# Elevator falling and bouncing back from a spring

• jolly_math
In summary, the concept of an elevator falling and bouncing back from a spring involves a hypothetical scenario where an elevator is suspended by a spring and experiences a downward force, causing it to fall and bounce back. This is caused by the principles of gravity and elasticity, but it is not possible in real life due to safety mechanisms and the limitations of the spring's force. The height of the elevator affects the speed and magnitude of its fall and bounce back, and while there are no real-life applications for this concept, understanding it can help improve safety mechanisms for elevators and other devices.

#### jolly_math

Homework Statement
The cable of a 4000-lb elevator snaps when the elevator is at rest at the first floor so that the bottom is a distance d = 12.0 ft above a cushioning spring whose force constant is k = 10,000 lb/ft. A safety device clamps the guide rails, removing 1000 ft-lb of mechanical energy for each 1.00 ft that the elevator moves.
(a) Find the speed of the elevator just before it hits the spring.
(b) Find the distance that the spring is compressed.
(c) Find the distance that the elevator will bounce back up the shaft.
(d) Calculate approximately the total distance that the elevator will move before coming to rest. Why is the answer not exact?
Relevant Equations
U(x) (gravity) = mgh
U(x) (spring) = (1/2)kx^2
KE = (1/2)mv^2
I don't understand the difference between part c and d. After compressing the spring, the elevator bounds back and moves before coming to rest in both cases. What is the difference? Thank you.

Hi,

Well, after c, the thing is still up in the air and will start to go down again, etc, until all energy has been dissipated.

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jolly_math