how fast must a cyclist climb a 6 degree hill to maintain a power output of 0.25

Question

how fast must a cyclist climb a 6 degree hill to maintain a power output of 0.25hp? neglect work done by friction and assume the mass of cyclist plus bicycle is 68kg.
So i know you can say the force would be mgsintheta but isnt that in correct because there would be another force (force applied, not just mgsintheta wouldnt there be)?

Explanation / Answer

as soon as cyclist starts up the incline >. it experiences (mg sin 6) force acting on it pulling it down-along. this is a constant force >. will continue to face till it reaches (h) destination. if he can exert exactly same force (mg sin 6) up-along while driving, the net force on it will become ZERO ma = up = F (applied, up) - mg sin 6 (down) ma = o = mg sin 6 - mg sin 6 a = 0 >> v = constant throughout motion Power output = Force * constant speed P = mg sin 6 * v 0.25*746 watt = 70*9.8 sin 6 * v v = should be driving = 0.25*746 / 70*9.8 sin 6 v = 2.6 m/s

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