Question

A uniform metal chain of length $2\text{ m}$ and mass $4\text{ kg}$ is lying on a horizontal table with $30\%$ of its length hanging down over the edge. The work done in pulling the hanging part back onto the table is ($g = 10\text{ m/s}^2$):

• A. 3.6 J
• B. 1.8 J
• C. 7.2 J
• D. 5.4 J Correct Answer: (A) 3.6 J

Hint:

For chain-pulling tasks on an edge, memorize this direct algebraic shortcut formula: $\mathbf{W = \frac{m g L n^2}{2}}$, where $n$ is the fractional percentage hanging down. Plugging our values directly into the expression gives: $\frac{4 \times 10 \times 2 \times (0.3)^2}{2} = 40 \times 0.09 = \mathbf{3.6\text{ J}}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem applies the Work-Energy Theorem to an extended body with a distributed mass distribution. When an object is not a single point mass, tracking changes in gravitational potential energy requires evaluating its Center of Mass (COM). The work needed to pull the dangling link segment back onto the flat table surface equals the increase in its gravitational potential energy. This is measured by determining how high the center of mass of that specific hanging section must be lifted.

Step 2: Key Formula or Approach:
1. Hanging Length ($L_h$) and Hanging Mass ($m_h$): $$ L_h = \eta L \quad \text{and} \quad m_h = \eta m $$ Where $\eta$ represents the fractional percentage hanging down over the edge ($30\% = 0.3$). 2. Center of Mass of the Hanging Segment ($h_{\text{com}}$): For a completely uniform, straight hanging chain section, its center of mass is located exactly at its geometric midpoint: $$ h_{\text{com}} = \frac{L_h}{2} $$ 3. Work Done Formula via Potential Energy Increase: $$ W = m_h \cdot g \cdot h_{\text{com}} = (\eta m) \cdot g \cdot \left(\frac{\eta L}{2}\right) = \frac{\eta^2 m g L}{2} $$

Step 3: Detailed Explanation:
Let's extract the parameters and compute the values step-by-step: - Total length of the uniform chain ($L$) = $2\text{ m}$ - Total mass of the uniform chain ($m$) = $4\text{ kg}$ - Hanging fraction ($\eta$) = $30\% = 0.3$ - Acceleration due to gravity ($g$) = $10\text{ m/s}^2$ 1. Calculate the physical metrics of the hanging part: - Length hanging down: $$ L_h = 0.3 \times 2\text{ m} = 0.6\text{ m} $$ - Mass hanging down: $$ m_h = 0.3 \times 4\text{ kg} = 1.2\text{ kg} $$ 2. Locate the Center of Mass for the hanging part: The section hangs vertically from the table's edge down to a depth of $0.6\text{ m}$. Its center of mass lies exactly halfway down this span: $$ h_{\text{com}} = \frac{0.6\text{ m}}{2} = 0.3\text{ m} $$ This means pulling the entire hanging loop back onto the flat tabletop is equivalent to lifting its concentrated mass center upward by a distance of $0.3\text{ m}$. 3. Calculate the total work required ($W$): $$ W = m_h \times g \times h_{\text{com}} $$ $$ W = 1.2\text{ kg} \times 10\text{ m/s}^2 \times 0.3\text{ m} $$ $$ W = 12 \times 0.3 = 3.6\text{ J} \text{ ???} $$ Let's check our direct multiplication steps again carefully to ensure complete arithmetic accuracy: $$ W = 1.2 \times 10 \times 0.3 = 12 \times 0.3 = 3.6\text{ J} $$ Wait, let's re-verify our shortcut formula: $$ W = \frac{\eta^2 m g L}{2} = \frac{(0.3)^2 \times 4 \times 10 \times 2}{2} = 0.09 \times 4 \times 10 = 3.6\text{ J} $$ Let's re-verify the option values provided in the text: (A) 3.6 J, (B) 1.8 J, (C) 7.2 J, (D) 5.4 J. The calculation yields 3.6 J, which corresponds directly to option (A).

Step 4: Final Answer:
The work done in pulling the hanging part back onto the table is $3.6\text{ J}$, matching option (A).