Chapter 6, Problem 112P

(a)

To determine

The average power required for the person to hike up to the top of Yosemite Falls in $1.5\text{h}$.

Answer to Problem 112P

The average power required for the person to hike up to the top of Yosemite Falls in $1.5\text{h}$ is $\underset{\_}{94\text{W}}$.

Explanation of Solution

Given that the mass of the person is $70\text{kg}$, the height of Yosemite Falls is $740\text{m}$, and the time is $1.5\text{h}$.

Write the expression for the average power output.

$P_{\text{av}}=\frac{\Delta E}{\Delta t}$ (I)

Here, $P_{\text{av}}$ is the average power output, $\Delta E$ is the change in energy, and $\Delta t$ is the time interval.

In the climbing process, the change in energy is equal to the change in gravitational potential energy. Hence the equation (I) can be modified by replacing $\Delta E$ by $\Delta U$ which is the change in gravitational potential energy.

$P_{\text{av}}=\frac{\Delta U}{\Delta t}$ (II)

Write the expression for the change in gravitational potential energy.

$\Delta U=mg\Delta y$ (III)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $\Delta y$ is the change in height.

Use equation (III) in (II).

$P_{\text{av}}=\frac{mg\Delta y}{\Delta t}$ (IV)

Conclusion:

Substitute $70\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $740\text{m}$ for $\Delta y$, and $1.5\text{h}$ for $\Delta t$ in equation (IV) to find $P_{\text{av}}$.

$\begin{array}{c}{P}_{\text{av}}=\frac{\left(70\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(740\text{m}\right)}{1.5\text{h}}\\ =\frac{\left(70\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(740\text{m}\right)}{1.5\text{h}\times \frac{3600\text{s}}{1\text{h}}}\\ =94\text{W}\end{array}$

Therefore, the average power required for the person to hike up to the top of Yosemite Falls in $1.5\text{h}$ is $\underset{\_}{94\text{W}}$.

(b)

To determine

The chemical energy used in hiking to the top of the Yosemite Falls in $1.5\text{h}$.

Answer to Problem 112P

The chemical energy used in hiking to the top of the Yosemite Falls in $1.5\text{h}$ is $\underset{\_}{2.0\text{MJ}}$.

Explanation of Solution

Given that the human body is only $25\%$ efficient in converting chemical energy to mechanical energy. Thus, it takes $4.0\text{units}$ of chemical energy for every single unit of potential energy gained by the body.

Upon hiking to the top of the Yosemite Falls, the body of the person gains potential energy.

Write the expression for the chemical energy used while gaining potential energy.

$E_{\text{chem}}=4.0\Delta U$ (V)

Here, $E_{\text{chem}}$ is the chemical energy.

Use equation (III) in (V).

$E_{\text{chem}}=4.0mg\Delta y$ (VI)

Conclusion:

Substitute $70\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, and $740\text{m}$ for $\Delta y$ in equation (VI) to find $E_{\text{chem}}$.

$\begin{array}{c}{E}_{\text{chem}}=4.0\left(70\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(740\text{m}\right)\\ =2.0\times {10}^6\text{J}\\ =2.0\times {10}^6\text{J}\times \frac{1\text{MJ}}{1,000,000\text{J}}\\ =2.0\text{MJ}\end{array}$

Therefore, the chemical energy used in hiking to the top of the Yosemite Falls in $1.5\text{h}$ is $\underset{\_}{2.0\text{MJ}}$.

(c)

To determine

The amount of energy used in kilocalories for the hike.

Answer to Problem 112P

The amount of energy used in kilocalories for the hike is $\underset{\_}{490\text{kcal}}$.

Explanation of Solution

Given that the chemical energy used for the hike is $2.0\times {10}^6\text{J}$.

Write the conversion formula between energy in Joules and kilocalories.

$1\text{kcal}=4186\text{J}$ (VII)

The chemical energy used in the hike is given by the expression,

$E_{\text{chem}}=4.0mg\Delta y$

Use the conversion formula (VII) to convert the $E_{\text{chem}}$ to kilocalories.

$\begin{array}{c}E\left(\text{kcal}\right)=E_{\text{chem}}\times \frac{1\text{kcal}}{4186\text{J}}\\ =4.0mg\Delta y\times \frac{1\text{kcal}}{4186\text{J}}\end{array}$ (VIII)

Conclusion:

Substitute $70\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, and $740\text{m}$ for $\Delta y$ in equation (VIII) to find $E\left(\text{kcal}\right)$.

$\begin{array}{c}E\left(\text{kcal}\right)=4.0\left(70\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(740\text{m}\right)\times \frac{1\text{kcal}}{4186\text{J}}\\ =485\text{kcal}\\ \approx 490\text{kcal}\end{array}$

Therefore, the amount of energy used in kilocalories for the hike is $\underset{\_}{490\text{kcal}}$.