A generator having power and creating work for a crane

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SUMMARY

The discussion focuses on calculating the lifting speed of a crane powered by a motor with an efficiency of 88% and a crane efficiency of 42%. Given a power supply of 5.5 kW and a load of 410 kg, the user initially calculated the lifting speed as 1.36 m/s. However, the correct answer provided in the reference material is 0.50 m/s. Key equations used include power calculations and efficiency formulas, emphasizing the relationship between force, velocity, and efficiency in mechanical systems.

PREREQUISITES
  • Understanding of mechanical power and efficiency calculations
  • Familiarity with basic physics concepts such as force and velocity
  • Knowledge of the equations for work, power, and mechanical advantage
  • Ability to perform unit conversions and basic algebraic manipulations
NEXT STEPS
  • Study the relationship between mechanical power, force, and velocity in detail
  • Learn about the principles of efficiency in mechanical systems
  • Explore the concept of mechanical advantage and its calculations
  • Review examples of crane operations and their power requirements
USEFUL FOR

Students studying physics, engineering students focusing on mechanics, and professionals involved in crane operations or mechanical design will benefit from this discussion.

Bassi
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Homework Statement


"A motor having an efficiencyof 88% operates a crane having an efficiency of 42%. With what constant speed doess the crane lift a 410-kg crate of machine parts if the power supplied to the motor is 5.5kW"
e'=42%
e=88%
m=410 kg
dr=?
de=?
t=?
g=9.8m/s square
fr=410x9.8=4018
p=5500 w
MA=?
IMA=?
wo=?
wi=?

Homework Equations


w=fd
p=w/t
ma=fr/dr
ima=de/dr
e=wo/wi then x100
e=ma/ima then x 100
wo=frdr
wi=fede
m=vd

The Attempt at a Solution


Im not sure if this is correct, but ill give it a shot:
p=w/d so p=frdr/t i think that since u r looking how fast the crane moves, you use 1 as t
so 5500=4018dr/1
so 5500/4018=dr
dr=1.36 so 1.36m/s
However the answer on the sheet says .50m/s so idont know? Please help as soon as possible.
-Bassi
 
Physics news on Phys.org
It might help you to know that mechanical Power = Force x Velocity at any instant. It should be clear what the minimum force required to lift the box is.
 

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