Question

An elevator and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 62 m. (Hint: if the elevator is slowing down and moving down then the upward force is greater)

Answer 1

12739.74 N

Step-by-step explanation:

m = Mass of load + Mass of elevator = 1200 kg

s = Displacement of the elevator = 62 m

g = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 10 m/s

v = Final velocity = 0

Equation of motion

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-10^2)/(2* 62)\\\Rightarrow a=-0.80645\ m/s^2`

Tension on the cable

`T=m(g-a)\\\Rightarrow T=1200(9.81-(-0.80645))\\\Rightarrow T=12739.74\ N`

Here, as the elevator is going down the acceleration due to gravity and the acceleration will be subtracted.

Hence, the tension in the supporting cable is 12739.74 N