# Homework Help: Finding Spring Constant w/ Conservation of Energy

1. Nov 15, 2014

### logan3

1. The problem statement, all variables and given/known data
An engineer is designing a spring to be placed at the bottom of an elevator shaft. 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.

${\vec s}_{elevator} = h$
$m = M$
$\vec F_{net} = 5.0 g$

2. Relevant equations
The elevator has it’s maximum velocity, and thus maximum acceleration and force, when the spring is most compressed. The sum of the forces at this point are:

${\vec F}_{net} = {\vec F}_{spring} - {\vec F}_W = Ma = M5.0g \Rightarrow {\vec F}_{spring} = M5.0g + {\vec F}_W = M5.0g + Mg = 6.0Mg$

When the spring is most compressed, the total displacement of the elevator will be the distance it fell plus the compression distance of the spring. The displacement of the elevator is h. This means that the lowest position is when the elevator has maximum compression on the spring.

${\vec s}_{net} = {\vec s}_{elevator} + {\vec s}_{spring} = h + {\vec s}_{spring}$

At this point, the gravitational PE is zero. During the fal it was converted to KE and as the elevator hit the spring the KE was converted to elastic PE. Thus, the sum of the energies is:

$KE_i + gPE_i + ePE_i = KE_f + gPE_f + ePE_f$, where $gPE$ and $ePE$ are the gravitational and elastic potential energy respectively.

$KE_i = 0 J$
$gPE_i = Mg(h + {\vec s}_{spring})$
$ePE_i = 0 J$
$KE_f = 0 J$
$gPE_f = 0 J$
$ePE_f = \frac {1}{2} k {\vec s}_{spring}^2$

3. The attempt at a solution
${\vec F}_{spring} = 6.0Mg = k {\vec s}_{spring} \Rightarrow {\vec s}_{spring} = \frac {6.0Mg}{k}$

$0 J + gPE_i + 0 J = 0 J + 0 J + ePE_f \Rightarrow gPE_i = ePE_f \Rightarrow Mg(h + {\vec s}_{spring}) = \frac {1}{2} k {\vec s}_{spring}^2$

$\Rightarrow k = \frac {{2Mg(h + {\vec s}_{spring})}}{{\vec s}_{spring}^2} = \frac {{2Mg(h + (\frac {6.0Mg}{k}))}}{(\frac {6.0Mg}{k})^2} = \frac {2Mgh}{\frac {36.0M^2g^2}{k^2}} + \frac {\frac {12.0M^2g^2}{k}}{\frac {36.0M^2g^2}{k^2}} = \frac {2Mghk^2}{36.0M^2g^2} + \frac {12.0M^2g^2k^2}{36.0M^2g^2k}$
$\Rightarrow k = \frac {hk^2}{18.0Mg} + \frac {k}{3.0} \Rightarrow 1 = \frac {hk}{18.0Mg} + \frac {1}{3.0} \Rightarrow \frac {hk}{18.0Mg} = \frac {2}{3} \Rightarrow k = \frac {12.0Mg}{h}$

Thank-you

2. Nov 15, 2014

### Staff: Mentor

Did you have a question about your solution? Because at a glance it looks to be a correct method and result.