Question

How much work, in Joules, does it take to dig up a cylindrically shaped hole of depth 2 meters and outer diameter 3 meters? How many servings of pasta are required to complete this work, assuming the energy from the pasta is converted with 5% efficiency to work? Assume that one serving of pasta contains 200 Cal, and the density of soil is 1,200 kg/m3.

Answer 1

The work done in lifting an object is quantified by its change in potential energy.

Here, consider that the potential energy is zero in the bottom of the hole.

The volume of soil removed from a small thickness of soil of dh is V = A*dh = r² * dh = 3.14*1.5*1.5*dh = 7.065*dh. m³

The mass of this much amount of soil = density*V = 1200*7.065*dh = 8478* dh kg.

The potential energy is given by PE = mgh

The work done for lifting dh amount of soil from a height h to h₀ is given by dW = mg(h₀-h)

Here, the bottom of the well is taken as zero height. So, we have to raise the soil to 2m.

dW = mg(2-h)

The total work done is then given by

Now the energy in one serving of pasta = 200 cal = 200*4.184 = 836.8 J

The energy obtained from one serving of pasta 836.8*5% = 836.8*0.05 = 41.84 J.

So, the no of servings of pasta needed = 166168.8/41.84 = 3971.529 servings