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## Homework Statement

67. How fast must a cyclist climb a 6 degrees hill to maintain a power output of 0.25 horse power? Neglect work done by friction and assume the mass of cyclist plus bicycle is 70 kg.

A. 2.6 m/s

## Homework Equations

W = mgh = Fdsin[tex]\theta[/tex]= [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]

## The Attempt at a Solution

(0.25 hp * 735.5 W * 3600 J) = 70 * 9.8 * h * sin(6) = [tex]\frac{1}{2}[/tex]*70v[tex]^{2}[/tex]

661950=71.707h =35v[tex]^{2}[/tex]

I don't get how the answer should be only 2.6 m/s.

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