Question

A 100 ft rope, weighing \(0.1 \mathrm{lb} / \mathrm{ft}\), hangs over the edge of a tall building. (a) How much work is done pulling the entire rope to the top of the building? (b) How much rope is pulled in when half of the total work is done?

Short answer

(a) 50000 ft-lb. (b) 69.5 feet pulled.

Step-by-step solution

Understand the problem

We need to calculate the work done in pulling up a 100 ft rope with a linear weight distribution. We will use the concept of work, which is the integral of force over distance, across the rope's length.

Establish the variables and equations

Let the length of rope already pulled be denoted by \( x \) (where \( x \) ranges from 0 to 100 feet). The weight of the remaining rope is \( (100 - x) imes 0.1 \) lb/ft. The force at any position \( x \) is the weight of the remaining rope, given by \( 10(100 - x) \) because the rope weighs 0.1 lb/ft.

Set up the integral for total work

The work done to move an infinitesimally small piece \( dx \) of the rope is \( 10(100 - x) \, dx \). The total work done is the integral from \( x = 0 \) to \( x = 100 \): \[W = \int_{0}^{100} 10(100 - x) \, dx\]

Solve the integral for total work

Calculating the integral:1. \[ W = 10 \int_{0}^{100} (100 - x) \, dx \] Split the integral: \[ = 10 \left( \int_{0}^{100} 100 \, dx - \int_{0}^{100} x \, dx \right) \]2. The integrals are: - \( \int_{0}^{100} 100 \, dx = 100x \big|_{0}^{100} = 10000 \) - \( \int_{0}^{100} x \, dx = \frac{x^2}{2} \big|_{0}^{100} = 5000 \)3. Substitute back: - \( W = 10 (10000 - 5000) = 50000 \) foot-pounds.

Half of the total work

To find when half of the total work is done, we solve for \( x \) such that work \( W = 25000 \). Substitute the formula for work done up to \( x \): \[ 10 \int_{0}^{x} (100 - x) \, dx = 25000 \].

Simplify and substitute in integral equation

Perform the integration:1. \[ 10 \left( \,100x - \frac{x^2}{2} \right) \big|_{0}^{x} = 25000 \]2. This simplifies to \( 1000x - 5x^2 = 25000 \).3. Divide by 5: \( 200x - x^2 = 5000 \). Rearrange to: \( x^2 - 200x + 5000 = 0 \).

Solve quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = -200 \), and \( c = 5000 \):1. \( x = \frac{200 \pm \sqrt{200^2 - 4 \times 1 \times 5000}}{2} \)2. \( x = \frac{200 \pm \sqrt{40000 - 20000}}{2} = \frac{200 \pm \sqrt{20000}}{2} \).3. \( x \approx \frac{200 \pm 141}{2} \). Use positive root: \( x \approx \frac{200 + 141}{2} = 170.5 \).

Interpret the solution

Since \( x = 170.5 \) is impossible (since it's over 100 feet), logically, we reverse the amount of rope left and find \( y \), where \( y = 100 - x \) for half the work done is correct. Solve for \( y \): \( y = 30.5 \), implying 69.5 feet have been pulled when half the work is done.

Key concepts

calculus integration

Calculus integration is a fundamental concept used to determine the work done in moving an object over a distance by summing the small amounts of force applied. In the context of this problem, integration helps us calculate work by considering tiny segments of the rope being pulled to the top of the building.

When we set up an integral, we're looking to sum these small pieces of work across the entire length of the rope, from 0 to 100 feet. The integral expression for this particular problem is given by \[ W = \int_{0}^{100} 10(100 - x) \, dx \] which represents the total work done in pulling up the entire rope.

To solve the integral, we break it down into manageable parts, taking each segment of work over a small distance \( dx \), reflecting how calculus allows us to handle continuously changing quantities. This technique helps us find the exact measure of work required.

physics of work

The physics of work involves understanding how force is applied over a distance. In this scenario, pulling a rope up a building connects directly to this concept. The work done, represented in foot-pounds or similar units, is calculated by multiplying the force exerted by the distance over which it is applied.

Here, we need to overcome the gravitational pull acting on the rope. As we pull the rope upward, the force required decreases because there is less rope to be lifted. Initially, when the entire 100-foot rope is hanging, the force required is at its maximum. With each foot of rope pulled, this force decreases proportionally.

Understanding work through calculus allows us to see how changing forces impact the overall effort needed to complete tasks like lifting an entire rope.

rope weight distribution

The concept of rope weight distribution is crucial in calculating the work needed to pull the rope. Each foot of the rope weighs 0.1 lb, which means the total weight of the rope when hanging is the sum of each section's weights.

As we pull the rope, however, the weight decreases. At any point \( x \) feet of rope already pulled, the weight remaining is \( 10(100 - x) \) pounds, reflecting the 0.1 lb/ft distribution. This linearly decreasing weight affects the force necessary at any point in the rope’s length.

Visualizing weight distribution can help us intuitively understand why less energy is required as more rope is pulled up.

quadratic equations in calculus

Quadratic equations come into play when solving for the distance of the rope pulled when half of the total work is done. In this case, the problem required finding an unknown distance \( x \) when the work done equals 25000 foot-pounds. From the integral, we developed the quadratic equation:\[ x^2 - 200x + 5000 = 0 \]
This quadratic equation can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -200 \), and \( c = 5000 \). Solving this provides a method to accurately find solutions for \( x \) even in complex physical situations.

The importance of this approach is highlighted in real-world applications, where simple direct calculations aren’t possible, and we rely on quadratic solutions to interpret sophisticated problems accurately.