Two workers are sliding 460 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 250 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor? VE ΑΣΦ pk = www ?
Two workers are sliding 460 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 250 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor? VE ΑΣΦ pk = www ?
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![### Physics Problem: Coefficient of Kinetic Friction
**Problem Statement:**
Two workers are sliding a 460 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 250 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
**Solution Space:**
- **Given Data:**
- Mass of the crate (m): 460 kg
- Force applied by the first worker (F₁): 450 N
- Force applied by the second worker (F₂): 250 N
- The crate moves with a constant speed, implying net force is zero (F_net = 0).
- **Equations:**
The forces acting horizontally cancel out with the force of kinetic friction (F_friction).
Since the speed is constant, the sum of the forces applied by both workers equals the frictional force:
\[ F_{total} = F₁ + F₂ \]
\[ F_{total} = 450\,N + 250\,N = 700\,N \]
The force of kinetic friction \( F_{friction} \) is also given by:
\[ F_{friction} = \mu_k \cdot N \]
where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.
The normal force \( N \) in this case is equal to the gravitational force acting on the crate:
\[ N = m \cdot g \]
where
\( g \approx 9.8\,m/s^2 \),
Thus,
\[ N = 460\,kg \cdot 9.8\,m/s^2 = 4508\,N \]
Combining these, we have:
\[ 700\,N = \mu_k \cdot 4508\,N \]
\[ \mu_k = \frac{700\,N}{4508\,N} \]
\[ \mu_k \approx 0.155 \]
**Answer:**
\[
\mu_k = 0.155
\]
**Detailed Explanation:**
This problem demonstrates how to determine the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf86c848-962e-42b8-b4eb-ec16b886dd8d%2F763d82ed-a7f3-4d32-8743-c85e82ad308d%2Fi0c0ay4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Coefficient of Kinetic Friction
**Problem Statement:**
Two workers are sliding a 460 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 250 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
**Solution Space:**
- **Given Data:**
- Mass of the crate (m): 460 kg
- Force applied by the first worker (F₁): 450 N
- Force applied by the second worker (F₂): 250 N
- The crate moves with a constant speed, implying net force is zero (F_net = 0).
- **Equations:**
The forces acting horizontally cancel out with the force of kinetic friction (F_friction).
Since the speed is constant, the sum of the forces applied by both workers equals the frictional force:
\[ F_{total} = F₁ + F₂ \]
\[ F_{total} = 450\,N + 250\,N = 700\,N \]
The force of kinetic friction \( F_{friction} \) is also given by:
\[ F_{friction} = \mu_k \cdot N \]
where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.
The normal force \( N \) in this case is equal to the gravitational force acting on the crate:
\[ N = m \cdot g \]
where
\( g \approx 9.8\,m/s^2 \),
Thus,
\[ N = 460\,kg \cdot 9.8\,m/s^2 = 4508\,N \]
Combining these, we have:
\[ 700\,N = \mu_k \cdot 4508\,N \]
\[ \mu_k = \frac{700\,N}{4508\,N} \]
\[ \mu_k \approx 0.155 \]
**Answer:**
\[
\mu_k = 0.155
\]
**Detailed Explanation:**
This problem demonstrates how to determine the
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