To maintain 1G of gravity, you have to accelerate at that rate.
if the mass of the craft is m, then you need to apply a force F = ma where a = 10 m/s²
if you continue that for t seconds, d = ½at²
d = ½10•t² = 5t² meters traveled
work = energy needed = Fd = ma(5t²) = 10m(5t²) = 50mt²
that tells you the energy needed.
velocity = at = 10t
momentum = mV = 10mt
If you are just throwing reaction mass out the back at velocity V₀, then the momentum of the fuel thrown out the back is m₀V₀
m₀V₀ = 10mt
so the mass of fuel used is
m₀ = 10mt/V₀
To try some numbers....
if you have a craft mass of 10000 kg, a time of 3600 seconds (1 hour), fuel velocity of 1e8 m/s (1/3 of light, pretty extreme)
then mass of fuel would be
m₀ = 10mt/V₀ = 10•10000•3600/1e8 = 3.6 kg
this is for every hour.
at some point, the mass thrown away would become a significant portion of the total mass, and the equations would change.
or course you would reach relativistic speeds at some point so the equations would change. v = at.
to reach c/2, t = 1.5e8/10 = 1.5e7 seconds or about 6 months.