College Physics
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![### Physics Problem: Understanding Potential Energy in a Compressed Spring
Two blocks are forced together compressing a spring \( S \) between them. Block A has a mass of 0.8 kg, and Block B has a mass of 3.0 kg. Then the system is released from rest on a level frictionless surface. The spring, which has negligible mass, is not fastened to either block, and drops to the surface after it has expanded. Block B acquires a speed of 2.20 m/s. How much potential energy was stored in the compressed spring?
#### Explanation:
This problem involves concepts from mechanics, specifically dealing with conservation of momentum and potential energy stored in a spring. Since the surface is frictionless, we can assume that no external forces act on the system other than the spring force. This allows us to use the principle of conservation of momentum and energy.
To find the potential energy stored in the compressed spring, we can follow these steps:
1. **Conservation of Momentum:**
- Before the system is released, the total momentum is 0 because both blocks are at rest.
- When the system is released, the momentum of Block A and Block B must be equal and opposite to keep the total momentum zero.
2. **Determine the Speed of Block A:**
- Let \( v_A \) be the speed of Block A after the spring releases.
- According to the conservation of momentum:
\[
m_A v_A = m_B v_B
\]
- Substituting the given values:
\[
0.8 \text{ kg} \cdot v_A = 3.0 \text{ kg} \cdot 2.20 \text{ m/s}
\]
- Solving for \( v_A \):
\[
v_A = \frac{3.0 \text{ kg} \cdot 2.20 \text{ m/s}}{0.8 \text{ kg}} = 8.25 \text{ m/s}
\]
3. **Conservation of Energy:**
- The potential energy stored in the spring is converted into the kinetic energy of both blocks.
- The total kinetic energy after the release can be given by:
\[
\frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v](https://content.bartleby.com/qna-images/question/070bdb73-a854-4da5-bed3-f81aa4a19aa3/6ffb4277-e602-4f48-b010-c9d77792615d/32fe0c5_processed.jpeg)
Transcribed Image Text:### Physics Problem: Understanding Potential Energy in a Compressed Spring
Two blocks are forced together compressing a spring \( S \) between them. Block A has a mass of 0.8 kg, and Block B has a mass of 3.0 kg. Then the system is released from rest on a level frictionless surface. The spring, which has negligible mass, is not fastened to either block, and drops to the surface after it has expanded. Block B acquires a speed of 2.20 m/s. How much potential energy was stored in the compressed spring?
#### Explanation:
This problem involves concepts from mechanics, specifically dealing with conservation of momentum and potential energy stored in a spring. Since the surface is frictionless, we can assume that no external forces act on the system other than the spring force. This allows us to use the principle of conservation of momentum and energy.
To find the potential energy stored in the compressed spring, we can follow these steps:
1. **Conservation of Momentum:**
- Before the system is released, the total momentum is 0 because both blocks are at rest.
- When the system is released, the momentum of Block A and Block B must be equal and opposite to keep the total momentum zero.
2. **Determine the Speed of Block A:**
- Let \( v_A \) be the speed of Block A after the spring releases.
- According to the conservation of momentum:
\[
m_A v_A = m_B v_B
\]
- Substituting the given values:
\[
0.8 \text{ kg} \cdot v_A = 3.0 \text{ kg} \cdot 2.20 \text{ m/s}
\]
- Solving for \( v_A \):
\[
v_A = \frac{3.0 \text{ kg} \cdot 2.20 \text{ m/s}}{0.8 \text{ kg}} = 8.25 \text{ m/s}
\]
3. **Conservation of Energy:**
- The potential energy stored in the spring is converted into the kinetic energy of both blocks.
- The total kinetic energy after the release can be given by:
\[
\frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v
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