F e. A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use g-10 N/kg. What is the magnitude of the friction force? 119.631 V o units n What is the magnitude of the normal force? units n
F e. A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use g-10 N/kg. What is the magnitude of the friction force? 119.631 V o units n What is the magnitude of the normal force? units n
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
![**Problem Statement:**
A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use \( g = 10 \, \text{N/kg} \).
**Questions:**
1. What is the magnitude of the friction force?
- Answer: \( 119.631 \, \text{N} \)
2. What is the magnitude of the normal force?
- Answer: [Blank in image]
**Diagram Explanation:**
The diagram depicts a block on a horizontal surface. A force \( F \) is applied at an angle \( \theta \) (29.9 degrees) to the right, above the horizontal.
**Key Concepts:**
- The block is in equilibrium in the horizontal direction since it moves with constant velocity.
- The horizontal component of the applied force counteracts the frictional force.
- The vertical component of the force affects the normal force acting on the block.
To find the magnitudes of the forces:
- **Friction Force (\( f \)):**
The horizontal component of force \( F \) is given by \( F \cdot \cos(\theta) \).
Since the block is moving at constant velocity, friction force \( f = F \cdot \cos(\theta) = 138 \cdot \cos(29.9°) \).
- **Normal Force (\( N \)):**
The normal force is affected by both the gravitational force and the vertical component of the applied force. It can be calculated using:
\( N = mg - F \cdot \sin(\theta) \).
These calculations lead to the determination of the respective forces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18608e9e-bffe-4b38-84b4-3b337a037743%2F1332db1c-4ce7-4c3e-a687-5864ed467640%2F9q1bq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use \( g = 10 \, \text{N/kg} \).
**Questions:**
1. What is the magnitude of the friction force?
- Answer: \( 119.631 \, \text{N} \)
2. What is the magnitude of the normal force?
- Answer: [Blank in image]
**Diagram Explanation:**
The diagram depicts a block on a horizontal surface. A force \( F \) is applied at an angle \( \theta \) (29.9 degrees) to the right, above the horizontal.
**Key Concepts:**
- The block is in equilibrium in the horizontal direction since it moves with constant velocity.
- The horizontal component of the applied force counteracts the frictional force.
- The vertical component of the force affects the normal force acting on the block.
To find the magnitudes of the forces:
- **Friction Force (\( f \)):**
The horizontal component of force \( F \) is given by \( F \cdot \cos(\theta) \).
Since the block is moving at constant velocity, friction force \( f = F \cdot \cos(\theta) = 138 \cdot \cos(29.9°) \).
- **Normal Force (\( N \)):**
The normal force is affected by both the gravitational force and the vertical component of the applied force. It can be calculated using:
\( N = mg - F \cdot \sin(\theta) \).
These calculations lead to the determination of the respective forces.
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 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

College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning

University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON

Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press

College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning

University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON

Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press

Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning

Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley

College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON