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Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 07: Newton's Third LawProblema 25
Capitolo 7, Problema 25

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of 6.0 kg, accelerates downward at (3/4)g. What is the mass of the other block?

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Step 1: Begin by identifying the forces acting on each block. For the 6.0 kg block, the forces are its weight (gravitational force) and the tension in the rope. The downward acceleration is given as (3/4)g, where g is the acceleration due to gravity.
Step 2: Write the equation of motion for the 6.0 kg block using Newton's second law: \( F = ma \). The net force acting on the block is \( F_{net} = m_1 g - T \), where \( m_1 \) is the mass of the block, \( g \) is the acceleration due to gravity, and \( T \) is the tension in the rope. Substitute the acceleration \( a = \frac{3}{4}g \) into the equation: \( m_1 g - T = m_1 \cdot \frac{3}{4}g \).
Step 3: Solve for the tension \( T \) in the rope from the equation derived in Step 2. Rearrange the equation: \( T = m_1 g - m_1 \cdot \frac{3}{4}g \). Simplify to find \( T = \frac{1}{4}m_1 g \).
Step 4: Write the equation of motion for the second block (unknown mass \( m_2 \)) using Newton's second law. Since this block is accelerating upward at \( \frac{3}{4}g \), the net force is \( T - m_2 g = m_2 \cdot \frac{3}{4}g \). Substitute \( T = \frac{1}{4}m_1 g \) into this equation: \( \frac{1}{4}m_1 g - m_2 g = m_2 \cdot \frac{3}{4}g \).
Step 5: Solve for \( m_2 \) by rearranging the equation from Step 4. Combine like terms and isolate \( m_2 \): \( \frac{1}{4}m_1 g = m_2 g + m_2 \cdot \frac{3}{4}g \). Factor out \( m_2 \) and simplify: \( m_2 = \frac{1}{4}m_1 \). Substitute \( m_1 = 6.0 \, \text{kg} \) to find the mass of the second block.

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Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, the forces acting on the blocks must be analyzed to determine the mass of the second block.
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Intro to Forces & Newton's Second Law

Tension in a Rope

In a system involving a pulley and a rope, tension is the force transmitted through the rope when it is pulled tight by forces acting at each end. Since the pulley is massless and frictionless, the tension is the same on both sides of the rope. Understanding how tension relates to the weights of the blocks and their acceleration is crucial for solving the problem.
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Calculating Tension in a Pendulum with Energy Conservation

Acceleration due to Gravity

Acceleration due to gravity (g) is the acceleration experienced by an object when it is in free fall near the Earth's surface, approximately 9.81 m/s². In this problem, the block's acceleration is given as 3/4g, which indicates that it is accelerating downward at a rate less than free fall. This value is essential for calculating the net force acting on the blocks and ultimately finding the mass of the second block.
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Acceleration Due to Gravity
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