Going to ball park this so please feel free to jump in and correct any errors.
Let's say for argument's sake, they do not use the motor till the final hill. Go with 2.2 lbs = 1kg as that makes calculations easy. Ideal situations yadda yadda. Just ball parking things here, not submitting a paper to the E Coyle school of dodgy science.
50 W for half an hour = 50 x 30 x 60 joules. ie 90kj of work done by the motor at the end of the stage. We can go full-*** and say it's a 2000 VAM hill and they cover 1000m during that 30 minute climb.
Let's make it a really simple stage design where along the flat the rider is sucked along by the peloton and coasts with a tailwind and an efficient team of domestiques around him. No additional energy needed to transport the motor as the predominant energy sink is aerodynamic and he's tucked in behind Adam Hansen or similar.
There is 1000m of elevation during the stage, to match the final MTF.
The energy required to raise the 1kg battery + motor system 1000m is
w = f x d = 1 x 9.8 x 1000 = 9.8kj
This work is required for the in-stage elevation gain, as well as the MTF at the end of the stage, so we ball park the final stage energy requirements to transport motor + battery to 2 x 9.8kj = 20kj.
At the end of the day, there is a nett gain of (90-20) = 70kj of work available to the rider.
As for when the break even point occurs, we have
90kj work - 10kj transport cost = 80kj =~ 8000m elevation gain.
A crazy TdF stage is around 5000m elevation gain, so the hypothetical motor installation is always going to give you a nett positive energy source.
Clearly this is worth doing.
The killer is when you look at power::weight. A top level cyclist is going to do around 6W/kg at the end of the stage, but with this motor, he is adding 50W/kg for 1kg of weight to his overal W/kg for that half an hour. If we take a 60kg climber, doing 6W/kg (360W) and give them 80kj of work for that half an hour, their overall W/kg goes up to 6.7W/kg.
Work done = 360W x 30 x 60 = 648 kj
Add the 80kj of the motor: 728kj
728kj work done in 30 minutes = 728000 / (30 x 60) = 404 W avg
404W / 60 = 6.7 W/kg
Let assume that instead of smashing the peloton, he sits in the wheels and conserves energy for the next day.
To match his non-motor enabled, 60kg compatriot doing 6W/kg, he only has to do
648kj - 80kj of work = 568kj
568kj of work done in 30 minutes = 315W avg
315W / 60 = 5.25W/kg = pretty darn cruisy.
I think.