A block of mass 2 kg is moving to the right with velocity V= 25 m/s. The block then compresses the spring until it finally stops. The spring is compressed by an amount of A X= 0.5 m. Use conservation of energy to find the spring constant.
A block of mass 2 kg is moving to the right with velocity V= 25 m/s. The block then compresses the spring until it finally stops. The spring is compressed by an amount of A X= 0.5 m. Use conservation of energy to find the spring constant.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![## Problem Statement
A block of mass 2 kg is moving to the right with a velocity \( V = 25 \, \text{m/s} \). The block then compresses the spring until it finally stops. The spring is compressed by an amount of \( \Delta X = 0.5 \, \text{m} \). Use conservation of energy to find the spring constant.
## Graph/Diagram
The image shows a rectangle (representing the block) on the left side, moving to the right. On the right side, there is a compressed spring that appears to stop the block. The spring is shown in its compressed state with a line indicating the point where the block comes to rest.
**Analysis Approach:**
1. **Initial Kinetic Energy of the Block**:
\[
KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \, \text{kg} \times (25 \, \text{m/s})^2 = 625 \, \text{J}
\]
2. **Final Potential Energy in the Spring**:
\[
PE_f = \frac{1}{2} k (\Delta X)^2
\]
3. **Conservation of Energy**:
The initial kinetic energy of the block is converted completely into potential energy of the spring:
\[
KE_i = PE_f \Rightarrow 625 \, \text{J} = \frac{1}{2} k (0.5 \, \text{m})^2
\]
4. **Solving for the Spring Constant \( k \)**:
Rearrange and solve for \( k \):
\[
625 = \frac{1}{2} k \times 0.25
\]
\[
625 = 0.125k
\]
\[
k = \frac{625}{0.125} = 5000 \, \text{N/m}
\]
In summary, using conservation of energy principles, the spring constant \( k \) is calculated to be 5000 N/m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa15ffa4-06ff-402c-ae2f-dda95408fa6d%2F3bc54b2a-f64b-4f88-bcb7-f3eef6bb5b38%2F0p16kkh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement
A block of mass 2 kg is moving to the right with a velocity \( V = 25 \, \text{m/s} \). The block then compresses the spring until it finally stops. The spring is compressed by an amount of \( \Delta X = 0.5 \, \text{m} \). Use conservation of energy to find the spring constant.
## Graph/Diagram
The image shows a rectangle (representing the block) on the left side, moving to the right. On the right side, there is a compressed spring that appears to stop the block. The spring is shown in its compressed state with a line indicating the point where the block comes to rest.
**Analysis Approach:**
1. **Initial Kinetic Energy of the Block**:
\[
KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \, \text{kg} \times (25 \, \text{m/s})^2 = 625 \, \text{J}
\]
2. **Final Potential Energy in the Spring**:
\[
PE_f = \frac{1}{2} k (\Delta X)^2
\]
3. **Conservation of Energy**:
The initial kinetic energy of the block is converted completely into potential energy of the spring:
\[
KE_i = PE_f \Rightarrow 625 \, \text{J} = \frac{1}{2} k (0.5 \, \text{m})^2
\]
4. **Solving for the Spring Constant \( k \)**:
Rearrange and solve for \( k \):
\[
625 = \frac{1}{2} k \times 0.25
\]
\[
625 = 0.125k
\]
\[
k = \frac{625}{0.125} = 5000 \, \text{N/m}
\]
In summary, using conservation of energy principles, the spring constant \( k \) is calculated to be 5000 N/m.
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