MasteringPhysics: HW 6- Chpt. 6- Microsoft Edge HW6-Cho Alternative Exercise 6.1
ID: 1773050 • Letter: M
Question
MasteringPhysics: HW 6- Chpt. 6- Microsoft Edge HW6-Cho Alternative Exercise 6.123 /myctitemView?assignment Problem ID=86010542 Resourc8s V a previous 18 of 19 next Alternative Exercise 6.123 Part A For a touring bicyclist the drag coefficient C (where fir-1; CApu2) is 1.00, the frontal area A is 0.463 m? the air density is 1.20 kg/m3 , and the coefficient of rolling friction is 4.40 10 The rider has mass 58.0 kg. and her bike has mass 12.3 kg To maintain a speed of 12.3 m/s on a level road, what must the rider's power output to the rear wheel be? Submit My Answers Give Up Part B For racing, the same rider uses a different bike with a coefficient of rolling friction 2.80x10-3 and mass 9.10 kg. She also crouches down, reducing her drag coefficient to 0.880 and reducing her frontal area to 0.369 m2 . What must her power output to the rear wheel be then to maintain a speed of 12.3 Submit My Answers Give Up Part C For the situation in part B, what power output is required to maintain a speed of 7.00 m/s? (For more on aerodynamic speed limitations for a wide variety of human-powered vehicles see "The Aerodyncs of Human-Powered Land Vehicles," Scientific American, December, 1983)Explanation / Answer
Assume no wind, grav. accel. = 9.8, 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 = 0.5 *A * C * * v^2 (fa=aero force, A=area, C =drag coeff, =density, v=velocity)
ff = m * g * cf (ff = friction force, m=mass, g=grav accel, cf = friction coeff)
power = v * (fa + ff)
a) fa = 42.03 N
ff = 3.03 N
P = 554.24 W
b) fa = 29.48 N
ff = 1.84 N
P = 385.24 W
c) fa = 9.55 N
ff = 1.84 N
P = 79.73 W