A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 45 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.7 m as shown on the graph. Determine the maximum compression x of the outer springs. Deceleration m/s² 4380 2920 1460 0 x, m 45 m/s 0 1 0.7 1.4
A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 45 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.7 m as shown on the graph. Determine the maximum compression x of the outer springs. Deceleration m/s² 4380 2920 1460 0 x, m 45 m/s 0 1 0.7 1.4
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![### Damping Mechanism of a Triple-Spring Bumper System
**Problem Statement:**
A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 45 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.7 m as shown on the graph. Determine the maximum compression \( x \) of the outer springs.
**Illustration Description:**
The setup comprises a large mass approaching the bumper system at a velocity of 45 m/s. The bumper system includes an arrangement of three springs - two outer springs and a central spring. The central spring engages its deceleration effect when the compression of the system exceeds 0.7 meters.
A deceleration graph is provided:
- **X-axis (horizontal)**: Represents the compression, \( x \), in meters (m).
- **Y-axis (vertical)**: Represents the deceleration, in meters per second squared (m/s²).
**Graph Details:**
The deceleration graph is linear, with the following key points:
- At **0 to 0.7 meters compression**:
- The deceleration is constant and remains at 0 m/s².
- At **0.7 meters compression**:
- The deceleration begins to increase.
- At **1.4 meters compression**:
- The deceleration reaches approximately 4380 m/s².
The graph implies a direct proportionality between compression beyond 0.7 meters and the rate of deceleration.
**Objective:**
To find the maximum compression \( x \) of the outer springs, as experienced by the mass traveling at 45 m/s.
**Conclusion:**
The answer box is provided for students and users to input their calculated value of \( x \) (in meters), indicating the maximum compression of the outer springs as affected by the motion of the mass.
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**Answer:**
\[ x = \_\_\_\_ \text{ m} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F043ffaaf-193d-4248-b0cb-e437da87418d%2F8ca94c17-ef4c-404f-9ea2-b5e8b89b4ef4%2F3jdquqo_processed.png&w=3840&q=75)
Transcribed Image Text:### Damping Mechanism of a Triple-Spring Bumper System
**Problem Statement:**
A bumper, consisting of a nest of three springs, is used to arrest the horizontal motion of a large mass which is traveling at 45 m/s as it contacts the bumper. The two outer springs cause a deceleration proportional to the spring deformation. The center spring increases the deceleration rate when the compression exceeds 0.7 m as shown on the graph. Determine the maximum compression \( x \) of the outer springs.
**Illustration Description:**
The setup comprises a large mass approaching the bumper system at a velocity of 45 m/s. The bumper system includes an arrangement of three springs - two outer springs and a central spring. The central spring engages its deceleration effect when the compression of the system exceeds 0.7 meters.
A deceleration graph is provided:
- **X-axis (horizontal)**: Represents the compression, \( x \), in meters (m).
- **Y-axis (vertical)**: Represents the deceleration, in meters per second squared (m/s²).
**Graph Details:**
The deceleration graph is linear, with the following key points:
- At **0 to 0.7 meters compression**:
- The deceleration is constant and remains at 0 m/s².
- At **0.7 meters compression**:
- The deceleration begins to increase.
- At **1.4 meters compression**:
- The deceleration reaches approximately 4380 m/s².
The graph implies a direct proportionality between compression beyond 0.7 meters and the rate of deceleration.
**Objective:**
To find the maximum compression \( x \) of the outer springs, as experienced by the mass traveling at 45 m/s.
**Conclusion:**
The answer box is provided for students and users to input their calculated value of \( x \) (in meters), indicating the maximum compression of the outer springs as affected by the motion of the mass.
---
**Answer:**
\[ x = \_\_\_\_ \text{ m} \]
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