Question

A large spring is placed at the bottom of an elevator shaft to minimize the impact in case the elevator cable breaks. A loaded car has mass 480kg, and its maximum height above the spring is 11.8m. In order to minimize the shock, the maximum acceleration of the car after hitting the spring is 4g. Find the spring constant, k.

Answer 1

F = kx
ma = kx
480kg * (4*9.81) = k * 11.8m
k = 1596.20339

Answer 2

2990 N/m

Answer 3

2 equations:

Work spring = work of gravity: 0.5kx^2 = mgh
Force spring = Force of gravity + Force of car: kx = mg + m(4g)

Use the force equation to find that kx = 5mg, then substitute that into the work equation to get 0.5(5mg)x = mgh. Do some math and get x = h / 1.5. Substitute this answer back into the force equation: k = 5mg / (11.8 / 1.5) = 23520 / 7.867; k = 2989.7 N/m.

Answer 4

EDIT: h = 11.8 + x, x being the displacement of the spring once the car lands on it. Otherwise you'll get 2.5 instead of 1.5 when solving for x.