Question

A worker on a scaffolding 75 ft above the ground needs to lift a 500 lb bucket of cement from the ground to a point 30 ft above the ground by pulling on a rope weighing 0.5 lb/ft. Choose the integral needed to calculate how much work is required?

Answer 1

Answer:

W = 15900 ft.lb

Step-by-step explanation:

From the given information.

Consider y to be the distance from the ground towards the bucket of the cement.

At height y, suppose the bucket is lifted by Δy.

Then, the workdone can be = 500Δy + 0.5(75 - y) Δy

where;

The bucket of the cement =  500Δy

The remaining of the rope =  0.5(75 - y) Δy

However; the workdone (W) can be calculated as:

$W = \int^{30}_{0} \ 500 \ dy + \int^{30}_{0} \ 0.5 (75-y ) \ dy$

$W = 500 \times 30 + 0.5 \bigg(75 \times 30 - \dfrac{1}{2} \times 30^2 \bigg )$

W = 15000 + 0.5 ( 2250 - 450)

W = 15000 + 900

W = 15900 ft.lb

Answer 2

The required work is $15,900 \ ft-lb$.

Integral Calculus:

Integral Calculus is the branch of calculus where we study integrals and their properties. Integration is an essential concept which is the inverse process of differentiation. Both the integral and differential calculus are related to each other by the fundamental theorem of calculus.

Let $x$ be the distance from the ground to the bucket of cement.

At the height $x$, if the bucket is lifted by $\Delta x$, the work done is,

$500\Delta x+0.5\left ( 75-x \right )\Delta x$.

(See the attached figure below).

The $500\Delta x$ term is due to the bucket of cement; the $0.5\left ( 75-x \right )\Delta x$ term is due to the remaining cable.

So, the total work, $W$, required to lift the bucket is,

$W=\int_{0}^{30}500dx+\int_{0}^{30}0.5\left ( 75-x \right )dx \\ =500.30+0.5\left ( 75.30-\frac{1}{2}30^{2} \right ) \\ =15,900 \ ft-lb.$

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