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.

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 $P=300\mathrm{W}$.

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

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.

role="math" localid="1667895890812" $$\begin{gathered} t_{\mathrm{biking}}=\frac{1\mathrm{km}}{15\mathrm{km}/\mathrm{h}} \\ =\frac{1}{5}\mathrm{h} \end{gathered}$$

The energy used to cover a distance of while walking.

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

The energy used to cover a distance of while biking.

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

Take the ratio of both the energy.

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

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).