How energy efficient is an aircraft take off?
You can make a decent approximation with very basic numbers. Take for example the CFM56-7B (which is incidentally also used on the 737NG).


Maximum thrust specific fuel consumption $TSFC$ at takeoff: ~10.9 g/s/kN
Maximum thrust $F$ at takeoff: 121 kN
Jet fuel specific energy $e$:  43 MJ


Combining these figures gives a fuel burn of 1.4 kg/s at takeoff, or 57 MJ of chemical energy released every second ($P=57MW$).

To get an airplane of mass $m$ to a certain speed $v$, it needs $$E=\frac{1}{2}mv^2$$ of kinetic energy. It does so in $$t=\frac{v}{a} = \frac{vm}{F}$$ seconds (remember, $F=ma$). The efficiency $\eta$ is useful work done $E$ divided by input energy ($Pt$),
$$\eta =\frac{mv^2}{2Pt} = \frac{vF}{2P} = \frac{v}{2\cdot TSFC \cdot e}$$

Immediately, we see why TSFC is an important figure. Plugging in some numbers, let's say $V_R =145\mathrm{kts} \approx 75\mathrm{m/s}$, we get a total efficiency of $\eta \approx 8\%$ until rotation.

Note that this calculation assumes constant TSFC while in reality it is very much a function of $v$ (indeed, with this calculation you could have $\eta&gt;1$ if $v\to\infty$ which is clearly impossible).