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Acceleration of the loop when its speed is half of its terminal speed is (A) \(\frac{g}{2}\) (B) \(g e^{-2}\) (C) \(g e^{-1 / 2}\) (D) \(g e^{-4}\)

Short Answer

Expert verified
Answer: (A) \(\frac{g}{2}\)

Step by step solution

01

Understand the terminal speed concept

When an object is falling freely under gravity, it experiences a resistive force (air resistance) which opposes its motion. Initially, the gravitational force acting downwards is greater than the resistive force, so the object keeps accelerating. However, as the object's speed increases, so does the resistive force. At a certain point, the resistive force becomes equal to the gravitational force, and at that moment, the object is said to have reached its terminal speed, at which it will continue to travel without any further acceleration.
02

Write down the equations for gravitational force and resistive force

In this problem, we have a loop falling under gravity, experiencing a resistive force. 1. Gravitational force: \(F_{g} = mg\), where m is the mass of the object and g is the acceleration due to gravity. 2. Resistive force: \(F_{R} = k v\), where k is a proportionality constant and v is the velocity of the object.
03

Calculate the terminal speed

At terminal speed, the gravitational force is equal to the resistive force: \(F_{g} = F_{R}\) or \(mg = kv\). Solving for the terminal speed (v): \(v_T = \frac{mg}{k}\), where v_T is the terminal speed.
04

Write down the equations for half terminal speed

We are given that the loop's speed is half of its terminal speed. So, the loop's current speed, v, is: \(v = \frac{1}{2}v_T = \frac{1}{2} \cdot \frac{mg}{k} = \frac{mg}{2k}\).
05

Calculate the acceleration at half terminal speed

The net force acting on the loop at half terminal speed is the difference between the gravitational force and the resistive force: \(F_{net} = F_{g} - F_{R} = mg - k \cdot \frac{mg}{2k}\). Now, we know that \(F_{net} = ma\), where a is the acceleration. Solving for the acceleration (a) at half terminal speed: \(a = \frac{F_{net}}{m} = g - \frac{k}{2k} \cdot g = g \left(1 - \frac{1}{2}\right) = \frac{g}{2}\). Hence, the acceleration of the loop when its speed is half of its terminal speed is: (A) \(\frac{g}{2}\)

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental concept that often comes up in physics, especially when discussing motion under natural influences like gravity. The gravitational force acting on an object near the Earth's surface is described by the equation: \[ F_g = mg \]Here,
  • \(m\) represents the mass of the object, which indicates how much matter it contains.
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\) on Earth.
This force acts downwards towards the center of the Earth and is responsible for causing objects to fall when dropped. It's important to note that each object, regardless of its mass, feels this gravitational attraction, which is why objects fall when released. Understanding gravitational force is key to analyzing how objects move when no other forces or only resistive forces, like air resistance, are acting upon them.
Resistive Force
Resistive force is another critical concept, especially when discussing motion through a medium like air or water. Resistive force, often called air resistance or drag, acts opposite to the direction of an object's motion. It's calculated using the formula: \[ F_R = k v \]Where:
  • \(k\) is a proportionality constant that depends on various factors, such as the shape and surface texture of the moving object and the medium's density.
  • \(v\) is the velocity of the object moving through the medium.
The resistive force increases with increasing velocity. At first, when the object is moving slowly, this force is relatively small compared to the gravitational force. But as the velocity increases, the resistive force also increases until it eventually balances the gravitational force. This balance is what establishes terminal velocity, where the net force on the object becomes zero, and it stops accelerating.
Acceleration Calculation
Calculating acceleration involves understanding the relationship between net force, mass, and acceleration itself, as outlined by Newton's second law of motion. In the specific scenario of falling objects, when the speed is a fraction of the terminal speed, the force balance must be considered carefully. The net force \(F_{\text{net}}\) is the difference between the gravitational force \(F_g\) and the resistive force \(F_R\). The formula is:\[ F_{\text{net}} = F_g - F_R \]Given that \(F_{\text{net}} = ma\), we solve for acceleration \(a\) as follows:\[ a = \frac{F_{\text{net}}}{m} = g - \frac{k}{2k} \cdot g = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} \]This indicates that at half the terminal speed, the object's acceleration is \(\frac{g}{2}\). This calculation shows how the balance of forces results in a decreased acceleration as velocity approaches terminal velocity. The process illuminates not only how forces interact but also how an object's speed affects its acceleration in a resistive medium.

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Most popular questions from this chapter

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