# Work problem

## 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.

Last edited:

Related Introductory Physics Homework Help News on Phys.org
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?

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.

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