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Calculus IChapter 6: Application of IntegralsSection 6-6

Work — Question 7

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Question 7

A 1010-meter rope hangs vertically from the edge of a cliff. The rope has variable linear weight density given by ρ(x)=2+0.2xN/m,\rho(x)=2+0.2x \quad \text{N/m}, where xx is the distance (in meters) measured downward from the top of the cliff.

Find the work required to pull the entire rope to the top of the cliff.

See the diagram in the original worksheet below.

Initial configuration; xx is measured down from the top.

Original worksheet page 1: question and worked solution for 6-6-007
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Question 7 – Solution

1. Label each segment by its initial depth.

Let xx be distance downward from the cliff top, as in the question. The rope occupies 0≤x≤100\le x\le10 meters; ρ(x)\rho(x) describes the initial weight distribution.

2. Find the segment weight.

For a short piece of length dxdx,dF=ρ(x)dx=(2+0.2x)dx.dF=\rho(x)\,dx=(2+0.2x)\,dx.The units are newtons because ρ\rho is given in N/m.

3. Multiply by the distance lifted.

The segment at depth xx is lifted xx meters:dW=x(2+0.2x)dx,W=∫010x(2+0.2x)dx.dW=x(2+0.2x)\,dx,\qquad W=\int_0^{10}x(2+0.2x)\,dx.

4. Expand and integrate.

W=∫010(2x+x25)dx=[x2+x315]010.W=\int_0^{10}\left(2x+\frac{x^2}{5}\right)dx=\left[x^2+\frac{x^3}{15}\right]_0^{10}.

5. Evaluate and state the work.

W=100+100015=100+2003=5003J≈166.667J.W=100+\frac{1000}{15}=100+\frac{200}{3}=\boxed{\frac{500}{3}\ \mathrm{J}}\approx166.667\ \mathrm{J}.

Original worksheet page 2: question and worked solution for 6-6-007

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