Question

Built around 2600 BCE, the Great Pyramid of Giza in Egypt is 485 ft high (due to erosion, its current height is slightly less) and has a square base of side length 755.5 ft. Find the work W needed to build the pyramid if the density of the stone is estimated at 150 lb/ft3. (Round your answer to three decimal places.)

Answer 1

To find the work needed to build the Great Pyramid of Giza, we can use the formula W = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. First, we need to find the mass of the pyramid using the density of the stone. Then, we can calculate the work by multiplying the mass, gravity, and height.

Step-by-step explanation:

To calculate the work needed to build the Great Pyramid of Giza, we can use the formula:

W = mgh

Where W is the work, m is the mass, g is the acceleration due to gravity, and h is the height.

First, we need to find the mass of the pyramid. Given that the density of the stone is estimated at 150 lb/ft^3, we can calculate the volume of the pyramid:

Volume = base area * height

Base area = side length * side length

Then, we can calculate the mass:

Mass = density * volume

Next, we can calculate the work:

Work = mass * gravity * height

Plugging in the values, we get:

Work = (150 lb/ft^3) * (755.5 ft * 755.5 ft * 485 ft) * (32.174 ft/s^2)

Finally, we can calculate the work needed to build the pyramid:

Work = 1.462 x 10^10 lb-ft

Answer 2

Answer:1.69*10^12 J

Step-by-step explanation:

From figure above, using triangle ratio

485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485

Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3

Force slice = 150*755.5^2.y^2.∆y/485^2

From figure 2 in the attachment work done for elementary sclice

Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2

= (150*755.5^2*y^2)(485-y)∆y/485

To calculate the total work we integrate from y=0 to y= 485

Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485

Integrating the above

W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485

W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)

Work done 1.69*10^12joules