Answer:
A. 1,260 lb/ft
B. 315 lb/ft
Explanation:
(a) firs we divide the rope into n sub-intervals . each sub-interval has length Δx=[tex]\frac{60}{n}[/tex].
next we pick any ith sub-interval, [tex]X_{i}^*[/tex] ∈ [tex][X_{i-1}, X{i}][/tex]
the work required to move [tex]X_{i}^*[/tex] to the top of the building is given by
[tex]W_{i} = F_{i}d_{i}[/tex] = [(0.7)(Δx)][[tex]X_{i}^*[/tex]]
thus work required to pull the rope to the top of the building is
W = ∑ [tex]W_{i}[/tex] = [tex]\lim_{n \to \infty}[/tex]∑ [tex]0.7X_{i}^*[/tex]Δx
= [tex]\int\limits^{60}_0 {0.7x} \, dx = [0.7\frac{x^2}{2} ]_0^{60}[/tex]
= 1,260 lb/ft
(B) To find the work to pull half of the rope to the top of the building, we simply integrate using the limit 0 to 30
[tex]W=\int\limits^{30}_0 {0.7x} \, dx = [0.7\frac{x^2}{2} ]_0^{30}[/tex]
= 315 lb/ft
