Question

a construction worker is about to haul up a 60 - m length of hanging rope. how much work will it take if the rope weighs 0.73 n/m? set up the integral that gives the work done. (give an exact answer, using radicals as needed.)

Explanation:

Step1: Recall work - formula for variable - force

The work done $W$ by a variable force $F(x)$ over an interval $[a,b]$ is given by $W=\int_{a}^{b}F(x)dx$. When lifting a rope, the force required to lift a small length of the rope is equal to the weight of the part of the rope that still needs to be lifted.
Let the length of the rope be $L = 60$ m and the linear - density of the rope $\lambda=0.7$ kg/m. The weight of an infinitesimal length $dx$ of the rope at a distance $x$ from the top (where the rope is being pulled up) is $F(x)=g\lambda(L - x)$, where $g = 9.8$ m/s². Here, $a = 0$ and $b = 60$.

Step2: Set up the integral

Substitute $F(x)$ into the work - integral formula. We have $W=\int_{0}^{60}g\lambda(60 - x)dx$, where $g = 9.8$ m/s² and $\lambda=0.7$ kg/m.

Answer:

$W=\int_{0}^{60}(9.8\times0.7)(60 - x)dx=\int_{0}^{60}6.86(60 - x)dx$