Objects with masses m₁ = 13.0 kg and m₂ = 8.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m₂ falls 1.00 m in 1.56 s, determine the coefficient of kinetic friction between m, and the table. m Step 1 The free-body diagrams of the two objects in this system are shown below. Note that the accelerations of the two objects have the same magnitude, a, with the acceleration of the object of mass m, directed horizontally to the right and the acceleration of the object of mass m₂ directed vertically downward. and, solving for a, a= My ax=+a ay=0 M L ₁18 m = 0+ m/s². T Since m₂ is observed to drop downward 1.00 m in 1.56 s when released, the magnitude of the acceleration is found by solving the following equation for ay. Ay = Voyt + aye² From the diagram, we know a, -a. Substituting the given value and paying close attention to the signs, we have +(-²)(-a) (₁) ² [ m₂ = 1₂8
Objects with masses m₁ = 13.0 kg and m₂ = 8.0 kg are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m₂ falls 1.00 m in 1.56 s, determine the coefficient of kinetic friction between m, and the table. m Step 1 The free-body diagrams of the two objects in this system are shown below. Note that the accelerations of the two objects have the same magnitude, a, with the acceleration of the object of mass m, directed horizontally to the right and the acceleration of the object of mass m₂ directed vertically downward. and, solving for a, a= My ax=+a ay=0 M L ₁18 m = 0+ m/s². T Since m₂ is observed to drop downward 1.00 m in 1.56 s when released, the magnitude of the acceleration is found by solving the following equation for ay. Ay = Voyt + aye² From the diagram, we know a, -a. Substituting the given value and paying close attention to the signs, we have +(-²)(-a) (₁) ² [ m₂ = 1₂8
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)...
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![**Physics Problem: Analyzing a Two-Mass System**
**Problem Statement:**
Objects with masses \( m_1 = 13.0 \, \text{kg} \) and \( m_2 = 8.0 \, \text{kg} \) are connected by a light string that passes over a frictionless pulley as shown in the figure. If, when the system starts from rest, \( m_2 \) falls 1.00 m in 1.56 s, determine the coefficient of kinetic friction between \( m_1 \) and the table.
---
**Step 1: Free-Body Diagrams**
The free-body diagrams of the two objects in this system are shown below. Note that the accelerations of the two objects have the same magnitude, \( a \), with the acceleration of the object of mass \( m_1 \) directed horizontally to the right and the acceleration of the object of mass \( m_2 \) directed vertically downward.
- **Diagram 1: Mass \( m_1 \)**
- Forces acting:
- \( n \) (Normal force, perpendicular to the surface)
- \( f \) (Force of friction, to the left)
- \( T \) (Tension, to the right)
- \( w_1 = m_1g \) (Weight, downward)
- Coordinates:
- \( a_x = +a \), \( a_y = 0 \)
- **Diagram 2: Mass \( m_2 \)**
- Forces acting:
- \( T \) (Tension, upward)
- \( w_2 = m_2g \) (Weight, downward)
- Coordinates:
- \( a_x = 0 \), \( a_y = -a \)
Since \( m_2 \) is observed to drop downward 1.00 m in 1.56 s when released, the magnitude of the acceleration is found by solving the following equation for \( a_y \):
\[
\Delta y = v_{0y}t + \frac{1}{2}a_yt^2
\]
From the diagram, we know \( a_y = -a \). Substituting the given value and paying close attention to the signs, we have:
\[
(1.00 \, \text{m}) = 0 + \left](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30f79184-3047-455a-b3d8-f2ad47623cdf%2F5aa9a495-f0e3-44a2-918e-5ee7602d2398%2F74uia7r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Analyzing a Two-Mass System**
**Problem Statement:**
Objects with masses \( m_1 = 13.0 \, \text{kg} \) and \( m_2 = 8.0 \, \text{kg} \) are connected by a light string that passes over a frictionless pulley as shown in the figure. If, when the system starts from rest, \( m_2 \) falls 1.00 m in 1.56 s, determine the coefficient of kinetic friction between \( m_1 \) and the table.
---
**Step 1: Free-Body Diagrams**
The free-body diagrams of the two objects in this system are shown below. Note that the accelerations of the two objects have the same magnitude, \( a \), with the acceleration of the object of mass \( m_1 \) directed horizontally to the right and the acceleration of the object of mass \( m_2 \) directed vertically downward.
- **Diagram 1: Mass \( m_1 \)**
- Forces acting:
- \( n \) (Normal force, perpendicular to the surface)
- \( f \) (Force of friction, to the left)
- \( T \) (Tension, to the right)
- \( w_1 = m_1g \) (Weight, downward)
- Coordinates:
- \( a_x = +a \), \( a_y = 0 \)
- **Diagram 2: Mass \( m_2 \)**
- Forces acting:
- \( T \) (Tension, upward)
- \( w_2 = m_2g \) (Weight, downward)
- Coordinates:
- \( a_x = 0 \), \( a_y = -a \)
Since \( m_2 \) is observed to drop downward 1.00 m in 1.56 s when released, the magnitude of the acceleration is found by solving the following equation for \( a_y \):
\[
\Delta y = v_{0y}t + \frac{1}{2}a_yt^2
\]
From the diagram, we know \( a_y = -a \). Substituting the given value and paying close attention to the signs, we have:
\[
(1.00 \, \text{m}) = 0 + \left
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