Question

A 60-m-long, 9.4-mm-diameter rope hangs free from a ledge. The density of the rope is \(55 \mathrm{g} / \mathrm{m}\). How much work is needed to pull the entire rope to the ledge?

Short answer

Answer: The total work needed to pull the rope to the ledge is 671.73 Joules.

Step-by-step solution

Calculate the rope's mass and mass per unit length

We are given the density of the rope, which is 55 g/m, and the diameter of the rope which is 9.4 mm. We can calculate the mass per unit length of rope by first finding the rope's cross-sectional area, then using the density to find the mass. We know the diameter of the rope is 9.4 mm, so the radius is half of that, which is 4.7 mm or 0.0047 m. The cross-sectional area of the rope, A, can be found by using the formula for the area of a circle: \(A = \pi r^2 = \pi (0.0047)^2 = 6.938 \times 10^{-5} m^2\) Now we can use the density to find the mass per unit length, λ: \(\lambda = \rho A = 55 \frac{g}{m} * 6.938 \times 10^{-5} m^2 = 0.00381 \frac{kg}{m}\)

Divide the rope into small segments

We want to find the work needed to lift each small segment of the rope to the ledge, so let's divide the rope into small segments with length d\(x\). We'll consider each segment to have a constant mass dm. \(dm = \lambda dx\)

Calculate the work done on each segment

To find the work done on each segment, dw, we need to find the force acting on the segment, \(F = m \cdot g\), and the distance that the force is acting, which is the height of the segment, x. The work done on each segment is then given by: \(dw = F \cdot x = dm \cdot g \cdot x\) Since we know \(dm = \lambda dx\), we can substitute this into the equation: \(dw = (\lambda dx) \cdot g \cdot x\)

Integrate to find the total work done

Now we want to find the total work done, W, by adding up the work done on all the small segments of the rope. We do this by integrating the expression for dw with respect to x over the length of the rope (from 0 to 60 m): \(W = \int_{0}^{60} (\lambda gx) dx\) Using the value for λ that we calculated in Step 1, and taking g as 9.81 m/s², we get: \(W = \int_{0}^{60} (0.00381 \cdot 9.81 \cdot x) dx\) Integrating the expression with respect to x: \(W = \frac{1}{2} \cdot 0.00381 \cdot 9.81 \cdot x^2 \Big|_0^{60}\) \(W = \frac{1}{2} \cdot 0.00381 \cdot 9.81 \cdot (60)^2 - \frac{1}{2} \cdot 0.00381 \cdot 9.81 \cdot (0)^2\) \(W = 671.73 J\) So the total work needed to pull the entire rope to the ledge is 671.73 Joules.

Key concepts

Integration in Calculus

Integration is a fundamental concept in calculus, helping us to find quantities such as areas under curves, total accumulations, and more. It is particularly used to sum up small, infinitesimal quantities to find a total.

In the context of the problem, integration helps us find the total work done when pulling the rope. The rope is divided into tiny segments, and for each tiny piece, we calculate the work required to lift it. This is combined using integration, which aggregates the small segments' contributions to provide a total amount of work.

To make this calculation, we use the definite integral of the work function from the bottom of the rope to the top of the ledge. The formula used in the solution is:

• \( W = \int_{0}^{60} (\lambda g x) \, dx \)
• Here, \(x\) is the height of a segment, \(\lambda\) is the mass per unit length, and \(g\) is the acceleration due to gravity.

This method effectively turns many small calculations into one manageable equation using calculus.

Physics in Calculus

Physics often uses calculus to describe how things change over time or something affected by forces. Calculus provides the tools to analyze continuously changing systems.

In physics, force is a crucial concept described somewhat simply as mass times acceleration (\(F = ma\)). In problems involving work, such as lifting a rope or moving an object, work is evaluated as force times distance (\(W = Fd\)).

For our rope, the force involves

• mass \(dm\) of a segment, obtained from its density \(\lambda\) and length \(dx\),
• the gravitational acceleration \(g\).

Together, they contribute to the force equation for each and every small part of the rope drawn upwards. Calculating physics problems involving calculus gives a precise glimpse of the natural world, ensuring accuracy in defining these quantities.

Definite Integral Applications

Definite integrals are vital for calculating exact amounts where changes occur over a range or interval. They allow us to efficiently sum variable quantities such as work, mass, or charge distribution.

In work-related scenarios like our rope problem, the definite integral sums up many tiny work contributions into a single number—the total work. Here, the bounds of the integral (0 to 60) represent the starting and ending points of the rope.

Key points about using definite integrals:

• They manage continuous changes over a specific interval, determining total sums like total distance or total energy.
• They provide a precise solution, especially critical in physics and engineering for problem-solving.

This is why calculus's definite integrals are a powerful tool for handling real-world physics problems, allowing precise and accurate measurements in work-related scenarios as we've explored in this exercise.