Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Identify the force present and explain whether work is done when (a) a positively charged particle moves in a circle at a fixed distance from a negatively charged particle; (b) an iron nail is pulled off a magnet.

Short Answer

Expert verified
In scenario A, the force acting on the positively charged particle is the electrostatic force, which is attractive and directed towards the negatively charged particle. No work is done because the angle between the force and displacement is 90 degrees, resulting in a cosine of 0. In scenario B, the force acting on the iron nail is the magnetic force, which is attractive and directed towards the magnet. Work is done because the force and displacement are in the same direction, resulting in a cosine of 1.

Step by step solution

01

(Scenario A: Identify the Force)

In the first case, a positively charged particle moves in a circle at a fixed distance from a negatively charged particle. The force acting on the positively charged particle is the electrostatic force between the two charged particles. This force is attractive and always directed towards the negatively charged particle.
02

(Scenario A: Work Done)

For work to be done, there must be a component of the force acting in the direction of the particle's displacement. According to the definition of work, \( W = F \cdot d \cdot cos(\theta) \), where \(W\) is work, \(F\) is force, \(d\) is the displacement, and \(\theta\) is the angle between force and displacement. In a circular path, the displacement is always tangent to the circle, and the electrostatic force is always directed towards the negatively charged particle (i.e., towards the center of the circle). Thus, the angle between the force and displacement is 90 degrees (\(\theta = 90^\circ\)). The cosine of 90 degrees is 0, so the work done is 0. In this case, no work is done as the positively charged particle moves in a circle around the negatively charged particle.
03

(Scenario B: Identify the Force)

In the second case, an iron nail is pulled off a magnet. The force acting on the nail, in this case, is the magnetic force from the magnet. This force is attractive and acts towards the magnet.
04

(Scenario B: Work Done)

To determine if work is done, we need to check if there is a component of the magnetic force acting in the direction of displacement. In this case, the force and displacement are in the same direction (opposite to the magnet's attraction). The angle between the force and displacement is 0 degrees (\(\theta = 0^\circ\)). The cosine of 0 degrees is 1, and according to the work formula \( W = F \cdot d \cdot cos(\theta) \), the work done will be positive as both the force and displacement are in the same direction. So, in this case, work is done when an iron nail is pulled off a magnet.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Electrostatic Force
When exploring the wonders of physics, electrostatic force emerges as a fundamental interaction between objects with charge. Imagine two people on opposite ends of a party holding a rope; if one pulls, the other feels the tug. Similarly, charged particles exert forces on one another, attracting or repelling based on whether their charges are opposite or identical, respectively.

Such forces are at play when, for instance, a positively charged particle is locked in a dance around a negatively charged partner. The force that keeps it tethered, ensuring the circular motion, is the same electrostatic force; a persistent pull towards the negatively charged center. It’s this celestial tango that showcases one of nature's four fundamental interactions, keeping the structure of atoms to the pathways of charged particles in a cyclotron.
Magnetic Force
Imagine the magnetic force as an invisible artist, capable of orchestrating the path of certain materials like iron, cobalt, or nickel. This force is what you encounter when playing with magnets; it's what allows them to stick to your fridge or couple together in your hands.

Evident when pulling an iron nail from a magnet's embrace, the magnetic force is behind the invisible pull you feel. This force exhibits properties similar to the electrostatic force, yet is distinct in its interaction with materials and the way it's generated—often by moving electric charges or inherent in the material properties of magnets. Demonstrating this in practice offers insight into electromagnetism, an essential element of both classical and modern physics.
Circular Motion
Circular motion unfolds splendidly in nature and technology, from the electrons whirling around the nucleus to the rides at amusement parks. It's the kind of motion where any object travels along the perimeter of a circle or a path that can be curved into a circular shape.

At the heart of circular motion lies a key requirement: there must be a force that continuously acts perpendicular to the direction of the moving object's velocity. This is often called the centripetal force, which in our mentioned scenario of charged particles, is provided by the electrostatic attraction. Understanding the principles of circular motion illuminates ideas across vastly different scales, from particles in an atom to planets orbiting stars.
Work-Energy Principle
The work-energy principle is a cornerstone of physics, bridging the gap between forces acting on objects and the energy changes resulting from their motion. Often stated simply as 'work done equals change in energy,' it is a powerful way of understanding why and how objects move.

In essence, if you push a book across a table, you are doing work on it, and you're imparting energy to it. The principles at play highlight a harmonious relationship: when work is done on an object, energy is transferred, transforming into kinetic or potential energy. This principle underpins energy conservation and offers a lens to predict the outcome of physical scenarios, like a roller coaster's thrilling dive or an arrow launched from a bow.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

From the following data for three prospective fuels, calculate which could provide the most energy per unit volume:

Identify the force present and explain whether work is being performed in the following cases: (a) You lift a pencil off the top of a desk. (b) A spring is compressed to half its normal length.

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{aligned} $$

Using values from Appendix \(\mathrm{C},\) calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

You are given \(\Delta H\) for a process that occurs at constant pressure. What additional information do you need to determine \(\Delta E\) for the process?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free