Work Done on 500N Barbell 0.5m Above Floor

  • Thread starter Thread starter Dalip Saini
  • Start date Start date
  • Tags Tags
    Work Work done
Click For Summary

Homework Help Overview

The problem involves calculating the work done on a 500N barbell held 0.5m above the floor for a duration of 100 seconds. The discussion centers around the application of the work formula, W = Fdcos(phi), and the interpretation of the variables involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between force, distance, and the angle in the work equation. Questions arise regarding the definition of distance (d) in relation to the force (F) and the implications of vector directions.

Discussion Status

Participants are actively questioning the definitions and assumptions related to the work equation. There is a focus on clarifying the meaning of displacement and how it relates to the scenario presented, with no clear consensus reached yet.

Contextual Notes

There is an ongoing discussion about the interpretation of the angle in the work equation and the relevance of the barbell's position relative to the ground. Some participants express confusion about the implications of holding the barbell stationary versus moving it.

Dalip Saini
Messages
16
Reaction score
0

Homework Statement



You hold a 500N barbell 0.5m above a level floor for 100s . During this time, the amount of work you do on the barbell is


  • A

    1000J


  • B

    0J


  • C

    500J


  • D

    50000J

Homework Equations



W = Fdcos(phi)

The Attempt at a Solution


F= 500 N
d = 0.5 m
Since the difference in the vector direction is 0, I thought that all you had to do 500*0.5 = 250 J because cos(0) equals 1. But the answer is zero and I don't understand why.
 
Physics news on Phys.org
Dalip Saini said:
W = Fdcos(phi)
Equations aren't much use unless you know what all the variables in it represent.
What is d in this equation? Yes, I know it's a distance, but what distance in relation to F, exactly?
 
Its the distance of the barbell above the ground. So is that vector pointing down and the force is pointing up? But still cos(180) is still -1 not 0
 
Dalip Saini said:
Its the distance of the barbell above the ground.
That's not what I asked.
Forget this specific problem for the moment. You quoted a standard equation. How is d defined in that equation?
 
the displacement?
 
Dalip Saini said:
the displacement?
Displacement of what?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
11K
  • · Replies 13 ·
Replies
13
Views
6K
Replies
6
Views
4K