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

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A cyclist weighing 80 kg can coast down a 3.4-degree hill at 9.0 km/hr, where the gravitational force balances air resistance. When pedaling hard, the cyclist reaches 30 km/hr, increasing the air resistance proportionally to the speed. The discussion involves calculating the force of air resistance at both speeds to determine the power exerted while descending. By analyzing the forces involved, the cyclist's climbing speed can be derived using the same power output against the new resistance when going uphill. The key is to establish the relationship between speed, air resistance, and gravitational force to find the climbing velocity.
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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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