In the figure shown, stiffness of spring is k and mass of the block is m. The pulley is fixed. Initially the spring is relaxed. Now the system is released from rest. The

maximum elongation in the spring is

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Solution

The correct option is B

When the spring has maximum elongation, the mass is momentarily at rest. Now, decrease in gravitational P.E. = Increase in spring P.E.

$∴ m g x_{m} = \frac{1}{2} k x_{m}^{2} \Rightarrow x_{m} = \frac{2 m g}{k}$

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In the figure shown, stiffness of spring is k and mass of the block is m. The pulley is fixed. Initially the spring is relaxed. Now the system is released from rest. The maximum elongation in the spring is

The correct option is B When the spring has maximum elongation, the mass is momentarily at rest. Now, decrease in gravitational P.E. = Increase in spring P.E. ∴mg xm=12kx2m⇒xm=2mgk

In the figure shown, stiffness of spring is k and mass of the block is m. The pulley is fixed. Initially the spring is relaxed. Now the system is released from rest. The

maximum elongation in the spring is

The correct option is B

When the spring has maximum elongation, the mass is momentarily at rest. Now, decrease in gravitational P.E. = Increase in spring P.E.

$∴ m g x_{m} = \frac{1}{2} k x_{m}^{2} \Rightarrow x_{m} = \frac{2 m g}{k}$