A uniform chain of mass \(M\) and length \(L\) is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position.

Calculate the force exerted by the falling chain on the floor when a length of \(x\) has reached the floor. The answer is in the form of \(k \times \dfrac{Mgx}{L}\). Find the value of \(k\).

**Details and Assumptions:**

- Any portion that strikes the floor comes to rest.
- Assume that the chain does not form a heap on the floor.

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