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A heavy rope, 60 ft long, weighs 0.3 lb/ft and hangs over the edge of a building 130 ft high. (Let x be the distance in feet below the top of the building. Enter xi* as xi.) (a) How much work W is done in pulling the rope to the top of the building? Show how to approximate the required work by a Riemann sum. lim n→[infinity] n Δx i = 1 Express the work as an integral. 0 dx Evaluate the integral. ft-lb (b) How much work W is done in pulling half the rope to the top of the building? Show how to approximate the required work by a Riemann sum. lim n→[infinity] n Δx i = 1 Express the work as an integral. 0 dx Evaluate the integral. ft-lb

Answer :

fortuneonyemuwa

Answer:

Step-by-step explanation:

a)

Let x unit length of rope

Riemann sum

[tex]W= \lim_{n \to \infty} \sum\limits^n_{i=1}0.3x[/tex]

Weight = 0.3x

work done = 0.3xdx

The required region is 0<x<60

Thus, the required work is

[tex]W=\int\limits^{60}_0 {0.3x} \, dx \\\\=0.3[\frac{x^2}{2}]\imits^{60}_0\\\\=0.15(60)^2\\=(0.15)(3600)\\\\W=540ft-lb[/tex]

b)

Let x unit length of rope from bottom

The Riemann sum in two parts.

First half weight is = 0.3x

The second half weight is = (0.3)(60 - x)

Thus, the Riemann sum is

[tex]W= \lim_{n \to \infty} \sum\limits^n_{i=1}0.3x + \lim_{n \to \infty} \sum\limits^n_{i=1}0.3(60-x)[/tex]

The first weight is 0.3x when 0<x<30.

The second half weight is = 0.3(60-x) when 30<x<60

Thus, the work done is

[tex]0.3xdx + 0.3(60-x)dx[/tex]

Consider,

[tex]W=\int\limits^{30}_0 {0.3x} \, dx +\int\limits^{60}_{30} {(0.3)(60-x)} \, dx\\\\=[0.15x^2]^{30}_0-0.15[60-x^2]^{60}_{30}\\\\=(0.15)(30)^2-(0.15)(60-(60))^2+(0.15)(60-(30))^2 \\\\=135-0+135\\\\W=270ft-lb[/tex]

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