What is the power output needed to bike uphill at a steady speed of 4.5 m/s?

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To determine the power output needed for a cyclist to climb a 6.0-degree hill at a steady speed of 4.5 m/s, the cyclist's total mass is considered to be 60 kg. The initial calculation yielded a power output of 280 W based on gravitational force alone. However, it was later clarified that additional forces, such as friction or air resistance, must be accounted for when climbing uphill, which was not included in the initial assessment. This oversight led to the realization that the problem was more complex than it initially appeared. Therefore, the cyclist's actual power output required to maintain speed while climbing would be greater than the initial calculation.
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Homework Statement


A bicyclist coasts down a 6.0 degree hill at a steady speed of 4.5 m/s.
Assuming a total mass of 60 kg (bicycle plus rider), what must be the cyclist's power output to climb the same hill at the same speed?

Homework Equations


Newton's 1st law
P=W/t

The Attempt at a Solution


P=\frac{W}{t}=F \cdot v

Set x-axis along incline...
F-mg\sin\theta=0
F=mg\sin\theta
P=mg\sin\theta*4.5 = 276.6 W

To 2 sigfig = 280 W

Is this correct? I'm being told it is the wrong answer...
 
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It looks right.
Who told you that is not?
 
The site where I submit my homework =/ I can't find any problem with it either, so I don't know what's up - guess I'll go ask the professor tomorrow. Thanks for checking my work and please let me know if you do think of somewhere I went wrong.
 
Just an update... since the rider coasts down the hill but isn't accelerating, we must assume that there is a retardant force which must also be factored in when going uphill. This is where I went wrong. Kind of a trick question... blah!
 

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