The cable of the 1500 kg elevator cab shown the figure snaps when the cab is at

Question

The cable of the 1500 kg elevator cab shown the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.6m above a spring of spring constant k = 0.16 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 4.8 kN opposes the cab's motion. (Assume that the frictional force on the cab is negligible when the cab is stationary.) Find the speed of the cab just before it hits the spring 6.89 m/s Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression)X = m Find the distance that the cab will bounce back up the shaft, m Using conservation of energy, find the approximate total distance that the cab will move before coming to rest, m

Explanation / Answer

a)

Given that,

mass of the cable is , m=1500 kg,

The frictional force is Fk=4.8 kN=4800 N acting upwards
F=mg=1500 kg*10 m/s2

=15000N acting downwards
Resultant force downwards = 15000-4800=10200N;

F=mgFk=ma

a=gFf/m

a=7.4m/s2

v2f=v2i+2as

vf=sqrt(2(7.4m/s2)(3.6m))

so vf=7.2m/s
b)

Uginitial = Us + Ugfinal
   = 0.5kx^2 + mgh
10200 N(3.6 m + natural length of spring, L) = 0.5(150000)x^2 + 10200(L - x)
50320 + 10200L = 75000x^2 + 10200L - 10200x
50320 = 75000x^2 - 10200x
0 = 75000x^2 - 10200x - 50320
x = 0.81477m
c)
Us = Ug
0.5kx^2 + mghi = mghf
0.5*150000*0.91477^2 + 10200(L) = 10200(h + L)
62760.3 = 10200h
h =6.152 m

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