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

Step-by-step 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

`(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=(6mg)/(k)`

Substitute x in equation 1

`0.5* k* ((6mg)/(k))^2=mg(h+(6mg)/(k))`

`18((mg)^2)/(k)=mgh+6((mg)^2)/(k)`

`k=12\cdot (mg)/(h)`

`k=12* (7700* 9.8)/(32)`

`k=28.29 kN/m`