Calculating Work Done by a Variable Force in an Electric Elevator

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The discussion centers on calculating the work done by an electric elevator motor lifting a variable weight of cable. 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 calculations led to an incorrect work value of 648,000, while the correct answer is 72,900. Participants emphasize the importance of checking units and accounting for the variable force in the calculations. The correct approach involves integrating the force over the distance to accurately determine the work done.
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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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