Question

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.

Answer 1

Answer:
`k=(12Mg)/(h)`

Step-by-step 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

`F-Mg=M(a)`

`a=5g`

`F-Mg=5Mg`

`F=6Mg`

Suppose spring compresses to a distance x

thus
`kx=6Mg`

`x=(6Mg)/(k)`

Potential Energy is converted to Elastic Potential Energy

thus

`Mg(h+x)=(kx^2)/(2)`

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

`Mgh+(6M^2g^2)/(k)=(18M^2g^2)/(k)`

`12(M^2g^2)/(k)=Mgh`

`k=(12Mg)/(h)`