College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![The problem involves a pulley system with two masses, \( m_1 \) and \( m_2 \), connected by a light string over a frictionless pulley. The system experiences equal and opposite accelerations due to the masses. Using Newton’s 2nd law, determine the tension in the string when \( m_1 = 10 \, \text{kg} \) and \( m_2 = 13 \, \text{kg} \).
**Diagram Explanation:**
- The diagram shows a pulley system with the following components:
- Mass \( m_1 \) (labeled as 10 \, \text{kg}) hanging on the left side.
- Mass \( m_2 \) (labeled as 13 \, \text{kg}) hanging on the right side.
- Arrows indicating gravitational force (\( mg \)) downwards on both masses.
- Tension \( T \) in the string acts upward on both masses.
**Equations and Solution Steps:**
1. Newton's second law is applied to both masses:
\[
\sum F = ma \quad \Rightarrow \quad T - m_1g = m_1a
\]
\[
\sum F = ma \quad \Rightarrow \quad m_2g - T = m_2a
\]
2. Combine the equations to solve for tension \( T \) and acceleration \( a \):
\[
T = m_1(g + a)
\]
\[
m_2g - m_1g = (m_1 + m_2)a
\]
3. Simplifying:
\[
a = \frac{(m_2 - m_1)g}{m_1 + m_2}
\]
4. Substitute values for \( m_1 \) and \( m_2 \):
\[
m_1 = 10\, \text{kg}, \quad m_2 = 13\, \text{kg}
\]
5. Calculations:
\[
T = m_1(g + a)
\]
\[
T = 1.13 \, \text{N}
\]
**Answer Options:**
- A) 0.8 N
- B) 1.1 N (correct answer](https://content.bartleby.com/qna-images/question/38e39407-372d-4edd-bcc1-3a69e099571e/a3171ff5-b1b5-4537-b5d3-c62ad94bb220/is3ljmf_processed.jpeg)
Transcribed Image Text:The problem involves a pulley system with two masses, \( m_1 \) and \( m_2 \), connected by a light string over a frictionless pulley. The system experiences equal and opposite accelerations due to the masses. Using Newton’s 2nd law, determine the tension in the string when \( m_1 = 10 \, \text{kg} \) and \( m_2 = 13 \, \text{kg} \).
**Diagram Explanation:**
- The diagram shows a pulley system with the following components:
- Mass \( m_1 \) (labeled as 10 \, \text{kg}) hanging on the left side.
- Mass \( m_2 \) (labeled as 13 \, \text{kg}) hanging on the right side.
- Arrows indicating gravitational force (\( mg \)) downwards on both masses.
- Tension \( T \) in the string acts upward on both masses.
**Equations and Solution Steps:**
1. Newton's second law is applied to both masses:
\[
\sum F = ma \quad \Rightarrow \quad T - m_1g = m_1a
\]
\[
\sum F = ma \quad \Rightarrow \quad m_2g - T = m_2a
\]
2. Combine the equations to solve for tension \( T \) and acceleration \( a \):
\[
T = m_1(g + a)
\]
\[
m_2g - m_1g = (m_1 + m_2)a
\]
3. Simplifying:
\[
a = \frac{(m_2 - m_1)g}{m_1 + m_2}
\]
4. Substitute values for \( m_1 \) and \( m_2 \):
\[
m_1 = 10\, \text{kg}, \quad m_2 = 13\, \text{kg}
\]
5. Calculations:
\[
T = m_1(g + a)
\]
\[
T = 1.13 \, \text{N}
\]
**Answer Options:**
- A) 0.8 N
- B) 1.1 N (correct answer
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