Question

The food calorie, equal to $\mathbf{4186}\mathbf{J}$, is a measure of how much energy is released when the body metabolizes food. A certain fruit-and-cereal bar contains 140 food calories. (a) If a $\mathbf{65}\mathbf{kg}$hiker eats one bar, how high a mountain must he climb to “work off” the calories, assuming that all the food energy goes into increasing gravitational potential energy? (b) If, as is typical. Only $\mathbf{20}\mathbf{\%}$of the food calories go into mechanical energy, what would be the answer to part (a)? (Note: In this and all other problems, we are assuming that $\mathbf{100}\mathbf{\%}$of the food calories that are eaten are absorbed and used by the body. This is not true. A person’s “metabolic efficiency” is the percentage of calories eaten that are actually used: the body eliminates the rest. Metabolic efficiency varies considerably from person to person.)

Short answer

1. The height up to which the hiker climbs to “work off” the calories is$919.1\text{m}$.
2. Theheight up to which the hiker climbs to “work off” if only$20\%$of the food calorie goes into mechanical energy is$183.8\text{m}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The calorie of food is, $E_f=4186\text{J}$ .
• The mass of the hiker is, $m=65\text{kg}$.
• The number of calories contains in a certain fruit-and-cereal bar is, $n=140$ .
• The fraction of the food calories goes into mechanical energy is, $x=20\%$.

Significance of the gravitational potential energy 

Whenever an object lies at a specific altitude above the Earth’s surface, then the object offers an energy that is referred to as gravitational potential energy. The relation between the potential energy and the object’s mass is linear.

(a) Determination of theheight up to which the hiker climbs to “work off” the calories 

The relation of height up to which the hiker climbs to “work off” the calories is expressed as,

$$\begin{gathered} E=\left(E_f\right)\left(n\right) \\ mgh=\left(E_f\right)\left(n\right) \\ h=\left[\frac{\left(E_f\right)\left(n\right)}{mg}\right] \end{gathered}$$

Here, $E$is the energy contained by the hiker, $g$ is the gravitational acceleration whose value is $9.81{\text{m/s}}^{\text{2}}$and $h$ is the height up to which the hiker climbs to “work off” the calories.

Substitute all the known values in the above equation.

$$\begin{gathered} h=\left[\frac{\left(4186\text{J}\right)\left(140\right)}{\left(65\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}\right] \\ =919.1\frac{\text{J}·{\text{s}}^{\text{2}}}{\text{m}·\text{kg}}\left(\frac{1kg·m^2·s^{-2}}{1J}\right) \\ =919.1\frac{kg·m^2·s^{-2}·{\text{s}}^{\text{2}}}{\text{m}·\text{kg}} \\ =919.1\text{m} \end{gathered}$$

Thus, the height up to which the hiker climbs to “work off” the calories is $919.1\text{m}$.

(b) Determination of the height up to which the hiker climbs to “work off” if only 20%  of the food calorie goes into mechanical energy 

The relation of height up to which the hiker climbs to “work off” if only $20\%$of the food calorie goes into mechanical energy is expressed as,

$$\begin{gathered} E\text{'}=\left(E_f\right)\left(n\right)\left(x\right) \\ mgh\text{'}=\left(E_f\right)\left(n\right)\left(x\right) \\ h\text{'}=\left[\frac{\left(E_f\right)\left(n\right)\left(x\right)}{mg}\right] \end{gathered}$$

Here, $"E"$is the energy contained by the hiker when only $20\%$of the food calorie goes into mechanical energy, and $h$ is the height up to which the hiker climbs to “work off” if only $20\%$of the food calorie goes into mechanical energy.

Substitute all the known values in the above equation.

$$\begin{gathered} h\text{'}=\left[\frac{\left(4186\text{J}\right)\left(140\right)\left(0.20\right)}{\left(65\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)}\right] \\ =183.8\frac{\text{J}·{\text{s}}^{\text{2}}}{\text{m}·\text{kg}}\left(\frac{1kg·m^2·s^{-2}}{1J}\right) \\ =183.8\frac{kg·m^2·s^{-2}·{\text{s}}^{\text{2}}}{\text{m}·\text{kg}} \\ =183.8\text{m} \end{gathered}$$

Thus, the height up to which the hiker climbs to “work off” if only $20\%$of the food calorie goes into mechanical energy is $183.8\text{m}$.