/*! 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 102 In A block with mass \(m_{1}\) s... [FREE SOLUTION] | 91Ó°ÊÓ

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In A block with mass \(m_{1}\) slides on a frictionless horizontal table. It's connected to a block with mass \(m_{2}\) by a light cord that passes through a hole in the table, with \(m_{2}\) hanging vertically. (a) What should \(m_{2}\) be so that \(m_{1}\) slides in uniform circular motion in a circle with radius \(R\) and speed \(v\) ? Start with force diagrams for each block. (b) Evaluate your result for \(R=0.50 \mathrm{~m}, v=0.85 \mathrm{~m} / \mathrm{s}\) and \(m_{1}=0.25 \mathrm{~kg}\).

Short Answer

Expert verified
The required mass \( m_2 \) is approximately 0.0369 kg.

Step by step solution

01

Analyze Forces on Block 1

For block 1, which is sliding on the table, the tension in the cord provides the centripetal force needed for uniform circular motion. Therefore, the tension in the cord is the centripetal force: \( T = \frac{m_{1} v^{2}}{R} \).
02

Analyze Forces on Block 2

For block 2, the gravitational force acting on it is equal to the weight \( m_2 g \). This force causes the tension in the cord because it is hanging vertically. Thus, \( T = m_{2} g \).
03

Equate Tensions to Find m2

Set the expressions for tension from blocks 1 and 2 equal to each other: \( \frac{m_{1} v^{2}}{R} = m_{2} g \). Solve for \( m_2 \): \( m_{2} = \frac{m_{1} v^{2}}{R g} \).
04

Evaluate for Given Values

Substitute the values \( R = 0.50 \text{ m} \), \( v = 0.85 \text{ m/s} \), \( m_{1} = 0.25 \text{ kg} \), and \( g = 9.8 \text{ m/s}^2 \) into the formula: \( m_{2} = \frac{0.25 \times 0.85^2}{0.50 \times 9.8} \).
05

Compute the Result

Calculate \( m_2 \): First, calculate the numerator \( m_1 v^2 = 0.25 \times 0.7225 = 0.180625 \). Then, divide by the denominator \( Rg = 0.50 \times 9.8 = 4.9 \). Thus, \( m_2 = \frac{0.180625}{4.9} \approx 0.0369 \text{ kg} \).

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Key Concepts

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

Forces and Motion
Understanding forces and motion is crucial in solving problems involving objects in motion, such as the blocks in this exercise. In any physical scenario, forces are interactions that can cause changes in an object's motion. Forces are vectors, which means they have both a magnitude and a direction.

Forces in this exercise come from two main sources:
  • **Tension**: The force within the cord connecting the two blocks.
  • **Gravitational Force**: The pull of Earth's gravity on mass \( m_2 \).
When analyzing forces, it's common to use free-body diagrams to visualize them. These diagrams help clarify how forces interact with objects, making it easier to compute resulting motions.

In essence, understanding forces and motion is about recognizing how different forces interact to influence the movement of objects. For our problem, these interactions lead to one block causing the other to move in a circle.
Uniform Circular Motion
Uniform circular motion describes an object traveling in a circular path at a constant speed. This type of motion is fascinating because even though the speed is constant, the velocity isn't. Since velocity is a vector including direction, any change in direction (even at constant speed) means a change in velocity.

In our problem, block \( m_1 \) undergoes uniform circular motion on the frictionless table. The tension in the cord provides the necessary centripetal force to maintain this motion. The centripetal force is given by:\[ T = \frac{m_1 v^2}{R} \]where:
  • \( m_1 \) is the mass.
  • \( v \) is the speed.
  • \( R \) is the radius of the circle.
Centripetal force acts inward, towards the center of the circle, keeping the object in its circular path.

Remember, while the speed is constant, the direction change means acceleration is at play, specifically centripetal acceleration, which aligns with the force inward.
Tension and Gravitational Forces
Tension and gravitational forces play significant roles in mechanics, especially in problems like this one. Let's explore both to understand their impact better.

### Tension ForcesTension is the force conducted along the cord or rope in use. In this scenario, it acts to counter the gravitational pull and provide necessary centripetal force to block \( m_1 \). Tension forces can be decomposed from other forces applied to cords or ropes in a system, as seen in the equation:\[ T = \frac{m_1 v^2}{R} \]

### Gravitational ForcesThese are the forces of attraction between Earth and any mass, seen here as acting on block \( m_2 \). The equation for gravitational force is simple:\[ F_g = m_2 g \]where:
  • \( m_2 \) is the mass hanging.
  • \( g \) is the acceleration due to gravity (approx. \( 9.8 \text{ m/s}^2 \)).
In this setup, the gravitational force on \( m_2 \) creates tension in the cord, influencing the motion of both blocks.

Combining these forces sets the stage for dynamic balance, determining how these forces interact allows us to derive essential relationships for motion and tension throughout the system.

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