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- 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.