Calculating electric motor's weight lifting ability knowing power. What else do I need to know? So, my company works with electric motors, and my boss wanted me to calculate how much weight could be lifted by a motor rated at 750W.
I know that I can't just calculate the work done by it without any further specification of characteristics, but what I what to know is a couple of formulas that could get me there using  measurable values, for example, i can measure current consumed when lifting a 10kg weight.
Also, I'm open to suggestions, if you think there is an easier way to get where I want to, I am open to new ideas.
 A: There are two separate issues here. The maximum weight your motor will be able to lift depends on its torque. The rate at which your motor can lift the weight depends on its power.
The torque of the motor can be increased or decreased by running it through a gearbox, so in principle you could lift as big a weight as you want as long as you use a very low gear (and therefore lift the weight slowly).
The rate of lifting is easier to be precise about. If you lift a mass $m$ through a height $h$ in a time $t$ the power required is:
$$ W = \frac{mgh}{t} \tag{1} $$
or putting it another way, if you lift the mass at a velocity $v$ the power is:
$$ W = mgv \tag{2} $$
(because v = h/t).
So to take your example of a 10kg weight, to lift this at a speed of 1 m/s would require a power:
$$ W = 10 \times 9.81 \times 1 = 98.1W $$
which is well within the power of your 750W motor. Alternatively you could flip this around and ask how fast you could lift the 10kg weight if you ran the motor at full power. Rearranging equation (2) gives:
$$ v = \frac{W}{mg} = \frac{750}{10 \times 9.81} \approx 7.6 \text{m/s} $$
