Question

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

• A.
• B.
• C.
• D.

Answer 1

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.

$\therefore mg{x}_m=\frac{1}{2}k{x}_m^2\Rightarrow x_m=\frac{2mg}{k}$