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Chapter 6: Problem 54

A uniform chain of length \(\overline{L \text { and mass } M \text { is lying on a smooth table and one third of its length is }}\) 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) \(M g L\) (b) \(\frac{M g L}{3}\) (c) \(\frac{M g L}{9}\) (d) \(\frac{M g L}{18}\)

Short Answer

Expert verified
The work required to pull the hanging part of the chain onto the table is \(\frac{M g L}{9}\). This corresponds to option (c).

Step by step solution

01

Calculate the weight of the hanging part of chain

This can be done by multiplying the mass of the hanging part of the chain with the acceleration due to gravity. Since the chain is uniform, it has uniform mass distribution. So a third of its length will have a third of its mass. Therefore, the mass of the hanging part is \(M/3\) and its weight is \((M/3) * g\).
02

Compute the displacement or height

The hanging part of the chain is length \((1/3)L\), which is also the height (h) it is raised to be brought onto the table. Therefore, \(h = L/3\).
03

Calculate the work done

The work done is the product of force and displacement, and in this case, is equivalent to the product of weight and height. Hence, the work done, W is calculated as follows: \(W = (M/3) * g * (L/3) = M * g * L / 9 \).

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