Understanding Work: Calculating Work in Different Scenarios

  • Thread starter Thread starter physicos
  • Start date Start date
  • Tags Tags
    Work
Click For Summary

Homework Help Overview

The discussion revolves around understanding the concept of work in physics, specifically in scenarios involving holding and carrying a bag of groceries. The original poster presents two questions regarding the calculation of work done in these situations.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason through the work done while holding a bag still and while carrying it across a distance. Some participants confirm the reasoning for the first question but question the interpretation of work in the second scenario. There is also a discussion about the definitions of force (F) and distance (s) in the context of the work equation.

Discussion Status

The discussion is active, with participants exploring different interpretations of the work concept. Some guidance has been offered regarding the work equation, but there is no explicit consensus on the second question, as participants continue to seek clarification.

Contextual Notes

Participants are navigating through the definitions and implications of work in physics, with some confusion regarding the application of the work equation in different scenarios. There is an emphasis on understanding the conditions under which work is done.

physicos
Messages
46
Reaction score
1

Homework Statement


Here are two questions I have some trouble answering :
1- How much work do you do when you hold a bag of groceries while standing STILL?
2- How much work do you do when carrying the same bag on a distance d across the parking lot of the grocery store?

Homework Equations



We know that ΔK = W

The Attempt at a Solution


1- In the first question , the person stand still so V = 0 and thus ΔK=0 so the work W=0
2- In the second question, when carrying the bag we only have a gravitational work so W= Wg=0

IS IT CORRECT ?
 
Physics news on Phys.org
First question:
Yes you are right.There's another way to solve that.
##W=F.s## You are applying an equal and opposite force to that of gravity but it's not moving so ##W=F.0=0##

Second question:
I don't understand that W=Wg part.
The question asked how much work you do.
Use ##W=F.s## and answer!
 
what is F and s ??
 
physicos said:
what is F and s ??
"F" is the force and "s" is the distance in the direction of the force.
W is the work done.
 
Ah ok , I got it ! Thank you SO MUCH !
 
physicos said:
Ah ok , I got it ! Thank you SO MUCH !
So ... what is your answer http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif
 
Last edited by a moderator:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K