According to Riemann sum and expressing the work as an integral,400 ft/lb work is done in pulling the rope to the top of the building.
Given,
heavy rope length = 40ft
weighs = 0.5 lb/ft
building height = 90 ft
If we divide rope into small parts with length Δx, rope weighs 0.5 lb/ft force work on each section is x*i is a point in i the such intervals.
The work done on ith part is
x*i0.5Δx
So the total done is
[tex]\lim_{n \to \infty}[/tex]= ∑0.5x*iΔx expressing work as an integral
[tex]\int\limits^40_0 {(0.5x)} \, dx[/tex] = [0.5 × x²/2]
= [0.5 × 40²/2 - 0.5 × 0²/2]
= [0.5 × 1600/2 - 0]
= 0.5 × 800 - 0
= 400 - 0
= 400 ft/lb
The work done in lifting the pieces in upper half of rope is
x*i0.5Δx
The work done in lifting lower half of the rope is
0.5Δx × 20
10Δx
Total workdone,
[tex]\lim_{n \to \infty}[/tex]∑0.5x*iΔx + [tex]\lim_{n \to \infty}[/tex]∑10Δx
= [tex]\lim_{n \to \infty}[/tex]∑(0.5x*i + 10)Δx work as an integral
[tex]\int\limits^25_0 {0.5} \, dx[/tex] + [tex]\int\limits^50_25 {10} \, dx[/tex]
= [ 0.5 * x²/2] + [10x]
= [0.5 * 20²/2] +[10(40-20)]
= [0.5 * 400/2] + [10*20]
= 100 + 200
= 300 ft/lb
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