Calculations for a DC motor
Waiting for more detailed information but if it seems you measured:

Force to overcome static friction: \$F_{static} = ma = 2kg * 9.81m/s^2\$ = 19.62N

Force to overcome dynamic friction: \$F_{dynamic} = ma = 1.3kg * 9.81m/s^2 = 12.753N\$ 

20 inches = 0.508m

If you are applying just enough force to overcome dynamic friction the entire time (static friction is only at the start so let's ignore that for now), the work done is:

\$W = F \times d = 12.753N \times 0.508m = 6.478J\$

If you want that work to be done to occur over 10 seconds then the power required is:

\$P = \frac{W}{t} = \frac{6.478J}{10s} = 647.8mW\$ of continuous power

But there is also the high force required to overcome static friction at the beginning plus the friction losses in the leadscrew. I'm not sure how efficient leadscrews are but if they are anything like worm gears it's really bad. So I'm just going to guess 10% efficient.

So you need a motor that is capable of 
\$\frac{647.8mW}{10%} = 6.478W\$ of continuous power

And since the peak force required is 1.538x larger than the continuous torque, let's increase the continuous power by that much for peak power for a 10W motor.

\$P = \tau \times \omega = [N-m] \times [rad/s]
\$

So to convert the power into required torque and RPM you need information on your leadscrew and any potential gearboxes present. For the leadscrew, you need the gear reduction ratio or to measure it and figure it out.



NEEDS VERIFICATION

If no gearbox other than the lead-screw is present, then:

\$[Leadscrew Reduction Ratio] = \frac{[Circumferential Displacement(m)]}{[Linear Displacement(m)]}=\frac{[Angular Displacement(rad)] \times [Leadscrew Radius (m)]}{[Linear Displacement(m)]}\$

\$ \tau_{motorPeak} = 19.62N \times [Leadscrew Radius (m)] \times \frac{1}{[Leadscrew Reduction Ratio]}\$

\$ \tau_{motorContinuous} = 12.753N \times [Leadscrew Radius (m)] \times \frac{1}{[Leadscrew Reduction Ratio]}\$

\$RPM_{motor} = \frac{0.508m}{10s} \times [Leadscrew Reduction Ratio]\$

NOTE:  I am a bit uncertain about the equation for the lead screw reduction ratio. I came up with it on the spot. There's also probably a way to figure it out based on the thread angle and leadscrew radius but I don't feel like trying to figure that out.

I am also uncertain about how the \$[Leadscrew Radius (m)]^2\$ pops out in the torque equation which I was not expecting. It may not be correct.