An ideal spring having force constant k is suspended from the rigid support and a block of mass M is attached to
its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

\frac{4 M g}{k}

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\frac{2 M g}{k}

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\frac{M g}{k}

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\frac{M g}{2 k}

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Solution

The correct option is B $\frac{2 M g}{k}$
Potential energy $= \frac{1}{2} k x^{2} = M g x \Rightarrow x = \frac{2 M g}{k}$

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An ideal spring having force constant k is suspended from the rigid support and a block of mass M is attached to its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

The correct option is B 2MgkPotential energy =12kx2=Mgx⇒x=2Mgk

An ideal spring having force constant k is suspended from the rigid support and a block of mass M is attached to
its lower end. The mass is released from the natural length of the spring. Then the maximum extension in the spring is

The correct option is B $\frac{2 M g}{k}$
Potential energy $= \frac{1}{2} k x^{2} = M g x \Rightarrow x = \frac{2 M g}{k}$