Q97PP - University Physics with Modern Physics

Question

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

 

How many times greater is the kinetic energy of the person when biking than when walking? Ignore the mass of the bike. (a) 1.7; (b) 3; (c) 8; (d) 9

Step-by-Step Solution

Answer

The correct answer is option (d).

Step 1: Identification of given data

The mass of the person is m = 70 kg.

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

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

The metabolic power is P = 300 W.

Step 2: Concept/Significance of kinetic energy

The expression of kinetic energy is given by,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ 

Here, m is the mass of the body, and v is velocity of the body.

Step 3: Determine the number of times the kinetic energy of the person when biking than when walking

Take the ratio of kinetic energy of a person in cycling and walking.

$$\begin{gathered} \frac{K_{cycling}}{K_{cycling}}=\frac{{\displaystyle \frac{1}{2}}m{v^2}_{cycling}}{\frac{1}{2}m{v^2}_{walking}} \\ =\frac{{v^2}_{cycling}}{{v^2}_{walking}} \\ =\frac{{\left(15 km/h\right)}^2}{{\left(5 km/h\right)}^2} \\ =9 \end{gathered}$$                                                   

 

Simplify further.

 $K_{cycling}=9\left(K_{walking}\right)$                                                      

 

Therefore, the kinetic energy of a person in cycling is times greater than walking.

 

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

 

Hence, the correct answer is option (d).