---
title: "Three blocks of masses \\(m_1\\), \\(m_2\\), and \\(m_3\\) are connected in series on a rough horizontal table by two taut, light cords: Cord 1 connects Block 1 to Block 2, and Cord 2 connects Block 2 to Block 3.  A constant horizontal pulling force of magnitude \\(F\\) is applied to Block 3 to the right, causing the entire assembly to accelerate across the table, where the coefficient of kinetic friction between each block and the table is a uniform value \\(\\mu\\).  A student analyzes the assembly to determine the tension \\(T_2\\) in Cord 2.  Which of the following correctly gives \\(T_2\\) in terms of the given quantities, and explains why analyzing the two-block subsystem of Blocks 1 and 2 allows \\(T_2\\) to be determined without calculating the tension \\(T_1\\) in Cord 1?"
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url: "https://nerd-notes.com/ubq/124168/"
date_modified: "2026-09-28T13:30:52+00:00"
---

# Three blocks of masses \(m_1\), \(m_2\), and \(m_3\) are connected in series on a rough horizontal table by two taut, light cords: Cord 1 connects Block 1 to Block 2, and Cord 2 connects Block 2 to Block 3.

A constant horizontal pulling force of magnitude \(F\) is applied to Block 3 to the right, causing the entire assembly to accelerate across the table, where the coefficient of kinetic friction between each block and the table is a uniform value \(\mu\).

A student analyzes the assembly to determine the tension \(T_2\) in Cord 2.

Which of the following correctly gives \(T_2\) in terms of the given quantities, and explains why analyzing the two-block subsystem of Blocks 1 and 2 allows \(T_2\) to be determined without calculating the tension \(T_1\) in Cord 1?

Three blocks of masses \(m_1\), \(m_2\), and \(m_3\) are connected in series on a rough horizontal table by two taut, light cords: Cord 1 connects Block 1 to Block 2, and Cord 2 connects Block 2 to Block 3.

A constant horizontal pulling force of magnitude \(F\) is applied to Block 3 to the right, causing the entire assembly to accelerate across the table, where the coefficient of kinetic friction between each block and the table is a uniform value \(\mu\).

A student analyzes the assembly to determine the tension \(T_2\) in Cord 2.

Which of the following correctly gives \(T_2\) in terms of the given quantities, and explains why analyzing the two-block subsystem of Blocks 1 and 2 allows \(T_2\) to be determined without calculating the tension \(T_1\) in Cord 1?

![A horizontal line represents a tabletop. Resting on this line from left to right are three rectangular blocks labeled m_1, m_2, and m_3, with dimensions indicating m_1 is smallest, m_2 is intermediate, and m_3 is largest. A horizontal line segment labeled Cord 1 connects the right edge of m_1 to the left edge of m_2. A second horizontal line segment labeled Cord 2 connects the right edge of m_2 to the left edge of m_3. A single horizontal arrow points to the right from the right edge of m_3 and is labeled F. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790602252-PnIira.jpg)

- **A.** \(T_2 = \left(\dfrac{m_1 + m_2}{m_1 + m_2 + m_3}\right)\left(F - \mu(m_1 + m_2 + m_3)g\right)\); Cord 1 exerts an external backward force on Block 2, so \(T_1\) must be subtracted from \(T_2\) in the subsystem equation of motion.
- **B.** \(T_2 = \left(\dfrac{m_1 + m_2}{m_1 + m_2 + m_3}\right)F - \mu(m_1 + m_2)g\); Cord 1 exerts internal forces that cancel, but the friction on Blocks 1 and 2 directly reduces the tension transmitted through Cord 2.
- **C.** \(T_2 = \left(\dfrac{m_2}{m_1 + m_2 + m_3}\right)F\); Cord 2 is attached directly to Block 2, so it only accelerates \(m_2\), while \(T_1\) independently accelerates \(m_1\).
- **D.** \(T_2 = \left(\dfrac{m_1 + m_2}{m_1 + m_2 + m_3}\right)F\); Cord 1 exerts internal action-reaction forces that sum to zero for the two-block subsystem, and uniform friction decelerates all blocks equally so \(T_2\) is independent of \(\mu\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124168/*