(III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. 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 5.0 g when brought to rest. Let M be the total mass of the elevator and passengers.

The total mass of the elevator and passenger is M.

The initial height of the elevator is h before the cable breaks.

The spring constant is k.

The maximum acceleration is \(a = 5.0\;{\rm{g}}\).

The final potential energy stored in the spring equals the elevator's initial potential energy according to energy conservation.

Let the elevator compresses the spring by a distance \(x\).

Now the average force on the elevator by the spring is \(F = k\Delta x\).… (i)

Then force by the elevator on the spring, \(F = ma + mg\) … (ii)

Comparing equations (i) and (ii),

\(\begin{aligned}kx &= Ma + Mg\\kx &= \left( {M \times 5.0g} \right) + Mg\\x &= \frac{{6.0Mg}}{k}\end{aligned}\) … (iii)

From energy conservation, the elevator's potential energy loss is equal to the potential energy gain of the spring. Then,

\(\begin{aligned}\frac{1}{2}k{x^2} &= Mg\left( {h + x} \right)\\k{x^2} &= 2Mg\left( {h + x} \right)\end{aligned}\)…(iv)

Now using equations (iii) and (iv),

\(\begin{aligned}k{\left( {\frac{{6.0Mg}}{k}} \right)^2} &= 2Mg\left( {h + \frac{{6.0Mg}}{k}} \right)\\k \times \frac{{36{M^2}{g^2}}}{{{k^2}}} &= 2Mgh + \frac{{12{M^2}{g^2}}}{k}\\\frac{{36{M^2}{g^2}}}{k} - \frac{{12{M^2}{g^2}}}{k} &= 2Mgh\\\frac{{24{M^2}{g^2}}}{k} &= 2Mgh\end{aligned}\)

After further calculation,

After further calculation,

\(k = \frac{{12Mg}}{h}\).

Hence, the value of the spring constant is\(\frac{{12Mg}}{h}\).