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 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
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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.
Updated On: Jul 9, 2026
- \(\dfrac{4Mg}{k}\)
- \(\dfrac{2Mg}{k}\)
- \(\dfrac{Mg}{k}\)
- \(\dfrac{Mg}{2k}\)
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The Correct Option is B
Solution and Explanation
Concept:
At the position of maximum elongation, the block momentarily comes to rest.
Hence, the loss of gravitational potential energy is completely converted into elastic potential energy stored in the spring.
Step 1: Apply conservation of mechanical energy. Let the maximum elongation be \(x\). Initially, \[ K_i=0, \qquad U_{s,i}=0. \] At maximum elongation, \[ K_f=0, \qquad U_{s,f}=\frac12 kx^2. \] The mass falls through a distance \(x\), therefore the loss in gravitational potential energy is \[ Mgx. \] Hence, \[ Mgx=\frac12 kx^2. \]
Step 2: Solve for \(x\). \[ x\left(\frac12 kx-Mg\right)=0. \] Ignoring the trivial solution \(x=0\), \[ \frac12 kx=Mg. \] \[ x=\frac{2Mg}{k}. \]
Step 3: Write the final answer. \[ \boxed{x=\frac{2Mg}{k}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Apply conservation of mechanical energy. Let the maximum elongation be \(x\). Initially, \[ K_i=0, \qquad U_{s,i}=0. \] At maximum elongation, \[ K_f=0, \qquad U_{s,f}=\frac12 kx^2. \] The mass falls through a distance \(x\), therefore the loss in gravitational potential energy is \[ Mgx. \] Hence, \[ Mgx=\frac12 kx^2. \]
Step 2: Solve for \(x\). \[ x\left(\frac12 kx-Mg\right)=0. \] Ignoring the trivial solution \(x=0\), \[ \frac12 kx=Mg. \] \[ x=\frac{2Mg}{k}. \]
Step 3: Write the final answer. \[ \boxed{x=\frac{2Mg}{k}} \] \[ \boxed{\text{Answer = (B)}} \]
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