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Question

An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is:


A

4Mgk

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B

2Mgk

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C

Mgk

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D

Mg2k

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Solution

The correct option is B

2Mgk


Explanation for correct option

Step 1: Given

  1. The spring's spring constant is k.
  2. A block of mass M is attached to the spring, which is attached to the ceiling.

Step 2: Formula used

  1. U=12kx2 where k is the spring constant , U is the energy stored, x is the stretched distance
  2. P.E=Mgx where P.E is the potential energy, x is the stretched distance, g is the acceleration due to gravity and M is the mass

Step 3: Find the maximum extension

The spring constant is k.

A block of mass M is attached to the spring, which is attached to the ceiling.

The spring will stretch because of the weight of the block.

Let the weight at the spring's end cause it to be stretched by a distance x from its equilibrium position.

The block has a mass of Mg.

Since there is no longer any external force acting on the spring, it will now be stretched with a force of Mg.

By compressing to the equilibrium point, the spring will make an effort to keep its original shape.

The spring will experience a restorative force of,

F=kx......(i)

Where k is the spring constant

The energy stored in the string is

U=12kx2....(ii)

The potential energy stored in the block is,

P.E=Mgx.....(iii)

The energy stored in the spring must be equal to the potential energy of the block.

12kx2=Mgxx=2Mgk

Hence option B is correct


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