Question

A massless spring with spring constant \(k\) is fixed at its upper end. A block of mass \(M\) is attached to the lower end of the spring and released from rest in its unstretched position. The maximum elongation of the spring is

• A. \(\dfrac{4Mg}{k}\)
• B. \(\dfrac{2Mg}{k}\)
• C. \(\dfrac{Mg}{k}\)
• D. \(\dfrac{Mg}{2k}\)

Hint:

For a mass released from the natural length of a vertical spring, \[ Mgx=\frac12 kx^2. \] This gives the maximum extension \[ x_{\max}=\frac{2Mg}{k}. \] Note that the equilibrium extension is only \[ \frac{Mg}{k}, \] which is half the maximum extension.

Answer 1

The Correct Option is B

Concept: Energy conservation: loss in PE = gain in spring PE. \(Mgx = \frac12 k x^2 \Rightarrow x = 2Mg/k\).

Step 1: Write the final answer. \(\boxed{x=\frac{2Mg}{k}}\)