Question

The awe-inspiring Great Pyramid of Cheops was built more than 4500 years ago. Its square base, originally 230 m on a side, covered 13.1 acres, and it was 146 m high, with a mass of about 7×10⁹ kg. (The pyramid’s dimensions are slightly different today due to quarrying and some sagging.) Historians estimate that 20,000 workers spent 20 years to construct it, working 12-hour days, 330 days per year.

(a) Calculate the gravitational potential energy stored in the pyramid, given its center of mass is at one-fourth its height.

(b) Only a fraction of the workers lifted blocks; most were involved in support services such as building ramps (see Figure 7.45), bringing food and water, and hauling blocks to the site. Calculate the efficiency of the workers who did the lifting, assuming there were 1000 of them and they consumed food energy at the rate of 300 kcal/h. What does your answer imply about how much of their work went into block-lifting, versus how much work went into friction and lifting and lowering their own bodies?

(c) Calculate the mass of food that had to be supplied each day, assuming that the average worker required 3600 kcal per day and that their diet was 5% protein, 60% carbohydrate, and 35% fat. (These proportions neglect the mass of bulk and nondigestible materials consumed.)

Figure 7.45 Ancient pyramids were probably constructed using ramps as simple machines. (credit: Franck Monnier, Wikimedia Commons)

Short answer

(a) The gravitational potential energy stored in a pyramid is \(2.50 \times {10^{12}}{\rm{ J}}\).

(b) The efficiency of the workers is \(2.5\% \).

(c) The mass of food to be supplied each day is \(12520{\rm{ kg}}\).

Step-by-step solution

Step 1: Potential energy

The potential energy is defined as the work done in lifting a mass against gravity.

Mathematically,

\({\rm{PE}} = mgh\)

Here, m is the mass, g is the acceleration due to gravity, and h is the height.

Gravitational potential energy of the pyramid

(a)

The center of mass of the pyramid is,

\({h_{cm}} = \frac{h}{4}\)

Here, h is the height of pyramid \(\left( {h = 146{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} {h_{cm}} = \frac{{146{\rm{ m}}}}{4}\\ = 36.5{\rm{ m}}\end{aligned}\)

The gravitational potential energy of the pyramid is,

\({\rm{PE}} = mg{h_{cm}}\)

Here, m is the mass of the pyramid \(\left( {m = 7 \times {{10}^9}{\rm{ kg}}} \right)\), g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and \({h_{cm}}\) is the height of the center of mass of pyramid \(\left( {{h_{cm}} = 36.5{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} {\rm{PE}} &= \left( {7 \times {{10}^9}{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {36.5{\rm{ m}}} \right)\\ &= 2.50 \times {10^{12}}{\rm{ J}}\end{aligned}\)

Therefore, the required gravitational potential energy stored in a pyramid is \(2.50 \times {10^{12}}{\rm{ J}}\).

Efficiency of the workers

(b)

Since, it takes 20 years working \(330{\rm{ days}}/{\rm{year}}\) and \(12{\rm{ hr}}/{\rm{day}}\) to build the pyramid. The time taken to build the pyramid is,

\(\begin{aligned} t &= \left( {20{\rm{ years}}} \right) \times \left( {\frac{{330{\rm{ days}}}}{{1{\rm{ year}}}}} \right) \times \left( {\frac{{12{\rm{ hr}}}}{{1{\rm{ day}}}}} \right)\\ &= 79200{\rm{ hr}}\end{aligned}\)

The work input of a worker is,

\({W_{in}} = Rt\)

Here, R is the rate at which the energy is consumed \(\left( {R = 300{\rm{ kcal}}/{\rm{hr}}} \right)\), and t is the time taken to build the pyramid.

Putting all known values,

\(\begin{aligned} {W_{in}} &= \left( {300{\rm{ kcal}}/{\rm{hr}}} \right) \times \left( {79200{\rm{ hr}}} \right)\\ &= 2.376 \times {10^7}{\rm{ kcal}} \times \left( {\frac{{4184{\rm{ J}}}}{{1{\rm{ kcal}}}}} \right)\\ &= 9.94 \times {10^{10}}{\rm{ J}}\end{aligned}\)

Since 1000 workers were engaged, the total work input is,

\(W = 1000{W_{in}}\)

Putting all known values,

\(\begin{aligned} W &= 1000 \times \left( {9.94 \times {{10}^{10}}{\rm{ J}}} \right)\\ &= 9.94 \times {10^{13}}{\rm{ J}}\end{aligned}\)

The efficiency of the workers is,

\(\begin{aligned} \eta &= \frac{{{\rm{Useful work done}}}}{{{\rm{total work input}}}} \times 100\% \\ &= \frac{{{\rm{PE}}}}{{{W_{in}}}} \times 100\% \end{aligned}\)

Putting all known values,

\(\begin{aligned} \eta &= \frac{{2.50 \times {{10}^{12}}{\rm{J}}}}{{9.94 \times {{10}^{13}}{\rm{ J}}}} \times 100\% \\ &= 2.5\% \end{aligned}\)

Therefore, the required efficiency of the workers is \(2.5\% \).

Mass of food required per day

(c)

The energy content of 1 g of the diet is,

\(E = {E_p} \times p + {E_c} \times c + {E_f} \times f\)

Here, \({E_p}\) is the energy content of the protein \(\left( {{E_p} = 4.0{\rm{ kcal}}/{\rm{g}}} \right)\), p is amount of protein in diet \(\left( {p = 5\% } \right)\), \({E_c}\) is the energy content of the carbohydrate \(\left( {{E_c} = 4{\rm{ kcal}}/{\rm{g}}} \right)\), c is the amount of carbohydrates in diet \(\left( {c = 60\% } \right)\), \({E_f}\) is the energy content of fat \(\left( {{E_f} = 9{\rm{ kcal}}/{\rm{g}}} \right)\), and f is the amount of fat in diet \(\left( {f = 9{\rm{ kcal}}/{\rm{g}}} \right)\).

Putting all known values,

\(\begin{aligned} E &= \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times 5\% + \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times 60\% + \left( {9{\rm{ kcal}}/{\rm{g}}} \right) \times 35\% \\ &= \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{5}{{100}} + \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{{60}}{{100}} + \left( {9{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{{35}}{{100}}\\ &= 5.75{\rm{ kcal}}/{\rm{g}}\end{aligned}\)

The mass of the food required to meet demand of \(3600{\rm{ kcal}}\) for one worker is,

\(\begin{aligned} m &= \frac{{3600{\rm{ kcal}}}}{{5.75{\rm{ kcal}}/{\rm{g}}}}\\ &= 626{\rm{ g}}\end{aligned}\)

As a result, the total mass of the food required to fulfil the energy demand of 20000 workers is,

\(\begin{aligned} M &= 20000 \times \left( {626{\rm{ g}}} \right)\\ &= \left( {1.252 \times {{10}^7}{\rm{ g}}} \right) \times \left( {\frac{{1{\rm{ kg}}}}{{{{10}^3}{\rm{ g}}}}} \right)\\ &= 12520{\rm{ kg}}\end{aligned}\)

Therefore, the required mass of food to be supplied each day is \(12520{\rm{ kg}}\).