IY A 0.25-kg block along a horizontal track has a speed of 1.5 m/s immediately before colliding with a light spring of force constant 4.6 N/m located at the end of the track. What is the spring's maximum compression if the track is frictionless? [SVF E08 05.35] 23. A. 0.2940 m D. 0.4351 m B. 0.2727 m E. 0.3497 m C. 0.3140 m F. 0.3087 m 1--1 NUTO A 2.9 kg mass is attached to a spring with constant 24_N/m and is moving horizontally on a frictionless surface. If the object has speed 1.4 m/s as it passes through the equilibrium position, then what is the maximum displacement of the object? 24. A. 0.4867_m D. 0.4348_m B. 0.1773_m F 0.257.
IY A 0.25-kg block along a horizontal track has a speed of 1.5 m/s immediately before colliding with a light spring of force constant 4.6 N/m located at the end of the track. What is the spring's maximum compression if the track is frictionless? [SVF E08 05.35] 23. A. 0.2940 m D. 0.4351 m B. 0.2727 m E. 0.3497 m C. 0.3140 m F. 0.3087 m 1--1 NUTO A 2.9 kg mass is attached to a spring with constant 24_N/m and is moving horizontally on a frictionless surface. If the object has speed 1.4 m/s as it passes through the equilibrium position, then what is the maximum displacement of the object? 24. A. 0.4867_m D. 0.4348_m B. 0.1773_m F 0.257.
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Question
![**Physics: Spring Compression Problems**
**Question 23:**
A 0.25-kg block along a horizontal track has a speed of 1.5 m/s immediately before colliding with a light spring of force constant 4.6 N/m located at the end of the track. What is the spring’s maximum compression if the track is frictionless?
**Options:**
A. 0.2940 m
B. 0.2727 m
C. 0.3140 m
D. 0.4351 m
E. 0.3497 m
F. 0.3087 m
---
**Question 24:**
A 2.9-kg mass is attached to a spring with a constant of 24 N/m and is moving horizontally on a frictionless surface. If the object has a speed of 1.4 m/s as it passes through the equilibrium position, then what is the maximum displacement of the object?
**Options:**
A. 0.4867 m
B. 0.1773 m
C. 0.3140 m
D. 0.4348 m
E. 0.257 m
---
### Explanation and Solution Approach:
For both questions, we need to use the conservation of energy principle. The kinetic energy of the moving block or mass is converted into the potential energy stored in the spring when the block or mass compresses the spring to its maximum extent.
**Formula:**
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]
where,
- \( m \) is the mass of the block,
- \( v \) is the speed of the block,
- \( k \) is the spring constant,
- \( x \) is the maximum compression of the spring (which we are solving for).
Solving for \( x \):
\[ x = \sqrt{\frac{m v^2}{k}} \]
For detailed step-by-step calculations, input the given values into the formula accordingly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9169dcce-1ad4-419d-a577-e4e403df2f68%2F7c19ab30-a837-4589-99b6-fd3759e7996a%2Fkf54n9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics: Spring Compression Problems**
**Question 23:**
A 0.25-kg block along a horizontal track has a speed of 1.5 m/s immediately before colliding with a light spring of force constant 4.6 N/m located at the end of the track. What is the spring’s maximum compression if the track is frictionless?
**Options:**
A. 0.2940 m
B. 0.2727 m
C. 0.3140 m
D. 0.4351 m
E. 0.3497 m
F. 0.3087 m
---
**Question 24:**
A 2.9-kg mass is attached to a spring with a constant of 24 N/m and is moving horizontally on a frictionless surface. If the object has a speed of 1.4 m/s as it passes through the equilibrium position, then what is the maximum displacement of the object?
**Options:**
A. 0.4867 m
B. 0.1773 m
C. 0.3140 m
D. 0.4348 m
E. 0.257 m
---
### Explanation and Solution Approach:
For both questions, we need to use the conservation of energy principle. The kinetic energy of the moving block or mass is converted into the potential energy stored in the spring when the block or mass compresses the spring to its maximum extent.
**Formula:**
\[ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \]
where,
- \( m \) is the mass of the block,
- \( v \) is the speed of the block,
- \( k \) is the spring constant,
- \( x \) is the maximum compression of the spring (which we are solving for).
Solving for \( x \):
\[ x = \sqrt{\frac{m v^2}{k}} \]
For detailed step-by-step calculations, input the given values into the formula accordingly.
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