Lift 1 kg weight using wind power from a fan

Don't bother with so many conversions. You loose energy each time you do that.
Would be best to take the output directly from the generator which is usually AC, but could also be pulsed DC.
You can rectify this to a clean DC value, and use a boost converter (switching power supply) to get a higher voltage out than you put in.

Please note that doing so will decrease the amperage, and while voltage will provide the speed for the motor to turn, you will need enough current to actually do the work. Yes voltage and current function in relation to V=IR, but in real life, if you attempt to pull too much current the voltage will drop.
So, at this time I would strongly urge you to start out small, and get some measurements from a generator you plan to use.

 

Why convert to DC and back? I don't understand. Surely all you need is a propeller on a shaft, with a cog and chain or a wheel and cord, to raise the weight?

If you really want to convert to electricity and back, then as Gryd3 said, just power the motor from the generator.

In either case, a ratchet mechanism would stop the weight from falling when power isn't available, and a cluch or a switch would control the start and stop of the motion.

 

Is it a correct assumption that the fan will be just a small table fan? In that case the wind turbine will be low power and the motor will be even lower power. So the motor will need a gearbox to increase the torque. Is the mass on wheels, or does the motor need to overcome sliding friction also? If the object of this project is to learn about power conversion loss, then your plan will succeed. However, if the object is to raise the weight, then would it not be easier to couple the wind turbine shaft directly to the gearbox?

 

If you choose the electrical route then do you need to consider varying wind speed which will have an effect on the available power.
Adam

 

As a physics student you should know that "work" is moving a certain weight a certain height for example.
Then the rate of doing that work is "horsepower".
I would consider that in your calculations also.