Question

If the elevator cable breaks when the elevator is at a height h above the top of the spring, calculate the value that the spring constant k should have so that passengers undergo an acceleration of no more than 7.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

Answer 1

Answer:

$k=\frac{24Mg}{h}$

Explanation:

We have to know the maximum compression of the spring in order to know the total height at which the elevator falls. According to Newton's second law:

$\sum F=ma\\kx-Mg=M(7g)\\kx=8Mg\\x=\frac{8Mg}{k}$

The law of the conservation of energy states that:

$MgH=\frac{kx^2}{2}\\Mg(h+x)=\frac{k(\frac{8Mg}{k})^2}{2}\\Mgh+Mg(\frac{8Mg}{k})=\frac{64M^2g^2}{2k}\\Mgh=\frac{32M^2g^2}{k}-\frac{8M^2g^2}{k}\\k(Mgh)=24M^2g^2\\k=\frac{24Mg}{h}$