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.

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nuuk nuuk · Answer 1 · 2019-11-15T08:06:06+01:00

Answer:k=28.29 kN/m

Explanation:

Given

mass [tex]m =7700 kg[/tex]

height from which Elevator falls [tex]h=32 m[/tex]

Let x be the compression in the spring

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

[tex]\frac{kx^2}{2}=mg(h+x)[/tex]----------1

also maximum acceleration is 5g

thus

[tex]mg-kx=ma[/tex]

here [tex]a=-5g[/tex]

[tex]kx=mg-m(-5g)=6mg[/tex]

[tex]x=\frac{6mg}{k}[/tex]

Substitute x in equation 1

[tex]0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})[/tex]

[tex]18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}[/tex]

[tex]k=12\cdot \frac{mg}{h}[/tex]

[tex]k=12\times \frac{7700\times 9.8}{32}[/tex]

[tex]k=28.29 kN/m[/tex]