3 ms·
Hmm.. the acceleration part. 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 accelerat
by jagrsw 4y ago
Hmm.. the acceleration part.
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?
- topaz0 4y agoCommon 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.
- aoeusnth1 4y agoIf 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.
- wiredfool 4y agoWork (change in energy) is F*d, so the force you're applying to go from 10->20m/s happens over a larger distance if it's the same force.