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Chapter 4: Problem 433

A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on to the table is (A) MgL (B) \([(\mathrm{MgL}) /(3)]\) (C) \([(\mathrm{MgL}) /(9)]\) (D) \([(\mathrm{MgL}) /(18)]\)

Short Answer

Expert verified
The short answer is: (D) \(\frac{(\mathrm{MgL})}{(18)}\)

Step by step solution

01

Identify the length and mass of the hanging part of the chain

Since one-third of the chain is hanging off the edge of the table, the length of the hanging part of the chain is (1/3)L, and the mass of the hanging part of the chain is (1/3)M.
02

Calculate the potential energy of the hanging part of the chain

The potential energy of the hanging part of the chain can be calculated using the following formula: Potential Energy (PE) = mgh where m is the mass of the object (in this case, the hanging part of the chain), g is the acceleration due to gravity, and h is the height of the object (in this case, the length of the hanging part of the chain). Since the chain is uniform, the center of mass of the hanging part will be at the midpoint of the hanging length i.e., \(\frac{1}{2}\left(\frac{1}{3}L\right) = \frac{L}{6}\) The mass of the hanging part is \(\frac{1}{3}M\), so the potential energy (PE) can be calculated as: PE = \(\frac{1}{3}M \times g \times\frac{L}{6}\) PE = \(\frac{MgL}{18}\)
03

Determine the work required

Since the potential energy of the hanging part of the chain is equal to the work done in pulling the chain onto the table, the work required is: Work Required = PE = \(\frac{MgL}{18}\) Based on the given options, the correct answer is: (D) \(\frac{(\mathrm{MgL})}{(18)}\)

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