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 m ass of cyclist plus bicycle is 68 kg.
1 answer:
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
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