# Friction Problem

by MellowOne
Tags: friction, solved
 P: 44 1. The problem statement, all variables and given/known data 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 2. Relevant equations Sigma F = ma Ffr = cv 3. 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?
 HW Helper P: 1,662 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.
 P: 44 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.
HW Helper
P: 1,662
Friction Problem

 Quote by MellowOne 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?
 P: 44 Yeah, I must have forgotten something cause I did it again and I got the right answer. Thanks a lot.

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