Question

An engineer is designing an ideal spring to be placed at the bottom of an elevator shaft in vertical position. If the elevator cable breaks at the height $h$ above the upper end of the spring, then calculate the required value of the spring constant so that the passenger undergoes an acceleration of not more than $5g$, when elevator is brought to rest. Let $M$ be the mass of the elevator and passengers (Given acceleration due to gravity $=g$).

• A. $\frac{12Mg}{h}$
• B. $\frac{15Mg}{2h}$
• C. $\frac{20Mg}{h}$
• D. $\frac{25Mg}{4h}$

Answer 1

The correct option is A $\frac{12Mg}{h}$

Let the spring is compressed by a distance $x$,before the lift stops,then
$Mg(h+x)=\frac{1}{2}k{x}^2\cdots \left(i\right)$
also
$kx-Mg=M\left(5g\right)\cdots \left(ii\right)$

$\Rightarrow x=\frac{6Mg}{k}$

substituting in equation (i)
$\Rightarrow Mgh+Mg\times \frac{6Mg}{k}=\frac{1}{2}k\frac{36M^2{g}^2}{k^2}$

$h=\frac{18Mg}{k}-\frac{6Mg}{h}$

$k=\frac{12Mg}{h}$