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 ?
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Paul W. Zitzewitz
Chapter5: Displacement And Force In Two Dimensions
Section: Chapter Questions
Problem 87A
Related questions
Topic Video
Question
![### 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
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
Glencoe Physics: Principles and Problems, Student…
Physics
ISBN:
9780078807213
Author:
Paul W. Zitzewitz
Publisher:
Glencoe/McGraw-Hill
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning