Alright, so I'm a little stuck on the last bit of this problem:(adsbygoogle = window.adsbygoogle || []).push({});

The cable of the 1,800 kg elevator cab 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. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

I was able to solve parts a-c without too much of a problem. I just can't get part 3.

Here are my solutions:

a) [tex]

\Sigma F=mg-F_{k}=ma[/tex]

[tex]a=g-\frac{F_{k}}{m}[/tex]

[tex]a=7.4m/s^{2}[/tex]

[tex]v_{f}^{2}=v_{i}^{2}+2as[/tex]

[tex]v_{f}=\sqrt{2(7.4m/s^{s})(3.7m)}[/tex]

so

[tex]

v_{f}=7.4m/s

[/tex]

b) [tex]\frac{1}{2}mv^{2}-(mg-F_{k})x-\frac{1}{2}kx^{2}=0[/tex]

Then I solved the quadratic equation and found

x=.90m

c) [tex]\frac{1}{2}kx^{2}=mgh+F_{k}h[/tex]

[tex]\frac{1}{2}kx^{2}=h(mg+F_{k})[/tex]

[tex]h=\frac{\frac{1}{2}kx^{2}}{mg+F_{k}}[/tex]

[tex]h=2.8m[/tex]

Now for d, I tried:

[tex]K=F_{k}x[/tex]

[tex]\frac{1}{2}mv^{2}=F_{k}x,

[/tex]

Since the only slowing force doing work on the elevator is friction. If I use the v from part a and solve for x, however, I get 11.2m. The solution in the back of the book is 15m. Any ideas?

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# Conservation of Energy - Elevator Problem.

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