Question

When you ride a bicycle at constant speed, nearly all the energy you expend goes into the work you do against the drag force of the air. Model a cyclist as having cross-section area $0.45m^2$ and, because the human body is not aerodynamically shaped, a drag coefficient of $0.90$.

a. What is the cyclist’s power output while riding at a steady $7.3m/s\left(16mph\right)$?

b. Metabolic power is the rate at which your body “burns” fuel to power your activities. For many activities, your body is roughly $25\%$efficient at converting the chemical energy of food into mechanical energy. What is the cyclist’s metabolic power while cycling at $7.3m/s$?

c. The food calorie is equivalent to $4190J.$How many calories does the cyclist burn if he rides over level ground at$7.3m/sfor1hr$?

Short answer

a. The output power of the cyclist is $102W$.

b. The metabolic power of the cyclist is $4.1\times {10}^{2}W$.

c. The amount of energy burnt by the cyclist is$3.5\times {10}^{2}cal$.

Step-by-step solution

Content Introduction

The output power of the cyclist due to drag force is $P=Fv$

The equation of the drag force is,

$F=\frac{1}{2}cp{v}^{2}A$

Here, $cpv$is the drag coefficient, $p$is density, $A$is cross section area.

Explanation (Part a)

The output power of the cyclist is

$$\begin{gathered} P=\frac{1}{2}\left(cp{v}^{2}A\right)v \\ P=\frac{1}{2}cp{v}^{3}A \\ P=\frac{1}{2}(0.90)(1.3kg/m^3){(7.3m/s)}^3(0.45m^2)P=102.4W \end{gathered}$$

Explanation (Part b)

The efficiency of body converting chemical energy into mechanical energy is

$$\begin{gathered} e=\frac{P}{P_f} \\ {P}_f=\frac{P}{e} \\ {P}_f=\frac{102.4W}{25\%} \\ {P}_f=4.10\times {10}^{2}W \end{gathered}$$

Explanation (Part c)

The metabolic power of the cyclist is

$$\begin{gathered} P_f=\frac{E}{t} \\ E=P_{f}t \\ E=(4.10\times {10}^{2}W)\left[\right(1h\left)\right(\frac{3600s}{1h}\left)\right] \\ E=3.5\times {10}^{5}cal \end{gathered}$$