/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 A block \(A\) of mass \(2 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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A block \(A\) of mass \(2 \mathrm{~kg}\) is placed over another block \(B\) of mass \(4 \mathrm{~kg}\) which is placed over a smooth horizontal ffoor. The coefficient of friction between \(A\) and \(B\) is \(0.4\). When a horizontal force of magnitude \(10 \mathrm{~N}\) is applied on \(A\), the acceleration of blocks \(A\) and \(B\) are a. \(1 \mathrm{~ms}^{-2}\) and \(2 \mathrm{~ms}^{-2}\), respectively. b. \(5 \mathrm{~ms}^{-2}\) and \(2.5 \mathrm{~ms}^{-2}\), respectively. c. Both the blocks will moves together with acceleration \(1 / 3 \mathrm{~ms}^{-2}\) d. Both the blocks will move together with acceleration \(5 / 3 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
d. Both blocks will move together with acceleration \(5/3 \text{ ms}^{-2}\).

Step by step solution

01

Analyze the Forces on Block A

First, identify the forces acting on block A. These include the applied force of magnitude 10 N and the frictional force due to block B. Since the maximum static friction between A and B is required to keep block A from slipping over block B, calculate this frictional force using the formula for friction: \[ f_{ ext{friction}} = ext{coefficient of friction} \times ext{normal force} \]. Given the mass of block A is 2 kg, the normal force is equal to its weight, \(2 \times 9.8 = 19.6\) N. The maximum static friction is thus \(0.4 \times 19.6 = 7.84\) N.
02

Determine Acceleration If Blocks Move Together

For blocks A and B to move together, the applied force must be less than or equal to the maximum static friction. The net force available to accelerate both blocks is the applied force minus any internal frictional force constraints. Since 10 N is greater than 7.84 N, blocks A and B cannot move together without slipping. If they were to move together, calculate the combined acceleration by using the formula: \[ F = m_{ ext{total}} \times a \], where \( m_{ ext{total}} = m_A + m_B = 2 + 4 = 6 \) kg. This would yield an acceleration: \[ a = \frac{10}{6} = \frac{5}{3} \text{ ms}^{-2} \]. This only applies if they move together without slipping.
03

Resolve Movement of Block A and Block B

Since the applied force exceeds the maximum static frictional force, block A will slip over block B. The acceleration of block A is determined by subtracting the maximum frictional force from the applied force and using Newton's second law: \[ F_{ ext{net on A}} = 10 - 7.84 = 2.16 \text{ N} \] \[ a_A = \frac{2.16}{2} = 1.08 \text{ ms}^{-2} \]. Block B will only move due to the friction force acting on it, hence: \[ a_B = \frac{7.84}{4} = 1.96 \text{ ms}^{-2} \]. However, these calculations show that initially, the system might appear to allow them to move with \(5/3 \text{ ms}^{-2} \) before slipping prevails.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Laws of Motion
Understanding the interaction between forces and motion is essential in mechanics. Newton’s Laws of Motion provide the framework for analyzing this relationship.

1. **First Law (Law of Inertia):** An object at rest stays at rest, and an object in motion remains in motion at a constant velocity unless acted upon by a net external force. This law explains why static objects don't move until a force is applied and why kinetic motion stays constant unless external forces intervene.

2. **Second Law (Law of Acceleration):** The acceleration of an object is directly proportional to the net external force acting on it and inversely proportional to its mass. It can be expressed as: \[ F = m \cdot a \] where \( F \) is the net force applied, \( m \) is the mass of the object, and \( a \) is its acceleration. This is crucial for calculating how forces affect the motion of an object.

3. **Third Law (Action and Reaction):** For every action, there is an equal and opposite reaction. This principle explains interactions such as friction where forces work between surfaces in contact.

By utilizing these laws, we can predict the behavior of objects under various forces, which we apply to solve mechanics problems as shown in the given exercise.
Static and Kinetic Friction
Friction is the resistive force that acts between two surfaces in contact. It can significantly impact the movement of objects. In the exercise:
  • **Static Friction** acts between two stationary objects to prevent relative motion. It has a threshold, known as maximum static friction, which must be overcome to start motion. In our problem, this threshold is calculated as \( 7.84 \, \text{N} \).
  • **Kinetic Friction** takes over once the objects start moving relative to each other, typically less than static friction. It's responsible for maintaining movement but not covered explicitly in our problem as only initial motion was considered.
Understanding these forces involves:
  • **Formula Application:** The frictional force is calculated using the equation: \[ f_{friction} = \mu \times \text{normal force} \] where \( \mu \) is the coefficient of friction, and the normal force is usually the object's weight supported by the surface.
Static and kinetic frictions are essential concepts in physics to understand how objects resist and undergo motion under different conditions.
Acceleration Calculations
Acceleration is the rate of change of velocity of an object. It is a vital concept when evaluating how forces influence motion. Here is how it applies to the exercise:

To find the acceleration of each block:
  • First, determine the net force acting on each object. For block \( A \), we subtract the maximum static friction from the applied force, which results in a net force of \( 2.16 \, \text{N} \). This is then applied using Newton's second law as: \[ a_A = \frac{F_{net}}{m_A} = \frac{2.16}{2} \approx 1.08 \, \text{ms}^{-2} \]
  • For block \( B \), its acceleration is purely due to the frictional force exerted by block \( A \): \[ a_B = \frac{f_{friction}}{m_B} = \frac{7.84}{4} \approx 1.96 \, \text{ms}^{-2} \]
Calculations like these show how forces translate into motions, revealing block interactions and accelerations under different conditions in mechanics problems. By understanding how to apply equations, students can tackle various dynamic scenarios effectively.

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Most popular questions from this chapter

A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a uniform angular speed \(2 \mathrm{rad} / \mathrm{s}\). The radius of the disc is \(50 \mathrm{~cm}\). Find the minimum coefficient of friction between disc and coin so that the coin ?oes not slip \(\left(g=10 \mathrm{~ms}^{-2}\right)\). a. \(0.1\) b. \(0.2\) c. \(0.3\) d. \(0.4\)

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There is a chain of length \(6 \mathrm{~m}\) and coefficient of friction \(\frac{1}{2}\). What will be the maximum length of chain which can be held outside of table without sliding a. \(2 \mathrm{~m}\) b. \(4 \mathrm{~m}\) c. \(3 \mathrm{~m}\) d. \(1 \mathrm{~m}\)

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