How much work was done by the tension in the cable?

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Homework Help Overview

The problem involves calculating the work done by the tension in a cable as a 1500-kg elevator moves upward at a constant speed over a vertical distance of 25 meters. The context is rooted in mechanics, specifically focusing on forces and work in a gravitational field.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between force, mass, and acceleration, questioning how to apply the formula for work given the elevator's constant speed. There is exploration of the implications of zero acceleration and the balance of forces acting on the elevator.

Discussion Status

Participants are actively engaging with the problem, clarifying the role of tension and gravitational force. Some guidance has been offered regarding the balance of forces, but multiple interpretations of the force dynamics are still being explored.

Contextual Notes

There is an ongoing discussion about the implications of constant speed on the forces involved, particularly regarding the assumption that the total force is zero due to the lack of acceleration.

brncsfns5621
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I think I have been staring at this question too long as I can not figure it out. I am sure the answer is simple:

A 1500-kg elevator moves upward with a constant speed through a vertical distance of 25m. How much work was done by the tension in the cable?

I know that W= (F cos 0)s, and that F= ma. What I can't figure out is the acceleration. It doesn't have one as it is moving with a constant speed. Any info on where to start?
 
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F=ma when F is the total force on the object. Here, the total force is 0, since the object isn't accelerating, and so the tension in the cable must exactly balance the force of gravity.
 
So F= 0 and T= mg?
 
Yea, if F is the total force on the object (you can always use F to stand for only one contribution to the force, it's just then F=ma won't hold).
 

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