(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

he cable of the 1,800 kg elevator cab in Fig. 8-51 snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 4.4 kN opposes the cab's motion. (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 (above the point of maximum compression) that the cab will bounce back up the shaft.

2. Relevant equations

3. The attempt at a solution

I got (a)...b and c are where I hit problems.

for b i used the equation....

mg(h+x) + F(h+x)(cos(pi))=.5kx^2

Now when I tried to solve for x i got an equation with huge numbers:

49012+13247x-0.075x^2

It just doesn't seem reasonable to solve for x here, and I know i must be doing something wrong I just cant see it.

for (c):

I used 1/2k(x)^2 - F(x+h) = (1800)(9.81)(x+h)

I would plug in x from c, its just x that I can't find.

Thanks!

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# Homework Help: Conservation of energy of an elevator cable

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