Work Exchange: Object 1 Pushes Object 2 - 15 J Work

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Homework Help Overview

The problem involves two objects, where object one pushes object two, and the original poster questions whether object two does work on object one as a result. The context is centered around the concept of work in physics, particularly in relation to forces and motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which work is done, with some focusing on the definitions of force and distance. The original poster expresses uncertainty about how to approach the problem and considers the possibility of negative work. Others suggest applying Newton's third law to the scenario.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided numerical examples to illustrate their points, while others are seeking clarification on fundamental concepts related to work and forces.

Contextual Notes

There is a lack of consensus on the specific definitions and implications of work in this scenario, and the original poster is navigating through these concepts without clear guidance on how to proceed.

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Homework Statement


Object one pushes on object 2 as the objects move together, like a bulldozer pushing a stone. Assume object 1 does 15 J of work on object 2. Does Object 2 do work of object 1? If possible determine how much work?

Homework Equations





The Attempt at a Solution


I was thinking it would do work on object 1, could someone just point me in the right direction on how to start this. I'm not real sure where to begin. I was thinking it might be negative work that the 2nd object does on number 1.?
 
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What is needed for an object to do work?
 
Force and distance moved?
 
student 1 said:
Force and distance moved?

What is the force that object 2 exerts on object 1?
 
I am pretty sure that the answer is 15J.

Lets try it with numbers (although they don't really matter.) Let's say the bulldozer applies 5 Newtons for 3 Meters (3*5=15). This means that if the rock was not present, it would take 15 Joules less to move those 3 Meters. You can think of this -15J as the work that the boulder does on the bulldoser. It is the extra energy that the bulldozer has to do for those 3 meters (or any distance for what it matters).
 
Basically it was just to apply Newton's third law.
 
Thanks a bunch!
 

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