A 70 - kg person walks at a steady pace of 5 - km/hr on a treadmill at a 5.0% grade. (That is, the vertical distance covered is 5.0% of the horizontal distance covered.) If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? (a) 300 W; (b) 315 W; (c) 350 W; (d) 370 W.

The mass of the person is m = 70 kg.

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

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

The metabolic power is P = 300 W.

The relationship between the power and work done 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, W is work done, and t is time.

The work done in the form of potential energy is given by,

W = mgh

Here, m is mass of the body, g is acceleration due to gravity, and h is height.

When the person is walking 5 km in one hour then the power required is 300 W. Therefore, $P_{horizontal}=300W$.

Since the vertical distance covered is equal to the 5% of the horizontal distance covered. Therefore,

$$\begin{gathered} v_{vertical}=0.05v_{horizontal} \\ =\left(0.05\right)\left(1.4m/s\right) \\ =0.07m/s \end{gathered}$$

Find the power required for the vertical climb as follows.

$$\begin{gathered} P_{vertical}=\frac{mg{h}_{vertical}}{t} \\ =mg{h}_{vertical} \\ =\left(70kg\right)\left(9.8m/s^2\right)\left(0.07m/s\right) \\ =48.02W \end{gathered}$$

Find the total power required as follows.

$$\begin{gathered} P_{total}=P_{horizontal}+P_{vertical} \\ =300W+48.02W \\ =348.02W \\ \approx 350W \end{gathered}$$

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

Hence, the correct answer is option (c).