If the elevator cable should break when the elevator is at a height h above the top of the spring, calculate the value that the spring stiffness constant k should have so that passengers undergo an acceleration of no more than 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

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nuuk nuuk · Answer 1 · 2019-11-07T11:50:09+01:00

Answer:[tex]k=\frac{12Mg}{h}[/tex]

Explanation:

Given

Mass of Elevator is M

maximum acceleration is 5g

Cable breaks at a height of h above spring

Let K be the spring constant

For Elevator

[tex]F-Mg=M(a)[/tex]

[tex]a=5g[/tex]

[tex]F-Mg=5Mg[/tex]

[tex]F=6Mg[/tex]

Suppose spring compresses to a distance x

thus [tex]kx=6Mg[/tex]

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

Potential Energy is converted to Elastic Potential Energy

thus

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

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

[tex]Mgh+\frac{6M^2g^2}{k}=\frac{18M^2g^2}{k}[/tex]

[tex]12\frac{M^2g^2}{k}=Mgh[/tex]

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