Work — Question 7

Question 7

A $10$-meter rope hangs vertically from the edge of a cliff. The rope has variable linear weight density given by $$\rho(x)=2+0.2x \quad \text{N/m},$$ where $x$ 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.

Question 7 – Solution

1. Label each segment by its initial depth.

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

2. Find the segment weight.

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

3. Multiply by the distance lifted.

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

4. Expand and integrate.

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

5. Evaluate and state the work.

$$W=100+\frac{1000}{15}=100+\frac{200}{3}=\boxed{\frac{500}{3}\ \mathrm{J}}\approx166.667\ \mathrm{J}.$$