What Mistakes Are in the Elevator Problem Calculation?

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SUMMARY

The discussion focuses on the calculation of the force exerted by the cable on a 1000kg elevator that decelerates from 6 m/s to a stop over a distance of 3 meters. The proposed solution incorrectly calculates the acceleration as 6.0 m/s², which is derived from the formula a = v₀² / 2d. The correct acceleration should be calculated as a = (v_f - v₀) / t, where t is the time taken to stop. Additionally, the gravitational force acting on the elevator must be considered, leading to a corrected tension force calculation of T = mg + ma, resulting in a total force of 8000N.

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Identify any errors in the solution to the following problem and provide a corrected solution if there are errors.
The problem A 1000kg elevator is moving down at 6 m/s. It slows to a stop in 3 m as it approaches the ground floor. Determine the force that the cable supporting the elevator exerts on the elevator as the elevator stops. Assume that g = 10N/kg.

Proposed solution The elevator at the right (picture) is the object of interest. It is considered a particle, and the forces that other objects exert on the elevator are shown in the free body diagram (T == F_e) The accelration of the elevator is:

a = v_0^2 / 2d = 6^2 / 2(3) = 6.0 m/s^2
The force of the cable on the elevator while stopping is:

T = ma = (1000kg) * (6.0 m/s^2) = 6000N
 
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Does the gravitational force cease to act on the elevator while it's decelerating?
 

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