E=(mv^2)/2 - so we put more energy accelerating the bike from 10-20m/s than 0-10m/s, no?
Yet a=F/m - which suggests the acceleration is proportional to force, which would suggest that applying force F for time t should speed you up 0-10m/s the same way as 10-20m/s?
I suspect the force applied to the pedals is not the force which is acting on the bike (counter-force of the ground-bike system) and this second force is somehow relatable to the current speed of the bike, no?
Common mistake: you are confusing energy and momentum, or power (rate of energy/work) and force (rate of momentum change). In fact, power at constant force is F.v (force time velocity), which solves your question.
If you don’t shift gears, then your pedals increase in speed as your bike accelerates. So your power consumption goes up with speed even though force is constant.
E=(mv^2)/2 - so we put more energy accelerating the bike from 10-20m/s than 0-10m/s, no?
Yet a=F/m - which suggests the acceleration is proportional to force, which would suggest that applying force F for time t should speed you up 0-10m/s the same way as 10-20m/s?
I suspect the force applied to the pedals is not the force which is acting on the bike (counter-force of the ground-bike system) and this second force is somehow relatable to the current speed of the bike, no?