What Power Output is Needed for a Bicyclist to Overcome Drag and Friction?

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SUMMARY

The discussion focuses on calculating the power output required for a touring bicyclist to overcome drag and friction while maintaining a speed of 11.7 m/s on a level road. Given parameters include a drag coefficient (C) of 1.04, a frontal area (A) of 0.462 m², a coefficient of rolling friction of 4.60×10−3, and a total mass (rider plus bike) of 64.1 kg. The initial calculation of total force resulted in a power output of 459.6 W, which was deemed incorrect, indicating the need for more precise calculations to achieve an accurate result.

PREREQUISITES
  • Understanding of physics concepts such as drag force and rolling friction
  • Familiarity with the formula for power output: Power = Force x Velocity
  • Knowledge of the drag equation: f of air = 0.5CApv²
  • Basic proficiency in unit conversions and calculations involving mass and speed
NEXT STEPS
  • Review the principles of fluid dynamics, specifically regarding drag forces on moving objects
  • Explore advanced calculations for power output in cycling, including variations in speed and terrain
  • Investigate the impact of different bike geometries on drag coefficients
  • Learn about the effects of rider position on aerodynamic efficiency
USEFUL FOR

This discussion is beneficial for cyclists, sports scientists, and engineers interested in optimizing cycling performance through understanding the physics of drag and friction.

makeAwish
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Homework Statement



For a touring bicyclist the drag coefficient C(where f of air = 0.5CApv^2) is 1.04, the frontal area A is 0.462 , and the coefficient of rolling friction is 4.60×10−3. The rider has mass 53.0kg, and her bike has mass 11.1kg.
Note: p is density of air (1.2kg/m^3)

To maintain a speed of 11.7m/s on a level road, what must the rider's power output to the rear wheel be?

The attempt at a solution

Total force = friction force + drag force
Power = total force x velocity = 459.6W

which is not correct.
 
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I would calculate it the same way, but with more precision. I got an answer close but possibly enough dissimilar?
 

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