What is the tension?

Question

An elevator in a tall building is allowed to reach a maximum speed of 3.3 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3.5 m if the elevator has a mass of 1320 kg including occupants?

Answer 1

use f = m*a
wight - tension = m * a
( 1320 * 9.8 ) - tension = 1320 * a
to finde a use
2*a*d = vf^2 - vi^2
a= ( 0 - 3.3^2)\( 2*3.5) = -1.56 mls^2

tension = 12936 - 2059.2 = 10876.8 n

Answer 2

This problem involves kinetic energy, work, and the principle of conservation of energy.

Kinetic Energy, KE=1/2mv^2
=1/2*1320*3.3^2
=7187.4 Joules

Work=F*s where F is the force and s is the displacement equal to 3.5m.

Based on principle of conservation of energy, KE is converted to work:

KE=Work

7187.4=F*3.5
F=7187.4/3.5
=2054N upward direction

The tension in the cable will be equal to the weight of the elevator of 1320g minus the net force of 2054N. Thus

Tension=1320*9.8-2054
=10,882N upward direction

Another solution which is longer is to solve for the acceleration using the formula v^2-u^2=2as. In this formula v=0, u=3.3m/s, s=3.5m, and a is the acceleration which should come out negative. After you get the acceleration, compute for F using the formula F=ma.

Then proceed to compute for the tension by deducting F from the weight of the lift plus passengers.

Answer 3

tension is a force
deceleration is from 3.3m/s down to 0m/s
in a distance of 3.5 meter with a mass of 1320kg
you can calculate a you can get tension force