Conservation of Energy - Elevator Problem.

dhchemistry

Alright, so I'm a little stuck on the last bit of this problem:

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?

Delphi51

This can't be right. The first term has units of energy while the second term has units of force. Second term should be mgx - Fx.

dhchemistry

You're right. That's a typo. Sorry!

Delphi51

So, for part (b) x = 0.783 ?
In part (c), I think you need an additional term on the right for the spring energy after it goes up h and is stretched away from equilibrium: 1/2*k*(x-h)^2.
But I'm getting a final answer of 1.86 for h.
Seems to me 15 m is unreasonable - it would bounce only to 3.7 + .783 m if there was no friction.

dhchemistry

The answers to a, b, and c I have are the correct answers. 15m is also correct, I confirmed this with my professor.

As far as c goes, that term is unnecessary because the spring is not both giving and taking energy. It's only releasing its stored energy as it expands back to its original position.

Delphi51

Doesn't make sense to me. You have an elevator that falls 3.7 meters and then bounces back up 15 meters?

Yes it is. It takes in energy when it is being compressed, releases it when it springs back to the equilibrium point, then takes in some more as it is stretched beyond the equilibrium point.

If it stretched 15 m, it would break!!

dhchemistry

Not the spring, the elevator. How far total does the elevator travel until it comes to rest?

Delphi51

Oh, do you mean 15 m is the answer to part D rather than C?
Certainly D could be 15 m because the elevator will go up and down quite a few times before it stops, and those up and down distances will add up.

Do you have answers for parts B and C?

Delphi51

I guess I am wrong about that extra spring energy term. I was thinking the spring was attached to the elevator. Getting .783 for B and 2.08 for C now.

dhchemistry

B is .90 m and C is 2.8 m.

Delphi51

Okay, found a mistake, now getting 0.898 for B and 2.74 for C.
I can rest now! Sorry if I wasted time for you.

dhchemistry

No, it's fine.

dhchemistry

Okay, I have the solution.
The two forces reducing the energy of the elevator are the spring force and the force of friction.
d) First, find the compression of the spring at which the cab is at equilibrium.
[tex]mg=kd_{eq}[/tex]
[tex]d_{eq}=\frac{mg}{k}[/tex]

So [tex]d_{eq}=[/tex] .12

Then,
[tex]mg(d+d_{eq})=\frac{1}{2}kd_{eq}^{2}+F_{k}d_{t}[/tex]
[tex] d_{t}=\frac{mg(d+d_{eq})-\frac{1}{2}kd_{eq}^{2}}{F_{k}}[/tex]
[tex] d_{t}=\frac{(1800kg)(9.8m/s^{2})(3.7m+.12m)-\frac{1}{2}(150000)(.12m)^{2}}{4400N\cdot m}[/tex]

So
[tex]d_{t}\approx 15m[/tex]