An elevator and its load have a combined mass of 1962 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 7.5 m/s, is brought to rest with constant acceleration in a distance of 4.2 m.

Answer 1

v²-u²=2aS
v=0
u=-7.5 m/s
S=-4.2 m

Solve for a=acceleration in m/s²
a=7.5²/(-2×4.2)
=6.7 m/s²