Question

A uniform chain of length 3 meter and mass 3 kg overhangs a smooth table with 2 meter laying on the table. If k is the kinetic energy of the chain in joule as it completely slips off the table, then the value of k is _________.
(Take \(g = 10 \text{ m/s}^2\))

Hint:

For potential energy of extended objects like chains, always calculate the height of the Center of Mass (COM) of the specific part that is moving vertically.

Answer 1

Correct Answer: 40

Step 1: Understanding the Concept:
This problem can be solved using the principle of conservation of mechanical energy. Since the table is smooth, there is no friction, and the loss in potential energy of the chain as it slips off equals the gain in its kinetic energy.
Step 2: Key Formula or Approach:
1. Potential Energy of a part of the chain: \(U = -m_{part} g y_{cm}\), where \(y_{cm}\) is the depth of the center of mass of the hanging part from the table level.
2. Conservation of Energy: \(U_i + K_i = U_f + K_f\). Since it starts from rest, \(K_i = 0\), so \(K_f = U_i - U_f\).
Step 3: Detailed Explanation:
Given: Total length \(L = 3 \text{ m}\), Total mass \(M = 3 \text{ kg}\).
Mass per unit length \(\lambda = \frac{3 \text{ kg}}{3 \text{ m}} = 1 \text{ kg/m}\).

Initial State:
Length on table = 2 m, Hanging length \(h_1 = 3 - 2 = 1 \text{ m}\).
Mass of hanging part \(m_1 = \lambda h_1 = 1 \text{ kg}\).
Center of mass of the hanging part is at a distance \(d_1 = \frac{h_1}{2} = 0.5 \text{ m}\) below the table.
Initial Potential Energy \(U_i = -m_1 g d_1 = -(1)(10)(0.5) = -5 \text{ J}\).

Final State:
The entire chain has slipped off. Hanging length \(h_2 = 3 \text{ m}\).
Total mass hanging \(m_2 = 3 \text{ kg}\).
Center of mass of the chain is now at \(d_2 = \frac{h_2}{2} = 1.5 \text{ m}\) below the table.
Final Potential Energy \(U_f = -m_2 g d_2 = -(3)(10)(1.5) = -45 \text{ J}\).

Calculating Kinetic Energy (k):
By conservation of energy:
\[ k = U_i - U_f \]
\[ k = -5 - (-45) \]
\[ k = 40 \text{ J} \]
Step 4: Final Answer:
The value of \(k\) is 40.