Having some trouble understanding how to approach this question:

If this cola bottle was inverted, would the amount of work still be the same?

The amount of work found is in "Example 3, Lifting Liquids":

Practice 2. How much work is done lifting an injured person of mass 50 kg to the top of a 30-meter cliff using a stretcher of mass 5 kg and a 30-meter cable of mass 10 kg? Lifting Liquids Example 3. A cola glass (see margin figure) has dimensions given in the margin table. Approximately how much work do you do when you drink a cola glass full of water by sucking it through a straw to a point 3 inches above the top edge of the glass? Solution. The table partitions the water into 1-inch "slices": top of straw height radius 3 3.5 in 1.4 4.5 in 5.5 in 6.5 in 4.7 oz 4.1 0z 1.8 oz 2.2 oz The work needed to move each slice is approximately the weight of You might wonder why the displacement the slice times the distance the slice is moved. We can use the radius is not computed by taking the distance at the bottom of each slice to approximate the volume-and then the from the bottom of the straw up to the top weight -of the slice, and a point halfway up each slice to calculate the of the straw, but when computing work we need to use the net displacement. distance the slice is moved. For the top slice: Water's density is 62.5- = 0.57874 weight = (volume) (density) = > (1.6 in) (1in) (0.5787 0z ) 4.7 oz and the distance this slice travels is roughly 3.5 inches, so: W =F . d = (4.7 oz) (3.5 in) =: 16.4 oz-in For the next slice: weight = (volume) (density) = > (1.5in) (1in) (0.5787 03 4.1 oz and the distance this slice travels is roughly 4.5 inches, so: W = F . d = (4.1 oz) (4.5 in) = 18.4 oz-inThe work for the last two slices is (1.802) (5.5in) = 9.9 oz-in and (2.2oz) (6.5in) = 143 oz-in. The total work is then sum of the work needed to raise each slice of water: (16.4 0z-in) + (18.4 0z-in) + (9.9 0z-in) + (143oz-in) = 59.0 oz-in or about 0.31 ft-lbs