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