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: Find the mass per metre of the chain.
The chain is 2 m long and weighs 4 kg, so each metre weighs \[ \lambda = \frac{4}{2} = 2\text{ kg/m} \] This lets me handle just the hanging piece.

Step 2: Measure the hanging piece.
30% of 2 m hangs down, so the hanging length is $0.3\times 2 = 0.6\text{ m}$. Its mass is \[ m_h = \lambda \times 0.6 = 2\times 0.6 = 1.2\text{ kg} \]

Step 3: Find how far its middle must rise.
The weight of the hanging piece acts at its middle, which sits $0.6/2 = 0.3\text{ m}$ below the table top. To pull it up onto the table that middle point rises by 0.3 m.

Step 4: Work done equals the gain in height energy.
\[ W = m_h\, g\, h = 1.2 \times 10 \times 0.3 \] \[ W = 3.6\text{ J} \]
So the work needed is 3.6 J, which is option (A).
\[ \boxed{3.6\text{ J}} \]