Question:
In the arrangement shown,
\[
k_1=1500\,\text{N m}^{-1},
\qquad
k_2=500\,\text{N m}^{-1},
\]
\[
m_1=2\,\text{kg},
\qquad
m_2=1\,\text{kg}.
\]
The potential energy stored in the system of springs in equilibrium is (spring masses negligible and \(g=10\,\text{ms}^{-2}\)).
In the arrangement shown,
\[
k_1=1500\,\text{N m}^{-1},
\qquad
k_2=500\,\text{N m}^{-1},
\]
\[
m_1=2\,\text{kg},
\qquad
m_2=1\,\text{kg}.
\]
The potential energy stored in the system of springs in equilibrium is (spring masses negligible and \(g=10\,\text{ms}^{-2}\)).
Show Hint
In a vertical spring system, first determine the tension in each spring. The upper spring supports all masses below it, while a lower spring supports only the masses hanging beneath it. Then use
\[
U=\frac12 kx^2.
\]
Updated On: Jul 29, 2026
- \(0.5\ \text{J}\)
- \(0.4\ \text{J}\)
- \(1.2\ \text{J}\)
- \(5.6\ \text{J}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
At equilibrium, each spring stores elastic potential energy
\[
U=\frac12 kx^2.
\]
First find the extension of each spring using force balance.
Step 1: Find the extension of spring \(k_2\). For mass \(m_2\), \[ T_2=m_2g. \] \[ T_2=(1)(10)=10\ \text{N}. \] Hence, \[ x_2=\frac{T_2}{k_2} =\frac{10}{500} =\frac1{50}\ \text{m}. \] \[ x_2=0.02\ \text{m}. \]
Step 2: Find the extension of spring \(k_1\). Spring \(k_1\) supports both masses. Therefore, \[ T_1=(m_1+m_2)g. \] \[ T_1=(2+1)10=30\ \text{N}. \] Thus, \[ x_1=\frac{T_1}{k_1} =\frac{30}{1500} =\frac1{50}\ \text{m}. \] \[ x_1=0.02\ \text{m}. \]
Step 3: Calculate the energy stored in each spring. For spring \(k_1\), \[ U_1 = \frac12 k_1x_1^2. \] \[ = \frac12(1500)(0.02)^2. \] \[ = 0.3\ \text{J}. \] For spring \(k_2\), \[ U_2 = \frac12 k_2x_2^2. \] \[ = \frac12(500)(0.02)^2. \] \[ = 0.1\ \text{J}. \]
Step 4: Find the total potential energy. \[ U=U_1+U_2. \] \[ U=0.3+0.1. \] \[ U=0.4\ \text{J}. \] Therefore, \[ \boxed{U=0.4\ \text{J}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Find the extension of spring \(k_2\). For mass \(m_2\), \[ T_2=m_2g. \] \[ T_2=(1)(10)=10\ \text{N}. \] Hence, \[ x_2=\frac{T_2}{k_2} =\frac{10}{500} =\frac1{50}\ \text{m}. \] \[ x_2=0.02\ \text{m}. \]
Step 2: Find the extension of spring \(k_1\). Spring \(k_1\) supports both masses. Therefore, \[ T_1=(m_1+m_2)g. \] \[ T_1=(2+1)10=30\ \text{N}. \] Thus, \[ x_1=\frac{T_1}{k_1} =\frac{30}{1500} =\frac1{50}\ \text{m}. \] \[ x_1=0.02\ \text{m}. \]
Step 3: Calculate the energy stored in each spring. For spring \(k_1\), \[ U_1 = \frac12 k_1x_1^2. \] \[ = \frac12(1500)(0.02)^2. \] \[ = 0.3\ \text{J}. \] For spring \(k_2\), \[ U_2 = \frac12 k_2x_2^2. \] \[ = \frac12(500)(0.02)^2. \] \[ = 0.1\ \text{J}. \]
Step 4: Find the total potential energy. \[ U=U_1+U_2. \] \[ U=0.3+0.1. \] \[ U=0.4\ \text{J}. \] Therefore, \[ \boxed{U=0.4\ \text{J}} \] \[ \boxed{\text{Answer = (B)}} \]
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