Answer:
[tex]k=\frac{24Mg}{h}[/tex]
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:
[tex]\sum F=ma\\kx-Mg=M(7g)\\kx=8Mg\\x=\frac{8Mg}{k}[/tex]
The law of the conservation of energy states that:
[tex]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}[/tex]
The value of the spring constant must be [tex]\frac{24Mg}{h}[/tex]
When the elevator pusses against the spring, there is a compression/displacement in the spring and the spring applies a restoring force against the compression.
Let the compression of the spring be x,
then the restoring force will be:
F = -kx
where k is the spring constant
The required acceleration of the elevator is a = 7g, where g is the acceleration due to gravity.
So the equation of motion of the elevator can be written as:
kx - Mg = Ma
here M is the mass of the elevator
kx - Mg = M(7g)
[tex]x=\frac{8Mg}{k}[/tex]
Now if we take the bottom position of the elevation as zero gravitational potential energy, then from conservation of energy:
(h+x) will be the initial height with respect to the bottom, also at the bottom and the tom the velocity is zero, so at the bottom, there is only potential energy of the spring, thus:
[tex]Mg(h+x)=\frac{1}{2}kx^2\\\\Mgh+\frac{8M^2g^2}{k}=\frac{1}{2}k\frac{64M^2g^2}{k^2}\\\\k=\frac{24Mg}{h}[/tex]
Learn more about spring constant:
https://brainly.com/question/10455854?referrer=searchResults
