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

`k = 9876 N/m`

Step-by-step explanation:

As per energy conservation we know that

initial total gravitational potential energy = final spring potential energy

so we have

`mg(h + x) = (1)/(2)kx^2`

also we know that maximum acceleration will be 5.3 g

so it is given as

`a = (k)/(m) x`

so we have

`x = (ma)/(k) = (5.3 mg)/(k)`

`mg(h + (5.3 mg)/(k)) = (1)/(2)k((5.3mg)/(k))^2`

`mg(h + (5.3 mg)/(k)) = (14.045(mg)^2)/(k)`

`h + (5.3mg)/(k) = (14.045 mg)/(k)`

`h = (8.745mg)/(k)`

`k = (8.745 (1439)(9.81))/(12.5)`

`k = 9876 N/m`