For a touring bicyclist the drag coefficient C (where Fair=12CAv^2) is 1.03, the

Question

For a touring bicyclist the drag coefficient C (where Fair=12CAv^2) is 1.03, the frontal area A is 0.469 m^2, the air density is 1.20 kg/m^3 ,and the coefficient of rolling friction is 4.70×10^3. The rider has mass 51.0 kg, and her bike has mass 12.1 kg.

Part A

To maintain a speed of 12.2 m/s on a level road, what must the rider's power output to the rear wheel be?

Part B

For racing, the same rider uses a different bike with a coefficient of rolling friction 2.50×103 and mass 9.30 kg . She also crouches down, reducing her drag coefficient to 0.930 and reducing her frontal area to 0.370 m2 . What must her power output to the rear wheel be then to maintain a speed of 12.2 m/s?

Part C

For the situation in part B, what power output is required to maintain a speed of 5.50 m/s ?

Explanation / Answer

M =51 kg , m=12.1 kg, =1.2 kg/m^3, C = 1.03,A =0.469 m^2, ur =4.7*10^-3

Part A: v =12.2 m/s

air force Fa =12CAv^2

friciton force = Ff =ur*N = ur*mg

Power P = (Fa+Ff).v

P = ((12*1.03*0.469*1.2*12.2*12.2)+(4.7*10^-3*(51+12.1)*9.8))*12.2

P = 12666.9 W

(b)

Power P = (Fa+Ff).v

P = ((12*0.93*0.37*1.2*12.2*12.2)+(0.93*(51+9.3)*9.8))*12.2

P = 15702.4 W

(c)

Power P = (Fa+Ff).v

P = ((12*0.93*0.37*1.2*5.5*5.5)+(0.93*(51+9.3)*9.8))*5.5

P = 3847 W

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