4p@@'()*+,-./0123456789:;<=BLå kBhc Work$kBhcPIIn physics, work is defined as motion through a distance against a force.kBhc workkBhc  = kBhc forcekBhc  * kBhcdistancekBhcÿIf the force is constant throughout the motion, the calculation is very simple. In many applications, however, the force is not constant. In that case, we can take advantage of the fact that over very small distances the force is approximately constant. TkBhcuno compute the total work in moving over a certain distance, we can use an integral to add up the bits of work.kBhc W = etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhcExamples from the textkBhcd]In class I showed examples 3 and 4 from the text. You should review those examples carefully.kBhc5.How much work does it take to build a pyramid?kBhcÿThe great pyramid at Giza is a pyramid with a square base of side length 755 feet and a height of 481 feet. The pyramid is constructed from roughly 2.3 million stones weighing an average of 5000 pounds each. How much energy did it take to pile up the stonkBhces in the pyramid?# kBhc­¦The best way to appoach this problem is to solve it in the abstract first, and then mix in the specific measurements. Consider a pyramid with a square base of length kBhcbkBhc and a height kBhchkBhc&. The pyramid has a density of kBhcÜkBhcZS pounds per cubic foot. Suppose we make a thin slice through the pyramid at height kBhcykBhckd above the base. Since the base of the pyramid is square, the slice will be a square of side length kBhcxkBhc…~. Since all of the material in that slice is at the same height, the amount of work needed to lift that slice into position is" kBhcÜkBhc Grap"kBhcvolume of slicekBhc kBhc y = ÜkBhc GrapEkBhcxopxEkBhc2kBhc  dkBhcykBhc kBhcy!kBhc:3A simple argument with similar triangles shows that kBhc x = carF$kBhcbkBhchkBhc GrapkBhc h - ykBhc:3Thus, the total work needed to build the pyramid isetnIkBhcÜkBhc carFTkBhcbopxEkBhc2kBhchopxEkBhc2kBhc Grap.kBhchkBhc  - kBhcyopxEkBhc2kBhc kBhcykBhcykBhchkBhc0kBhc  = ÜkBhc carFTkBhcbopxEkBhc2kBhchopxEkBhc2kBhc etnIÆkBhchopxEkBhc2kBhc kBhc y - 2 h yopxEkBhc2kBhc  + kBhcyopxEkBhc3kBhcykBhchkBhc0kBhc kBhc = Ü carFTkBhcbopxEkBhc2kBhchopxEkBhc2kBhc Grap\carF<kBhchopxEkBhc2kBhc2kBhc kBhcyopxEkBhc2kBhc kBhc-kBhc carF&kBhc 2 hkBhc3kBhc kBhcyopxEkBhc3kBhc  + carF$kBhc1kBhc4kBhc kBhcyopxEkBhc4kBhc|sbuSkBhc0opxEkBhchkBhc  =kBhc carF%kBhc1kBhc 12kBhc kBhcÜkBhc kBhcbopxEkBhc2kBhc kBhchopxEkBhc2kBhc The kBhcÜkBhcù factor here stands for weight per unit volume. The calculation of the volume of the pyramid is almost identical to the work calculation here. The only difference is that the volume calculation has no density term and is missing the extra factor of kBhcykBhc  in feR "(1)kBhc.kBhcvolume kBhc=kBhc etnIæcarFTkBhcbopxEkBhc2kBhchopxEkBhc2kBhc Grap.kBhchkBhc  - kBhcyopxEkBhc2kBhcykBhchkBhc0kBhc kBhc=kBhc carF$kBhc1kBhc3kBhc carFTkBhcbopxEkBhc2kBhchopxEkBhc2kBhc GrapkBhc y - hopxEkBhc3kBhc  |sbuSkBhc0opxEkBhchkBhc  = carF$kBhc1kBhc3kBhc kBhcbopxEkBhc2kBhc  hkBhczWe are now ready to bring in the specific numbers for this pyramid. The density is the total weight divided by the volume:kBhc Ü = carFÁkBhc 5000 Grap1kBhc 2.3 10opxEkBhc6carF$kBhc1kBhc3kBhc kBhcbopxEkBhc2kBhc kBhchkBhcPlugging this into feR (2)kBhc and setting kBhch = 481kBhc  giveskBhc workkBhc  = carF%kBhc1kBhc 12kBhc GrapÍcarFÁkBhc 5000 Grap1kBhc 2.3 10opxEkBhc6carF$kBhc1kBhc3kBhc kBhcbopxEkBhc2kBhc kBhchkBhc kBhcbopxEkBhc2kBhc kBhchopxEkBhc2kBhc  = carF]kBhc 5000 Grap1kBhc 2.3 10opxEkBhc6kBhc4kBhc  481kBhc  = kBhc1.38 10opxEkBhc 12kBhc ft-poundskBhc…~The unit of work in the metric system is called a Joule. One foot-pound equals 1.356 Joules, so in Joules the energy needed iskBhc 1.356 Grap3kBhc1.38 10opxEkBhc 12kBhc  = kBhc 1.87128 10opxEkBhc 12kBhc JouleskBhcXQBy way of comparison, the energy content in a liter of jet fuel is approximately kBhc35.5 10opxEkBhc6kBhcvo Joules. A Boeing 747 has a fuel capacity of 183,380 liters. Thus, the energy content in a fully loaded 747 is kBhc6.51 10opxEkBhc 12kBhc:3 Joules, or enough to pile up three great pyramids.