Cyclist of 80kg: Descending 3.4 degree Hill at 9.0 or 30 km/hr?

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SUMMARY

A cyclist weighing 80 kg can coast down a 3.4-degree hill at a speed of 9.0 km/hr, where the gravitational force balances the air resistance. When the cyclist exerts power to descend at 30 km/hr, the air resistance increases proportionally, calculated as 30/9 times the resistance at 9 km/hr. To determine the climbing speed on the same hill using the same power, one must analyze the forces involved, including the gravitational force and the new resistance force when ascending.

PREREQUISITES
  • Understanding of basic physics concepts such as force, gravity, and resistance
  • Knowledge of the relationship between speed and air resistance
  • Familiarity with equations of motion and force balance
  • Basic algebra skills for solving equations
NEXT STEPS
  • Calculate the gravitational force component acting on the cyclist on the hill
  • Determine the air resistance force at both 9.0 km/hr and 30 km/hr
  • Analyze the power exerted by the cyclist at 30 km/hr to find the climbing speed
  • Explore the effects of varying mass or hill angle on cycling performance
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Cyclists, physics students, and sports scientists interested in understanding the dynamics of cycling performance on inclines and the impact of speed on resistance forces.

Omanlew
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A bicylcist of mass 80 kg (including the bike) can coast down a 3.4 deg hill at a steady speed of 9.0 km/hr. Pumping hard, the cyclist can descend the hill at a speed of 30 km/hr. Using the same power, at what speed can the cyclist climb the same hill? Assume the force of air resitance is directly proportional to the speed v; that is Fsubfr=bv, where b is a constant.


Not really sure how to approach this..
 
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When he is coasting down the hill the component of gravity along the road balances the force of air resistance at 9km/hr. Figure out what that resistance force (F_fr) actually is since F_fr+F_g=0. When he's pumping hard at 30km/hr the resistance force is 30/9 times what it used to be. If the force he is exerting at 30km/hr is F_p then you now have F_p+F_g-(30/9)*F_fr. Figure out F_p. When he's going uphill F_g reverses. What's the new resistance force? Can you use that resistance force to figure the velocity?
 

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