Question

An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable should happen to break when the elevator is at a height h = 12.5 m above the top of the spring, calculate the value the spring constant k should have so that passengers undergo a maximum acceleration of 5.3g. Take the mass of the elevators plus passengers to be M = 1439 kg.

Answer 1

Answer:

$k = 9876 N/m$

Explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

$mg(h + x) = \frac{1}{2}kx^2$

also we know that maximum acceleration will be 5.3 g

so it is given as

$a = \frac{k}{m} x$

so we have

$x = \frac{ma}{k} = \frac{5.3 mg}{k}$

$mg(h + \frac{5.3 mg}{k}) = \frac{1}{2}k(\frac{5.3mg}{k})^2$

$mg(h + \frac{5.3 mg}{k}) = \frac{14.045(mg)^2}{k}$

$h + \frac{5.3mg}{k} = \frac{14.045 mg}{k}$

$h = \frac{8.745mg}{k}$

$k = \frac{8.745 (1439)(9.81)}{12.5}$

$k = 9876 N/m$