Friction bicycle homework Problem

AI Thread Summary
The discussion revolves around solving a physics homework problem related to friction experienced by a bicyclist coasting down a hill. The key calculations involve determining the constant c for friction, which is proportional to speed, and finding the average force required to maintain a higher speed. The solution process includes converting speeds and applying the equation for forces, leading to the realization that the mass of the cyclist and bicycle must be incorporated correctly. After some confusion, the correct approach yielded an accurate value for c, approximately 1 kg/s. The thread emphasizes the importance of correctly applying physics principles and unit conversions in solving such problems.
MellowOne
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[SOLVED] Friction Problem

Homework Statement


A bicyclist can coast down a 5° hill at a constant 8.0 km/h. Assume the force of friction (air resistance) is proportional to the speed v so that Ffr = cv.

(a) Calculate the value of the constant c.
________ kg/s

(b) What is the average force that must be applied in order to descend the hill at 18 km/h? The mass of the cyclist plus bicycle is 85 kg.
________N


Homework Equations


Sigma F = ma
Ffr = cv

The Attempt at a Solution


First i converted my 8 km/hr into 2.22 m/s. Then i went into the parallel forces which are Ff and Fwx. The sum of their forces is equal to 0 because its acceleration is 0 so Ff is equal to Fwx and Fwx is equal to sin5 times mg which equlaed .854m. Then i set that equal to 2.22c because Ff also equal that. Now I'm stuck because I don't have mass. I tried writing out all units and canceling out so that on kg/s was left, but that didn't work out.

Any ideas?
 
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You correctly found that cv = mg · sin 5º . Go ahead and use the mass given for the cyclist in part (b); I think it's not intended to be used there exclusively. The "resistance" coefficient (largely due to air drag) will depend on the mass of the rider and bicycle; a different rider would have a different value for c.
 
When using the mass given in part b, the answer came out to be approx 1 kg/s which wasn't correct. Is there something I missed or another way to approach this problem.
 
MellowOne said:
When using the mass given in part b, the answer came out to be approx 1 kg/s which wasn't correct. Is there something I missed or another way to approach this problem.

I used what you set up and found

c = (85 kg) · (9.81 m/sec^2) · (sin 5º) / (2.22 m/sec) .

Did you omit something?
 
Yeah, I must have forgotten something cause I did it again and I got the right answer. Thanks a lot.
 
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