71 Human Efficiency

Human EFFICIENCY

In the previous chapter we estimated a person walk 4.3 miles using the energy provided by bagel and you calculated many miles a person could bike on the energy in candy bar. Those values are types of  efficiency because we divided the quantity of the desired outcome (miles covered) by input required (amount of food). In this chapter we will focus on another type of efficiency known as the mechanical efficiency. When evaluating the the mechanical efficiency of the human body (or any machine) we are interested in useful work done by the body to lift, move, pull, or squeeze because that work to converts chemical potential energy into kinetic energy, or gravitational potential energy, or elastic potential energy, which are the themechanical energies [1]. We calculate the body’s mechanical efficiency as the ratio of useful work done by the body to chemical potential energy used (-\Delta P_c):

(1)   \begin{equation*} e_{body} = \frac{W_{by}}{-\Delta P_c} \end{equation*}

Notice the extra negative sign, which is there because change in chemical potential energy of the body is negative when the body does work. The extra negative makes sure the efficiency is a positive number. (\Delta P_c would be positive when you eat food). We often talk about efficiency in terms of percentage. To find percent efficiency, simply multiply the efficiency found using the previous equation by 100 %.

Everyday Example: Efficiency in Climbing Stairs

Based on the chart in the previous chapter titled Energy and Oxygen Consumption Rates for an average 76 kg male, climbing 116 stairs per minute requires 685 W of power. We can estimate the efficiency of the body at climbing stairs by calculating the potential energy change for climbing some stairs (the useful work) and dividing by the change in chemical potential energy of the body. First, some useful information: the averate step height in the U.S. is 7.5 in, or 0.19 m. 

In one minute 116 stairs are climbed so we can multiply by 0.19 m per stair to find that the height change in one minute is 22 m.

The change in gravitational potential energy is then:

(2)   \begin{equation*} mg\Delta h = (76\, \bold{kg} )(9.8\,\bold{m/s/s})(22\,\bold{m})= 16,386\,\bold{J} \end{equation*}

Now we find the chemical potential energy used in one minute by multiplying the 685 W of power, which is 685 J/s,  by the 60 seconds in one minute and we get: 41,100 J per minute.

Now we need to enter these into the efficiency formula:

(3)   \begin{equation*} e_{body} = \frac{W_{by}}{-\Delta PE_c} =\frac{16386\,\bold{J}}{41,100\,\bold{J}} = 0.39 \end{equation*}

Multiplying by 100% gives us an efficiency of 39%, so stair climbing is definitely one of the most mechanically efficient activities the body perform. The remaining 61% of the food energy used by the body becomes thermal energy that is transferred to the environment.

Limits on Efficiency

The mechanical efficiency of the body is limited because energy used for metabolic processes cannot be used to do useful work. Additional thermal energy generated during the chemical reactions that power muscle contractions along with friction in joints and other tissues reduces the efficiency of humans even further [2]. “Alas, our bodies are not 100 % efficient at converting food energy into mechanical output. But at about 25 % efficiency, we’re surprisingly good considering that most cars are around 20 %, and that an Iowa cornfield is only about 1.5 % efficient at converting incoming sunlight into chemical potential energy storage.”  [3]For an excellent discussion of human mechanical efficiency and comparisons with other machines and fuel sources, see MPG of a Human by Tom Murphy, the source of the previous quote.

Everyday Example: Chemical Energy Required to Lift a Patient

Jolene works with two other nurses to smoothly lift a patient that weighs 75 kg a distance of 0.20 m at constant speed. If the body is about 20% mechanically efficient, how much chemical potential energy did the nurses use? How much did the thermal energy of the nurses increase?

We start with the efficiency formula:

(4)   \begin{equation*} e_{body} = \frac{W_{by}}{-\Delta PE_c} \end{equation*}

We need to rearrange for change in chemical potential energy:

(5)   \begin{equation*} -\Delta PE_c = \frac{W_{by}}{e_{body}} \end{equation*}

The useful work done by the nurses was to lift the patient and increase the potential energy. We know to calculate change in potential energy as:

(6)   \begin{equation*} mg\Delta h = (75\, \bold{kg} )(9.8\,\bold{m/s/s})(0.20\,\bold{m})= 147\,\bold{J} \end{equation*}

Entering the value for the useful work done by the nurses along with the efficiency of 20% (e_{body} = 0.2) we can find the chemical potential energy used:

(7)   \begin{equation*} -\Delta PE_c = \frac{147\,\bold{J}}{0.2} = -735\,\bold{J} \end{equation*}

That sounds like a lot of chemical potential energy, but eating a single grape would give us more than 10x that much energy.

Intuitively, thermal energy of the nurses should have increased by the amount of chemical energy used that did not go to useful work. Let’s double check by examining the conservation of energy in our system of patient, Earth, and nurses.

(8)   \begin{equation*} 0 = \Delta KE_i + \Delta PE_g + \Delta TE + \Delta PE_c \end{equation*}

The works done by the nurses and by gravity are internal to our system so the net external work is zero. Nothing internal changes speed so \Delta KE_i = 0. Entering those values along with the \Delta PE_g and \Delta PE_c we just calculated:

(9)   \begin{equation*} 0 = 0 + \Delta TE + 147\,\bold{J} - 735\,\bold{J} \end{equation*}

Solving for the thermal energy:

(10)   \begin{equation*} \Delta TE = 735\,\bold{J} - 147\,\bold{J} = 588\,\bold{J} \end{equation*}

That thermal energy goes into warming up the body and then eventually leaves the body as heat exhausted to the environment.

Reinforcement Exercises

Examples: Changing Earth’s Kinetic Energy

A person of about 65 kg mass starts from rest and pushes off the ground to take a step and reach a velocity of 1.2 m/s at the end of the step. What is the change in kinetic energy for each component of the Earth-person system? If the body is 20% mechanically efficient, how much chemical energy did the person expend?

We will start with the Law of Conservation of Momentum applied to just two objects.

    \begin{equation*} m_{1,i}\bold{v_{1,i}} +m_{2,i}\bold{v_{2,i}} = m_{1,f}\bold{v_{1,f}} +m_{2,f}\bold{v_{2,f}} \end{equation*}

The initial velocity of both objects is zero:

    \begin{equation*} 0 + 0 = m_{1,f}\bold{v_{1,f}} + m_{2,f}\bold{v_{2,f}} \end{equation*}

Now we subtract the initial momentum of the Earth (object #2) to the left side:

    \begin{equation*} - m_{2,f}\bold{v_{2,f}} = m_{1,f}\bold{v_{1,f}} \end{equation*}

Divide both sides by the mass of the earth and multiply both sides by (-1) to isolate the final velocity of the earth:

    \begin{equation*} \bold{v_{2,f}}= \frac{-m_{1,f}\bold{v_{1,f}}}{m_{2,f}} \end{equation*}

The mass of the earth is 5.6 \times 10^{24}\, \bold{kg} so we have:

    \begin{equation*} \bold{v_{2,f}}= \frac{(-65\,\bold{kg})(1.2\,\bold{m/s})}{5.6 \times 10^{24}\, \bold{kg}} = - 1.4\times 10^{-23} \, \bold{m/s} \end{equation*}

In units of m/s that’s a final Earth velocity of 14 after 22 zeros following the decimal or about the width of one atom per million years. The answer is negative because the Earth moves in the opposite direction as the person and we entered a positive value for their velocity.

Now let’s track the kinetic energy in this perfectly inelastic collision:

For both of these objects, the initial kinetic energy was zero. Let’s start with the Earth’s change in kinetic energy:

    \begin{equation*} \Delta KE_E = \frac{1}{2}m_Ev_E^2 - 0 \end{equation}

Entering values:

    \begin{equation*} \Delta KE_E = \frac{1}{2}(5.6 \times 10^{24}\, \bold{kg})(1.4\times 10^{-23} \, bold{m/s})^2 = 5.5\times 10^{-22}\,\bold{J} \end{equation}

Now for the change in kinetic energy of the person:

    \begin{equation*} \Delta KE_P = \frac{1}{2}(65\,\bold{kg})(1.2\,\bold{m/s})^2 = 47 \,\bold{J} \end{equation}

Notice that the change in kinetic energy of the person in this collision is 100 billion trillion times larger than of the Earth, so we see that when analyzing collisions we can often neglect the kinetic energy change of an object that is many, many orders of magnitude more massive than the other. In any case, at 20% mechanical efficiency the person had to spend 5x more chemical potential energy, (235 J), to do the work of transferring 47 J of energy into the kinetic form.

  1. OpenStax, College Physics. OpenStax CNX. May 13, 2019 http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@16.4 ↵
  2. "The maximum work and mechanical efficiency of human muscles, and their most economical speed" by A. V. Hill, National Library of Medicine, U.S. National institutes of Health ↵
  3. "MPG of a Human" by Tom Murphy, Do The Math, UCSD Physics Department ↵
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