Q95PP - University Physics with Modern Physics

Question

Energy of locomotion. On flat ground, a $70-\mathrm{kg}$ person requires about $300 \mathrm{W}$ of metabolic power to walk at a steady pace of $5-\mathrm{km}/\mathrm{hr} \left(1.4 \mathrm{m}/\mathrm{s}\right)$. Using the same metabolic power output, that person can bicycle over the same ground at $15-\mathrm{km}/\mathrm{h}$.

 

Based on the given data, how does the energy used in biking 1 km compare with that used in walking 1 km? Biking takes (a) 1/3 of the energy of walking the same distance; (b) the same energy as walking the same distance; (c) 3 times the energy of walking the same distance; (d) 9 times the energy of walking the same distance.

Step-by-Step Solution

Answer

The correct answer is option (a).

Step 1: Identification of given data

The mass of the person is $m=70 \mathrm{kg}$.

The velocity of walking is $v_{\mathrm{walking}}=5 \mathrm{km}/\mathrm{h}$.

The velocity of cycling is $v_{\mathrm{cycling}}=15 \mathrm{km}/\mathrm{h}$.

The metabolic power is $P=300 \mathrm{W}$.

Step 2: The relationship between the power and energy, and expression of time

The time is given by,

$\mathit{t}\mathbf{=}\frac{\mathbf{d}}{\mathbf{v}}$ 

Here, d is distance, and v is velocity.

 

The relationship between the power and work is given by,

$$\begin{gathered} \mathit{P}\mathbf{=}\frac{\mathbf{E}}{\mathbf{t}} \\ \mathit{E}\mathbf{=}\mathit{P}\mathit{t} \end{gathered}$$ 

Here, P, is power, E is energy, and t is t

Step 3: Determine the way by which the energy used in biking 1 km compare with that used in walking 1 km

Find the time taken to cover when walking at the speed of $v_{\mathrm{walk}}=5 \mathrm{km}/\mathrm{h}$ as follows.

                                                       $$\begin{gathered} t_{\mathrm{walk}}=\frac{1 \mathrm{km}}{5 \mathrm{km}/\mathrm{h}} \\ =\frac{1}{5}\mathrm{h} \end{gathered}$$

 

Find the time taken to cover when walking at the speed of $v_{\mathrm{biking}}=15 \mathrm{km}/\mathrm{h}$ as follows.

                                                      $$\begin{gathered} t_{\mathrm{biking}}=\frac{1 \mathrm{km}}{15 \mathrm{km}/\mathrm{h}} \\ =\frac{1}{5}\mathrm{h} \end{gathered}$$

 

The energy used to cover a distance of while walking.

                                                $$\begin{gathered} E_{walk}=P_{walk}\times T_{\mathrm{biking}} \\ =\left(300 W\right)\left(\frac{1}{5}h\right) \\ =60 \mathrm{W}·\mathrm{h} \end{gathered}$$

 

The energy used to cover a distance of while biking.

                                               $$\begin{gathered} E_{biking}=P_{biking}\times T_{\mathrm{biking}} \\ =\left(300 W\right)\left(\frac{1}{15}h\right) \\ =20 \mathrm{W}·\mathrm{h} \end{gathered}$$

 

Take the ratio of both the energy.

                                                  $$\begin{gathered} \frac{E_{biking}}{E_{biking}}=\frac{20 W·h}{60 W·h} \\ =\frac{1}{3} \\ E_{biking}=\frac{1}{3}{E}_{walk} \end{gathered}$$

 

So, the energy used in biking will be $\frac{1}{3}$ times of the energy used in walking the same distance.

 

Thus, the options (b), (c), and (d) are incorrect.

 

Hence, the correct answer is option (a).