Solving for Speed in Elevator Motor Operation

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SUMMARY

The discussion focuses on calculating the maximum speed of a 20-kilowatt motor operating an elevator weighing 10,000 N. The user initially struggles with the relationship between power, work, and time but realizes that power can be expressed as P = F * V. By substituting the known values, the user successfully derives the equation 20,000 = 10,000 * V, leading to the conclusion that the maximum speed of the elevator is 2 m/s.

PREREQUISITES
  • Understanding of basic physics concepts, specifically power and work.
  • Familiarity with the formula P = F * V.
  • Knowledge of units of measurement, particularly watts and newtons.
  • Basic algebra skills for solving equations.
NEXT STEPS
  • Research the principles of mechanical power in elevator systems.
  • Explore the relationship between force, velocity, and power in different contexts.
  • Learn about the efficiency of electric motors in lifting applications.
  • Investigate safety factors and limits in elevator motor operations.
USEFUL FOR

Engineers, physics students, and professionals involved in elevator design and operation, as well as anyone interested in understanding the mechanics of motor-driven systems.

demode
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The prompt:

"A 20-kilowatt motor operates an elevator weighing 10,000 N. What is the maximum speed at which the motor can raise the elevator?"

How can i solve for speed?

I know that power = work / time, and work = force * distance, but I cannot solve for work because I only know a force without a distance, and I cannot solve for power because I don't know work NOR time... What am i missing here?

**EDIT**

Wow I just realized that P = w/t = f*d/t = f * d/t = F*V

With that being said, I can use P = F*V, or in my case, 20000 = 10000*X... Right?
 
Last edited:
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Yup. That works quite well.
 
Haha, thanks mate..

I hate making posts on here and realizing literally 5 seconds afterwards that I understand it :)
 

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