The cable of the 1700 kg elevator cab shown the figure snaps when the cab is at

Question

The cable of the 1700 kg elevator cab shown the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance

(a) Find the speed of the cab just before it hits the spring.

(b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression).

(c) Find the distance that the cab will bounce back up the shaft.

(d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest.

Explanation / Answer

4400N acting upwards
17000N (1700kg * 10m/s/s) acting downwards
Resultant force downwards = 17000-4600=12400N; call this "mg". This just means instead of 10m/s/s as g, it is effectively reduced to 7.29411m/s/s

Uginitial = Us + Ugfinal
"mg"hinitial = 0.5kx^2 + "mg"hfinal
12400(3.9 + natural length of spring, L) = 0.5(160000)x^2 + 12400(L - x)
48360 + 12400L = 80000x^2 + 12400L - 12400x
48360 = 80000x^2 - 12400x
x = .7038...m using quadratic formula (other answer is negative) <--- part 1

Us = Ug
0.5kx^2 + "mg"hinitial = "mg"hfinal
0.5*160000*0.7038489^2 + 12400(L) = 12400(h + L)
62760.3 = 13600h
h = 4.5096m (which seems odd, but the spring is very "propelling" - either that or I've made an error somewhere, which is likely)

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