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

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An elevator motor producing 2000 W can lift a 1000 kg load by using the relationship between power, force, and velocity. The relevant equations include P = Fv and W = Fd, where force is calculated as F = mg. By substituting the values into the equations, the velocity can be determined. The discussion highlights the realization that distance over time (d/t) is equivalent to velocity, simplifying the problem. Ultimately, the calculations confirm how to derive the lifting speed from the given power and load.
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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.
 
P = \frac{Fd}{t}

F = ma (in Newtons)

F = W = mg (in Kg)

Hence...

P = \frac{mg \cdot d}{t}

Which leads one to see...

\frac{d}{t} is how fast it can lift the load.

CS
 
Opps...too late.
 

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