The 15 kg block A slides on the surface for which coefficient of kinetic friction is 0.3. The block has a velocity of 11 m/s when it is a distance, s, that is 4 m from the 10 kg block B. The unstretched spring has a stiffness of 1000 N/m. The coefficient of restitution is 0.6. The coefficient of friction is the same for both blocks. Determine the maximum compression of the spring due to the collision. WWW A B

Elements Of Electromagnetics
7th Edition
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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### Problem Statement

A 15 kg block A slides on a surface with a coefficient of kinetic friction of 0.3. The block has a velocity of 11 m/s when it is at a distance \( s \) of 4 meters from a 10 kg block B. The unstretched spring has a stiffness of 1000 N/m. The coefficient of restitution is 0.6. The coefficient of friction is the same for both blocks. Determine the maximum compression of the spring due to the collision.

### Diagram Explanation

- **Blocks and Surface**: There are two blocks labeled A and B positioned on a flat surface with block B being connected to a fixed wall by a spring with stiffness \( k \).
- **Block A**: 
  - Mass: 15 kg
  - Initial Velocity: 11 m/s
  - Distance \( s \): 4 meters from block B
- **Block B**: 
  - Mass: 10 kg
- **Spring**:
  - Stiffness \( k \): 1000 N/m
  - Initially unstretched
- **Friction**: 
  - Coefficient of kinetic friction: 0.3
  - Acts on both block A and block B equally
- **Coefficient of Restitution**: 0.6

### Tasks

- Determine the maximum compression of the spring due to the collision.

### Detailed Explanation

First, we need to calculate the initial kinetic energy of block A as it slides towards block B. We will account for the energy lost due to friction over the distance \( s \). Then, we will use the principles of conservation of momentum and energy to analyze the collision between the blocks and the spring compression.

#### Step-by-Step Solution

1. **Initial Kinetic Energy of Block A**:
   \[
   KE_{A, \text{initial}} = \frac{1}{2} m_A v^2 = \frac{1}{2} \times 15 \, \text{kg} \times (11 \,\text{m/s})^2 = 907.5 \,\text{J}
   \]

2. **Work Done by Friction on Block A**:
   \[
   W_{\text{friction}} = \mu \cdot m_A \cdot g \cdot s = 0.3 \cdot 15 \,\text{kg
Transcribed Image Text:### Problem Statement A 15 kg block A slides on a surface with a coefficient of kinetic friction of 0.3. The block has a velocity of 11 m/s when it is at a distance \( s \) of 4 meters from a 10 kg block B. The unstretched spring has a stiffness of 1000 N/m. The coefficient of restitution is 0.6. The coefficient of friction is the same for both blocks. Determine the maximum compression of the spring due to the collision. ### Diagram Explanation - **Blocks and Surface**: There are two blocks labeled A and B positioned on a flat surface with block B being connected to a fixed wall by a spring with stiffness \( k \). - **Block A**: - Mass: 15 kg - Initial Velocity: 11 m/s - Distance \( s \): 4 meters from block B - **Block B**: - Mass: 10 kg - **Spring**: - Stiffness \( k \): 1000 N/m - Initially unstretched - **Friction**: - Coefficient of kinetic friction: 0.3 - Acts on both block A and block B equally - **Coefficient of Restitution**: 0.6 ### Tasks - Determine the maximum compression of the spring due to the collision. ### Detailed Explanation First, we need to calculate the initial kinetic energy of block A as it slides towards block B. We will account for the energy lost due to friction over the distance \( s \). Then, we will use the principles of conservation of momentum and energy to analyze the collision between the blocks and the spring compression. #### Step-by-Step Solution 1. **Initial Kinetic Energy of Block A**: \[ KE_{A, \text{initial}} = \frac{1}{2} m_A v^2 = \frac{1}{2} \times 15 \, \text{kg} \times (11 \,\text{m/s})^2 = 907.5 \,\text{J} \] 2. **Work Done by Friction on Block A**: \[ W_{\text{friction}} = \mu \cdot m_A \cdot g \cdot s = 0.3 \cdot 15 \,\text{kg
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