Question

an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a spring at the bottom of the shaft. Determine the spring constant (in kN/m) necessary to bring the elevator and occupants to rest without exceeding an acceleration of 5 g's.

Answer 1

Answer:k=28.29 kN/m

Explanation:

Given

mass $m =7700 kg$

height from which Elevator falls $h=32 m$

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

$\frac{kx^2}{2}=mg(h+x)$----------1

also maximum acceleration is 5g

thus

$mg-kx=ma$

here $a=-5g$

$kx=mg-m(-5g)=6mg$

$x=\frac{6mg}{k}$

Substitute x in equation 1

$0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})$

$18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}$

$k=12\cdot \frac{mg}{h}$

$k=12\times \frac{7700\times 9.8}{32}$

$k=28.29 kN/m$