Here are questions about human diet. (a) A typical candy bar provides $280$calories (one “food” or “large” calorie is equal to $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}} \mathbf{J}$). How many candy bars would you have to eat to replace the chemical energy you expend doing $100$ sit-ups? Explain your work, including any approximations or assumptions you make. (In a sit-up, you go from lying on your back to sitting up.) (b) How many days of a diet of $2000$ large calories are equivalent to the gravitational energy difference for you between sea level and the top of Mount Everest, $8848$ m above sea level? (However, the body is not anywhere near $100\%$ efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

The given data can be listed below as,

• One large calorie is, $4.2\times {10}^3 \text{J}$.

• The number of calories provided by the candy is, $c=280\text{large} \text{calories}\times \frac{4.2\times {10}^3 \text{J}}{1\text{large} \text{calories}}=1.176\times {10}^6\text{J}$.

• The number of sit-ups is, $n=100$.

• The number of calories in a diet is, $E=2000$ large calories.

The joule of energy consumed by a person in the form of food is utilized to do work on the basis of the efficiency of that person.

Thegravitational potential energy is obtained by taking the product of the mass of the object, acceleration due to gravity, and height at which the object is kept. It is expressed as follows,

$U=mgh$.….. (1)

Here, $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height at which the object is kept

The average power is determined by taking the ratio ofgravitational potential energyto the time. It is expressed as follows,

$P=\frac{U}{t}$ …… (2)

Here, $U$ is the gravitational potential energy and $t$ is the time.

(a)

Assume that the average mass of the person is$70\text{kg}$and an average height of the person is $1.75\text{m}$.

It is known that while doing the sit-ups, half of the mass of the person is lifted. Also, while doing the sit-ups, the center of mass of the body of the person is located at one-fourth of the height of the person.

The expression for the energy needed for doing the sit-ups is,

$E_1=\frac{m}{2}g\frac{h}{4}$

Here, $m$is the mass of the person, $g$is the acceleration due to gravity, that is $9.8 {\text{m/s}}^2$, and $h$is the height of the person.

Substitute all the values in the above expression, we get,

$E_1=\frac{70kg}{2}\left(9.8m/s^2\right)\frac{1.75m}{4}$

$$\begin{gathered} =150.1.\left(1kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2}\right) \\ =150.1J \end{gathered}$$

Let the muscular efficiency of the person is $20\%$. So, $0.2\text{J}$is the amount of energy that will be available for the work.

Write the expression for the practical value of energy needed for doing the sit-ups,

$E_2=\frac{E_1}{\eta }$

Here, $\eta$is the amount of energy available.

Substitute all the values in the above expression, we get,

$$\begin{gathered} E_2=\frac{150J}{0.18} \\ =834J \end{gathered}$$

The expression for the number of candy bars can be calculated as,

$$\begin{gathered} N=\frac{\left(100\right)(834J)}{1.176\times {10}^{6}J} \\ =0.071 \end{gathered}$$

Thus, the number of candy bars is$0.071$.

(b)

Write the expression for the average power.

$P=\frac{E}{t}$

Here, $E$is the number of calories in the diet and $t$is the time that is one day.

Substitute all the values in the above expression, we get,

$$\begin{gathered} P=\frac{2000largecalories}{1day} \\ =\frac{2000largecalories\times {\displaystyle \frac{4.2\times {10}^{3}J}{1largecalories}}}{1day\times \left({\displaystyle \frac{24h}{1day}}\right)\times \left({\displaystyle \frac{3600s}{1h}}\right)} \\ =97.223J/s \end{gathered}$$

Substitute all the values in equation (1), we get,

$$\begin{gathered} U=\left(67kg\right)\left(9.8m/s^2\right)\left(8848m\right) \\ =5.8155\times {10}^{6}kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2} \\ =5.8155\times {10}^{6}J \end{gathered}$$

Substitute all the values in equation (2), we get,

$$\begin{gathered} 97.223J/s=\frac{5.8155\times {10}^{6}J}{t} \\ t=59816.092s\times \frac{1h}{3600s}\times \frac{1day}{24h} \\ =0.692days \end{gathered}$$

Thus, the number of days is $0.692$days.