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

Step 1: Define the initial state.

Let the initial position of the block with the unstretched spring be the reference level for gravitational potential energy.

$$x_i = 0$$ $$v_i = 0$$ $$E_i = 0$$

Step 2: Define the state at maximum extension.

At the maximum extension $x_{\max}$, the block momentarily comes to rest.

$$v_f = 0$$ $$E_f = \frac{1}{2}kx_{\max}^2 - Mgx_{\max}$$

Step 3: Apply conservation of mechanical energy.

The total mechanical energy of the system remains constant.

$$E_i = E_f$$ $$0 = \frac{1}{2}kx_{\max}^2 - Mgx_{\max}$$ $$Mgx_{\max} = \frac{1}{2}kx_{\max}^2$$

Step 4: Solve for the maximum extension.

Divide both sides by $x_{\max}$, since the maximum extension $x_{\max} \neq 0$.

$$Mg = \frac{1}{2}kx_{\max}$$ $$x_{\max} = \frac{2Mg}{k}$$

Answer: The maximum extension in the spring is $\dfrac{2Mg}{k}$.