How fast must a cyclist climb a 6.0º hill to maintain a power output of 0.25 hp? Neglect work done by friction, and assume the mass of cyclist plus bicycle is 68 kg.

Answer: 2.7m/s

Explanation:

cyclist fight only against the component of weight along the road (no friction), that is:

fx = Wsin6° = mgsin6°=68*9.8*sin6°=69.6N downward

so the force that moves the bike over the hill will be equal to fx but upward

Now we now that the power P = 0.25hp = 186.5W is:

P=f*v so v = P/fx = 186.5/69.6 = 2.7m/s