FIGURE P7.47 shows a 200 g hamster sitting on an 800 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs = μk = 0 is sprayed on the top surface of the block, causing the hamster to slide down. Friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?
Ch 07: Newton's Third Law
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 7, Problem 55
What is the acceleration of the 3.0 kg block in FIGURE CP7.55 across the frictionless table? Hint: Think carefully about the acceleration constraint.

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Identify the forces acting on the system: The 3.0 kg block is on a frictionless table, so there is no frictional force. The acceleration of the block is caused by the tension in the string, which is connected to another mass (not explicitly mentioned in the problem but implied by the hint). Assume the system involves a pulley and a hanging mass.
Define the acceleration constraint: Since the string is inextensible, the acceleration of the 3.0 kg block on the table must be equal in magnitude to the acceleration of the hanging mass. Let this common acceleration be denoted as \( a \).
Write Newton's second law for the 3.0 kg block: The net force on the block is the tension in the string \( T \). Using \( F = ma \), we have \( T = m_1 a \), where \( m_1 = 3.0 \; \text{kg} \).
Write Newton's second law for the hanging mass: The net force on the hanging mass is the difference between its weight \( m_2 g \) and the tension in the string \( T \). Using \( F = ma \), we have \( m_2 g - T = m_2 a \), where \( m_2 \) is the mass of the hanging block and \( g \) is the acceleration due to gravity.
Combine the equations to solve for \( a \): Substitute \( T = m_1 a \) from the first equation into the second equation. This gives \( m_2 g - m_1 a = m_2 a \). Rearrange to find \( a \): \( a = \frac{m_2 g}{m_1 + m_2} \). Plug in the known values for \( m_1 \), \( m_2 \), and \( g \) to calculate the acceleration.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 formula F = ma, where F is the net force, m is the mass, and a is the acceleration. Understanding this law is crucial for calculating the acceleration of the block when a force is applied.
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Intro to Forces & Newton's Second Law
Frictionless Surface
A frictionless surface is an idealized concept where no frictional forces oppose the motion of an object. In this scenario, the only forces acting on the block are those applied to it and its weight. This simplification allows for easier calculations of acceleration, as it eliminates the need to account for frictional forces that would otherwise slow the block down.
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Acceleration Constraint
The acceleration constraint refers to the relationship between the accelerations of connected objects in a system. In this case, if the block is part of a larger system (like being connected to another mass), the acceleration of the block may depend on the motion of that other mass. Understanding this constraint is essential for determining the correct acceleration of the block on the frictionless table.
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Intro to Acceleration
Related Practice
Textbook Question
Textbook Question
In FIGURE CP7.54, find an expression for the acceleration of m1. The pulleys are massless and frictionless. Hint: Think carefully about the acceleration constraint.
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Textbook Question
A house painter uses the chair-and-pulley arrangement of FIGURE P7.45 to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20 m/s².
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