Work Done on 500N Barbell 0.5m Above Floor

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The discussion revolves around calculating the work done on a 500N barbell held 0.5m above the floor for 100 seconds. The key equation for work is W = Fdcos(phi), where F is the force, d is the displacement, and phi is the angle between the force and displacement vectors. Despite initial calculations suggesting 250J, the consensus is that the work done is actually 0J because there is no displacement in the direction of the force while holding the barbell stationary. The confusion arises from understanding the definition of displacement in relation to the force applied. Ultimately, the work done is zero because the barbell is not moving.
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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.
 
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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?
 

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