Question

A \(50 \mathrm{~m}\) rope, with a mass density of \(0.2 \mathrm{~kg} / \mathrm{m}\), 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 work is done pulling in the first \(20 \mathrm{~m}\) ?

Short answer

(a) 2450 Joules, (b) 392 Joules.

Step-by-step solution

Understand Work and Mass Concepts

Work done is calculated as the force applied over a distance. The force needed to lift the rope is due to gravity, which is the mass of the rope segment times the acceleration due to gravity (9.8 m/s²). The mass density of the rope helps us find the mass of any segment of the rope we need to analyze.

Calculate Total Mass of the Rope

To find the total mass of the rope, multiply the mass density of the rope (0.2 kg/m) by its total length (50 m):\[ m = 0.2 \times 50 = 10 \text{ kg} \].

Determine Expression for a Small Element

To find work done, consider a small element of the rope at a distance \( x \) from the top. The force required to lift this element is its weight, which is \( 0.2 \times 9.8 \) acting over the distance \( x \). This element has:\[ \text{mass}= 0.2 \cdot \Delta x \] \[ \text{weight}= 0.2 \cdot 9.8 \cdot \Delta x \]

Setup Integral for Total Work

The total work done pulling the rope to the top is found by integrating the small elements along the entire length:\[ \text{Work} = \int_0^{50} 0.2 \times 9.8 \times x \, dx \]

Compute the Integral for Full Rope (Part a)

Calculate the integral:\[\text{Work} = 0.2 \times 9.8 \int_0^{50} x \, dx \= 1.96 \left[ \frac{x^2}{2} \right]_0^{50} \= 1.96 \left[ \frac{50^2}{2} - \frac{0^2}{2} \right] \= 1.96 \times 1250 = 2450 \text{ J}\]So, the work done in pulling the entire 50 m of the rope is 2450 Joules.

Setup Integral for Partial Work (Part b)

For the first 20 m of the rope, we set the limits of the integral from 0 to 20:\[ \text{Work} = \int_0^{20} 0.2 \times 9.8 \times x \, dx \]

Compute the Integral for First 20 m

Calculate the integral:\[\text{Work} = 0.2 \times 9.8 \int_0^{20} x \, dx \= 1.96 \left[ \frac{x^2}{2} \right]_0^{20} \= 1.96 \times \left( \frac{20^2}{2} \right) \= 1.96 \times 200 = 392 \text{ J}\]So, the work done pulling the first 20 m of the rope is 392 Joules.

Key concepts

Integral Calculus

In physics, integral calculus is an invaluable tool for determining work done over variable forces and distances, especially when dealing with non-constant forces. When applying this to mechanics problems, like pulling a rope, we use the concept of integration to summate the work done on infinitely small segments of the object. This relies on the understanding that the force applied might not be constant over the entire length. Here, the work done is obtained by integrating the force over a defined path or distance, essentially accumulating all these tiny contributions of force over each interval. By setting up an integral, like \( \int_0^{50} 0.2 \times 9.8 \times x \, dx \), we calculate the work required for lifting sections of the rope successively until reaching the full length of the rope.

Force and Gravity

When pulling the rope, the force needed aligns with the force of gravity acting downward on the rope's mass. Force due to gravity is a fundamental concept in mechanics, given by \( F = m \cdot g \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. In this problem, the force is calculated based on the rope's mass density and is shown by \( 0.2 \times 9.8 \times \Delta x \), representing the weight of each small segment of rope. This is integrated over the desired lengths. By understanding that gravity exerts a consistent downward force, calculating how much work is needed to lift each section becomes feasible through the step-by-step integration approach.

Mass Density

Mass density is crucial when dealing with problems involving objects like ropes or other extended bodies. Here, it represents the distribution of mass per unit length of the rope, given as \( 0.2 \text{ kg/m} \). In physics problems, knowing the mass density allows you to compute how much mass is within a section of material. For the rope, the overall mass is given by \( 0.2 \cdot 50 = 10 \text{ kg} \). This concept simplifies complex calculations by allowing us to derive the weight and thus the gravitational force acting along individual pieces of the rope, further enabling accurate calculations of the work done when lifting the rope.

Mechanics Problems

Mechanics involves the study of forces and their effects on motion. In this context, solving comprehension problems like pulling a rope up a building allows the conceptualization and application of fundamental physics laws. These problems require a sequential approach to problem-solving, applying Newton's laws and calculus to find solutions. Utilizing concepts like mass density, integral calculus, and forces, mechanics problems translate real-world scenarios into mathematical operations, offering practical examples of theoretical principles. Approaching these studies steadily builds problem-solving skills and provides a deeper understanding of physics applications, going beyond theoretical equations to tangible experiences.