Answer
a) $625$ $ft-lb$
b) $\frac{1875}{4}$ $ft-lb$
Work Step by Step
a)
The portion of the rope from $x$ ft to $(x+Δx)$ ft below the top of the building weighs $\frac{1}{2}Δx$ lb and must be lifted $x_i$ ft
So its contribution to the total work is $\frac{1}{2}x_i{Δx}$ $ft-lb$
The total work is
$W$ = $\lim\limits_{n \to {\infty}}{\Sigma_{i=1}^{n}}{\frac{1}{2}x_i{Δx}}$ = $\int_0^{50}\frac{1}{2}xdx$ = $\left[\frac{1}{4}x^{2}\right]_0^{50}$ = $\frac{2500}{4}$ = $625$ $ft-lb$
b)
When half the rope is pulled to the top of the building, the work to lift the top half of the rope is
$W_1$ = $\int_0^{25}\frac{1}{2}xdx$ = $[\frac{1}{4}x^{2}]_0^{25}$ = $\frac{625}{4}$ $ft-lb$
The bottom half of the rope is lifted $25$ ft and the work needed to accomplish
that is
$W_2$ = $\int_{25}^{50}\frac{1}{2}(25)dx$ = $\frac{x}{y}[x]_{25}^{50}$ = $\frac{625}{2}$ $ft-lb$
The total work done in pulling half the rope to the top of the building is
$W$ = $W_1$ + $W_2$ = $\frac{625}{4}$ + $\frac{625}{2}$ = $\frac{1875}{4}$ $ft-lb$
