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Cycling. For a touring bicyclist the drag coefficient is 1, the frontal area is 0.463 m^2, and the coefficient of rolling friction is 0.0045. The rider has a mass of 50 kg, and her bike has mass 12 kg.

a.) To maintain a speed of 12 m/s (about 27 mi/h) on a level road, what must the rider's power output to the rear wheel be?
b.) For racing, the same rider uses a different bike with a coefficient of rolling friction 0.0030 and mass 9 kg. She also crouches down, reducing her drag coefficient to 0.88 and reducing her frontal area to 0.366 m^2. What must her power output to the rear wheel be then to maintain a speed of 12 m/s.
c.) For the situation in part (b), what power output is required to maintain a speed of 6 m/s? Note the great drop in power requirement when the speed is only halved.

2006-09-04 03:31:59 · 2 answers · asked by just_askin' 1 in Science & Mathematics Physics

2 answers

Assume no wind, grav. accel. = 9.80665, and air density=1.2 kg/m^3, and friction is ideal, that is, independent of speed when speed>0. Note that if you ignore friction the power required due to aero drag is proportional to v^3. The formulas are:
fa = .5 *a * cd * rho * v^2 (fa=aero force, a=area, cd=drag coeff, rho=density, v=velocity)
ff = m * g * cf (ff=friction force, m=mass, g=grav accel, cd=friction coeff)
power=v * (fa + ff)
Answers (forces in N, power in W)::
fa-----------ff----------Power
a. 40.0032 2.736055 512.8711
b. 27.82771 1.735777 354.7619
c. 6.956928 1.735777 52.15623

2006-09-04 04:10:20 · answer #1 · answered by kirchwey 7 · 3 1

actually, that could desire to be MW (MegaWatts) because of the fact the unit of ability, a Watt, became named after a guy or woman, as became the Joule for potential, and as a effect could be capitalized. even even with the undeniable fact that the capitalization is in many cases forgotten while spelling out the unit, the emblem for the unit is often capitalized. At any cost... ability is potential in step with unit time, and a Watt is equivalent to a minimum of one Joule in step with 2nd. in case you burn fifty six MW for an hour (ability cases time = potential) how lots potential did you employ? ascertain all your instruments cancel!

2016-09-30 08:10:29 · answer #2 · answered by ? 4 · 0 0

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