Cyclist's Speed for 0.25 hp Output on 6° Hill

[cyspope]

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\theta= \frac{1}{2}mv^{2}

The Attempt at a Solution

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

661950=71.707h =35v^{2}

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

 

[godtripp]

I think your equations may be confusing you.

For instance what is h? h or (d) should be the distance that the cyclist is traveling.
so h=d sin (6) make sense?

now W = -\Delta U so if you call your initial position h=0 your final potential is mgh

now can you use d to solve for velocity?

 

[cyspope]

but how do I find the velocity to maintain 0.25 hp?

661950 = 35v^{2}
v^{2} = 18912
v = 137.524 m/s

this is not what I should get. I need to get 2.6 m/s.

 

[Mindscrape]

This may help you out

P=\frac{dW}{dt}=\frac{F\cdot ds}{dt} = F \cdot v

So you should figure out the force he needs to cancel out gravity, and from there get the velocity. Be careful of units, and it should be a straight shot.