Problem 10 - Basic Technical Mathematics with Calculus

Question

Find the work done in winding up (a) \(30 \mathrm{m}\) of a \(40-\mathrm{m}\) rope on which the force of gravity is \(6.0 \mathrm{N} / \mathrm{m},\) and \((\mathrm{b})\) all of the rope.

Step-by-Step Solution

Answer

(a) 2700 Joules, (b) 4800 Joules.

Step 1: Understand the Problem

We need to calculate the work done in winding up part or all of a rope vertically, where work is done against the force of gravity acting on the rope.

Step 2: Define Work Formula for a Continuous Load

The work done in winding up a segment of rope can be calculated using the integral of the force over the distance. The force is the weight of the segment, which varies as the length wound up increases. The formula is \( W = \int_{0}^{L} \text{force per unit length} \times \text{length} \, dy \).

Step 3: Calculate Work for 30m of Rope

For part (a), we only need to wind up 30m of the rope. Let \( L = 30 \) m. The force per unit length is given as 6.0 N/m. The work is computed as: \[W = \int_{0}^{30} 6.0y \, dy \therefore W = 6.0 \int_{0}^{30} y \, dy = 6.0 \left[ \frac{y^2}{2} \right]_{0}^{30} = 6.0 \times 450 = 2700 \, \text{Joules}.\]

Step 4: Calculate Work for Entire 40m of Rope

For part (b), we need to wind up the entire 40m rope. Let \( L = 40 \) m. The work is computed as: \[W = \int_{0}^{40} 6.0y \, dy \therefore W = 6.0 \int_{0}^{40} y \, dy = 6.0 \left[ \frac{y^2}{2} \right]_{0}^{40} = 6.0 \times 800 = 4800 \, \text{Joules}.\]

Step 5: Write Final Answers

For part (a), the work done to wind up 30m of rope is 2700 Joules. For part (b), the work done to wind up the entire 40m rope is 4800 Joules.

Key Concepts

Work and Energy

The concepts of work and energy are central to understanding many physics problems. Work is defined as the effort exerted when a force causes an object to move over a distance. The equation for work is given by the product of the force applied and the distance over which it acts:
\[ W = F imes d \]
Where:

• \( W \) is work.
• \( F \) is the force applied.
• \( d \) is the distance over which the force is applied.

When dealing with continuous systems, like a rope being wound, the concept is slightly expanded. The force varies with the distance as more rope is coiled, requiring the use of calculus to integrate the force over the length of the rope. This approach provides the total work done in continuously winding up the rope.

Integration in Calculus

Integration is a powerful tool in calculus that allows us to calculate quantities like areas, volumes, and, importantly, the total accumulation of physical properties like work. When dealing with a variable force such as the one acting on a rope, calculating work requires the integration of force over the distance moved.
The integration process helps to sum the infinitely small contributions of work done over each tiny segment of the rope. This is achieved through the formula:
\[ W = \int_{0}^{L} F(y) \, dy \]
Where:

• \( W \) is the total work done.
• \( F(y) \) is the force function dependent on the position \( y \).
• \( L \) is the total distance over which the force acts.

In our rope example, the force per unit length is constant, making the integration straightforward by multiplying the force per meter by the position \( y \) and integrating over the total length.

Force and Gravity

Understanding the forces involved and how gravity acts on objects is crucial in physics problems like this. Gravity acts on every piece of mass, including individual parts of a rope.
For a rope being wound vertically, the force due to gravity can be described in terms of a force per unit length. In the problem, we have a constant gravitational force of 6.0 N/m. This force measure indicates that for every meter of rope, 6 newtons of force must be counteracted to lift the rope upward.
This distributed force means that as more length of rope is wound, the force effectively increases, because each segment of rope must be lifted against gravity from its hanging position. Summing up all these small force contributions using integration provides the total work required to elevate the rope fully.