Calculating Work Done by a Variable Force in an Electric Elevator

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SUMMARY

The discussion focuses on calculating the work done by a variable force when lifting a cable in an electric elevator. The cable weighs 4.5 lb/ft, and 198 ft of cable is paid out when the elevator is at the first floor. The initial calculation for work done was incorrectly computed as 648,000, while the correct value is 72,900. The error stemmed from not properly accounting for the variable force exerted by the cable as it is lifted.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Knowledge of physics concepts related to work and force
  • Familiarity with variable force calculations
  • Basic principles of mechanics, including gravity's effect on mass
NEXT STEPS
  • Study the concept of variable force in physics, particularly in lifting scenarios
  • Learn how to perform integration for variable forces in calculus
  • Review the principles of work-energy in mechanics
  • Explore examples of work done by variable forces in real-world applications
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Students and professionals in physics, engineering, and mathematics who are interested in understanding the calculations involved in work done by variable forces, particularly in mechanical systems like elevators.

birdec
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An electric elevator with a motor at the top has a cable weight 4.5 lb/ft. When the car is at the first floor, 198 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor the to the top?

This is what I've done:

180=4.5k
k=40
force=40x

work = integral from 0 to 180 40x dx
then I took the anti derivative and it was (40x^2)/2 which ends up 20x^2
I then plugged in 180 for x
20(180)^2 = 648000

The correct answer is 72900. Any thoughts?
 
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Check your units. The work done in lifting against gravity is: W = Force X Distance = mass X g X distance. And you need to acount for the variable force as you have noticed.
 

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