What Mistakes Are in the Elevator Problem Calculation?

[jtm]

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

 

[Tide]

Does the gravitational force cease to act on the elevator while it's decelerating?

 

[quicknote]

There are no major errors in the solution provided. However, there are a few minor errors that could be corrected.

1. In the free body diagram, the force of the cable should be labeled as "T" instead of "F_e".
2. The acceleration of the elevator should be negative since it is slowing down. Therefore, the equation for acceleration should be written as: a = -v^2 / 2d.
3. The value for acceleration should be written as -6.0 m/s^2 instead of 6.0 m/s^2.
4. The final answer for the force of the cable should also be negative, indicating that it is acting in the opposite direction of the elevator's motion. Therefore, the correct value for T should be -6000N.

Corrected solution:

a = -v^2 / 2d = -(6 m/s)^2 / 2(3 m) = -6.0 m/s^2

T = ma = (1000 kg) * (-6.0 m/s^2) = -6000N

Therefore, the force of the cable on the elevator while stopping is -6000N.