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 3.4 m if the elevator has a mass of 1320 kg including occupants?

Answer 1

Answer:

2114 newtons

Explanation:

Assuming the tension in the cable constant, the deacceleration of the elevator is constant too. So, we can use the Galileo's kinematic equation:

$v^{2}=v_{0}^{2}+2a\varDelta x$

With v the final velocity that is zero because is the velocity when the elevator stops, vo the initial velocity that is the velocity the elevator is allowed to reach, a the acceleration and Δx the distance the elevator takes to stop. Solving for a:

$a=\frac{-v_{0}^{2}}{2 \varDelta x}=\frac{-(3.3)^{2}}{2 (3.4)}$

$a= -1.60\frac{m}{s^{2}}$

Now with the acceleration we can use Newton's second law to find the force (the tension) on the cable:

$F=T=ma$

with F=T the tension, and m the mass of the elevator including occupants, solving for T

$T=-(1320)(1.6)=-2114 N$