Q: 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 unstreched. Then the maximum extension in the springs is

(a)(4 Mg)/k

(b)(2 Mg)/k

(c)Mg/k

(d)Mg/(2 k)

Ans: (b)

Sol: Let x be the maximum extension in the spring .

Applying conservation of enery .

Loss in gravitational P.E = Gain in elastic P.E of the spring

$\large M g x = \frac{1}{2}k x^2$

$\large x = \frac{2 M g}{k}$