT)$, helping to increase the rate for E1cb.


Secondly, the breaking of $\ce{C-F}$ bond doesn't happen in the rate determining step, which also doesn't alter the rate of the reaction  in this case.



So, observing the above mentioned cases, it can be definitely concluded that  E1cb is the more preferable pathway to undergo $\beta$-elimination to form Styrene.