Problem 14 A 15 kg box is initially sliding across a frictionless floor with a constant speed of 2.5 m/s. Then, the box encounters a rough surface and travels for 12 meters before coming to a stop. What is the coefficient of kinetic friction between the box and rough surface?
Problem 14 A 15 kg box is initially sliding across a frictionless floor with a constant speed of 2.5 m/s. Then, the box encounters a rough surface and travels for 12 meters before coming to a stop. What is the coefficient of kinetic friction between the box and rough surface?
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![### Problem 14
A 15 kg box is initially sliding across a frictionless floor with a constant speed of 2.5 m/s. Then, the box encounters a rough surface and travels for 12 meters before coming to a stop. What is the coefficient of kinetic friction between the box and rough surface?
### Solution Explanation
1. **Given**:
- Mass of the box, \( m = 15 \, \text{kg} \)
- Initial velocity, \( v_i = 2.5 \, \text{m/s} \)
- Final velocity, \( v_f = 0 \, \text{m/s} \) (the box stops)
- Distance traveled on rough surface, \( d = 12 \, \text{m} \)
2. **Looking for**:
- The coefficient of kinetic friction, \( \mu_k \)
3. **Using the equation of motion**:
\[
v_f^2 = v_i^2 + 2a d
\]
Rearrange to find acceleration \( a \):
\[
0 = (2.5)^2 + 2a \cdot 12
\]
4. **Solve for acceleration** \( a \):
\[
-6.25 = 24a
\]
\[
a = -\frac{6.25}{24}
\]
5. **Using frictional force equation**:
- The frictional force \( F_f = \mu_k \cdot N \) where \( N = m \cdot g \) (normal force)
- Since \( F_f = m \cdot a \), substitute:
\[
\mu_k \cdot (15 \cdot 9.8) = 15 \cdot a
\]
Solve for \( \mu_k \):
\[
\mu_k = -\frac{a}{9.8}
\]
By solving these equations, the coefficient of kinetic friction \( \mu_k \) can be determined.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa68f721b-c195-47eb-a86c-5a0471e1fa0b%2F862d3d3c-e3b1-4407-b03b-e6a6ccb26681%2Fdpghsw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 14
A 15 kg box is initially sliding across a frictionless floor with a constant speed of 2.5 m/s. Then, the box encounters a rough surface and travels for 12 meters before coming to a stop. What is the coefficient of kinetic friction between the box and rough surface?
### Solution Explanation
1. **Given**:
- Mass of the box, \( m = 15 \, \text{kg} \)
- Initial velocity, \( v_i = 2.5 \, \text{m/s} \)
- Final velocity, \( v_f = 0 \, \text{m/s} \) (the box stops)
- Distance traveled on rough surface, \( d = 12 \, \text{m} \)
2. **Looking for**:
- The coefficient of kinetic friction, \( \mu_k \)
3. **Using the equation of motion**:
\[
v_f^2 = v_i^2 + 2a d
\]
Rearrange to find acceleration \( a \):
\[
0 = (2.5)^2 + 2a \cdot 12
\]
4. **Solve for acceleration** \( a \):
\[
-6.25 = 24a
\]
\[
a = -\frac{6.25}{24}
\]
5. **Using frictional force equation**:
- The frictional force \( F_f = \mu_k \cdot N \) where \( N = m \cdot g \) (normal force)
- Since \( F_f = m \cdot a \), substitute:
\[
\mu_k \cdot (15 \cdot 9.8) = 15 \cdot a
\]
Solve for \( \mu_k \):
\[
\mu_k = -\frac{a}{9.8}
\]
By solving these equations, the coefficient of kinetic friction \( \mu_k \) can be determined.
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