How Fast Can an Elevator Lift a 1000 kg Load with 2000 W of Power?

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    Thermodynamics
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Homework Help Overview

The discussion revolves around calculating the speed at which an elevator can lift a 1000 kg load given a motor power output of 2000 W. The subject area includes concepts from thermodynamics and mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different equations related to power, work, and force. There are attempts to rearrange formulas to find velocity, and questions arise about the relationship between force and mass.

Discussion Status

Participants are actively engaging with the problem, with some providing alternative equations and clarifying their understanding of the relationships between the variables. There is a recognition of overlapping concepts in the equations discussed.

Contextual Notes

Some participants express uncertainty about converting mass to force and the implications of using different equations. There is a mention of the need for clarity in definitions and assumptions related to the problem.

desichick07
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Thermodynamics -- simple question

Homework Statement



An elevator motor prouces 2000 W of power. How fast ( in m/s) can it life a 1000 kg load?

Homework Equations



W = Fd
P = W / t


The Attempt at a Solution



by rearranging the formula i got:
W = Fd / t

d / t = W / F

but I'm not sure how to get the force, since it needs to be in N and i am only given kg.
 
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P=FV (force times velocity) This is also a handy equation.

P=mg(v)

2000= (1000)(9.8)(v)

Solve for V.
 
Thanks a million.

But just out of curiosity is there a way to do it just using the equations W = Fd and P = W/t ?
 
Never mind i realized what you are doing is the same thing...d/t (distance/time) is velocity. You just replaced d/t with V. Wow that was a stupid question.
But thanks again for the help.
 
[tex]P = \frac{Fd}{t}[/tex]

[tex]F = ma[/tex] (in Newtons)

[tex]F = W = mg[/tex] (in Kg)

Hence...

[tex]P = \frac{mg \cdot d}{t}[/tex]

Which leads one to see...

[tex]\frac{d}{t}[/tex] is how fast it can lift the load.

CS
 
Opps...too late.
 

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