@gibbobaz: nobody misunderstood you. Your advice was wrong, and it's clear from your replies that you haven't done any significant work with electric motors. Please don't give advice unless you really know the facts.
The chart is showing peak power OUTPUT of 12W, not the peak power INPUT. Peak power output is near the middle of the torque range, about halfway between the no-load current (0.2A) and the stall current (5.5A). So about 2.85A.
The OP might want to design for Maximum Efficiency rather than Maximum Output Power. That is a somewhat lower torque, roughly 1/4 stall torque, and 3/4 no-load speed. If the desired RPM is "40 to 60", the OP will want a no-load speed of about 50 to 80. The 131:1 gearing gives a no-load speed of 76 RPM and a stall torque of 45 kg-cm. To get a 7 kg pull force you would use a pulley with a 6.5 cm radius. Oops: The operating torque is 1/4 the stall torque so 11.25 kg-cm. That means you need a pully radius of 1.6 cm or smaller to get the desired 7 kg pull.
Seems like the answer was on Pololu's website the whole time... @johnwasser @SteveThackery @JohnRob
I added below the graph that represents all the data, and the specs of the motor (item #1106) we were talking about - the 100:1 gearbox version. There are different graphs for different ratio's of this motor.
If I understand the graph correctly, if my current will be <=1A than the maximum RPM will be a little more than 80 and the torque will be a little more than 5kg.cm.
Please let me know if I got it right.
I've not used the graph with a current limited power supply before, but I believe you are right, yes. A 1A current limit will limit the torque to about 65kg.mm, at which point the motor speed will be about 80 rpm. Increasing the load on the motor any further will cause the motor rpm to drop quickly to zero, whilst the current will remain constant at 1A.
Just to be clear, are you definitely using a power supply with a 1A current limiter built in? Or is it simply a power supply that is rated at 1A?
I don't know. This is the adapter.
On the website I got it from it says that it has a max current of 1A, and that when the max current is being drawn the output voltage could be a bit less than 12V. I intend to put a 1A fuse if it matters.
Edit on edit: Do you think I should try the 1A PSU I already have or just get 3A one? I'm ordering all of the components from a website so if the 1A is really far fetched I'd just get the 3A now so I won't pay shipping costs twice, but if the 1A could work I prefer saving the money.
My worry is regarding the current spikes, when the motor first starts, will the 1A be enough...
@SteveThackery
Ahh, I see. I misunderstood that, no wonder everyone thought I was crazy 
Thanks for pointing that out.
The reason I ask is that if it is a properly designed current-limited PSU, then you don't need to worry about putting a fuse in the circuit - the PSU will be designed to prevent the current going above 1A. It should even be quite safe feeding continuously into a short circuit, although its over-temperature protection may kick in.
Anyway, remembering that the motor's starting torque will be higher than 1A, stick a slow-blow fuse in series with it and see how it goes. At this point we don't know enough about the mechanical arrangements to predict just how much torque you are going to need.
Raw power (in Watts) = force (in Newtons) X distance (in meters) / time (in seconds), all we know is the mass is 7kg, so...?
Well, this bit is easy. Assume we are lifting a 7kg mass and we have 12W of power at our disposal. How quickly can we lift the mass?
The force F from the mass in earth's gravity = 7kg * 9.81m/s = 68.67N
Starting with this fundamental equation: P = (F x d) / t
where P = power in watts
F = force in newtons
d = distance in metres
t = time in seconds
Rearrange for distance (i.e. the distance the weight is lifted per second)
d = (t * P) / F
Insert values:
d = (1 x 12) / 68.67 = 0.175m
So 12W will lift a 7kg weight at a rate of 175mm per second.
Note: this is theoretical! We don't know what the motor will actually produce with that particular power supply; we don't know if the weight remains constant; we don't know how much power will be lost in the bearings and other frictional losses. So we should consider the result to be a theoretical maximum speed.
Since the power output at the "Maximum Efficiency" torque is considerably less than the Maximum Power Output, it might be good to derate the power output by 50% or more. That is, use a power output of 6 or even 5 watts in the calculations. That will get you about 87mm/sec.
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