Falling elevator onto a spring.

[J-dizzal]

Homework Statement

The cable of the 2000 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 2.8 m above a spring of spring constant k = 0.30 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 2.4 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.)

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_59.jpg

Homework Equations

The Attempt at a Solution

I got part a, but part b is where I am stuck. i got an answer of .0573m I am not sure how to find distance without using the spring equation.

[/B]

 

[Nathanael]

x=(mg-F_f)/k is when the acceleration is zero, but you want the velocity to be zero not acceleration.

If the spring is compressed by an amount x, then how much work from gravity and friction is done? And if the velocity is zero, then what is the spring-potential-energy?

 

[J-dizzal]

x=(mg-F_f)/k is when the acceleration is zero, but you want the velocity to be zero not acceleration.

If the spring is compressed by an amount x, then how much work from gravity and friction is done? And if the velocity is zero, then what is the spring-potential-energy?

x amount would be w=1/2 kx² + 48160J.
If velocity is zero then spring potential energy is at the maximum.
Edit
work done = 17200d
max being = 48160J

 

[Nathanael]

x amount would be w=1/2 kx² + 48160J.
If velocity is zero then spring potential energy is at the maximum.
Edit
work done = 17200d
max being = 48160J

Gravity and friction still do work while the spring is being compressed, so it should depend on x.

 

[J-dizzal]

Gravity and friction still do work while the spring is being compressed, so it should depend on x.

would work be (19600+2400)x? because the spring is working against friction? But so is the gravity force.

 

[Nathanael]

would work be (19600+2400)x? because the spring is working against friction? But so is the gravity force.

I'm not sure what you mean by 'the spring works against friction.' Gravity pulls it down (positive work since it's falling) and the spring and friction push it up (negative work).

You had the work done by gravity and friction almost correct before when you put (mg-F_f)d but "d" is not just the distance above the spring, it also depends on how much the spring is compressed.

 

[J-dizzal]

I'm not sure what you mean by 'the spring works against friction.' Gravity pulls it down (positive work since it's falling) and the spring and friction push it up (negative work).

You had the work done by gravity and friction almost correct before when you put (mg-F_f)d but "d" is not just the distance above the spring, it also depends on how much the spring is compressed.

work=Fd=17200(2.8)=48160J.
48160=1/2 kx². but when I solve for x i get .566 its wrong

 

[Nathanael]

work=Fd=17200(2.8)=48160J.
48160=1/2 kx². but when I solve for x i get .566 its wrong

d is not 2.8, d depends on x

 

[J-dizzal]

d is not 2.8, d depends on x

d=2.8m was given. I am using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.

 

[Nathanael]

d=2.8m was given. I am using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.

2.8m is given as the distance between the spring and the bottom of the elevator... But the elevator falls farther than that doesn't it? If the elevator only fell 2.8m then the spring wouldn't be compressed.

 

[J-dizzal]

2.8m is given as the distance between the spring and the bottom of the elevator... But the elevator falls farther than that doesn't it? If the elevator only fell 2.8m then the spring wouldn't be compressed.

That is the distance I am trying to find. I was thinking if the energy of the elevator hits the spring with 48160J the spring must compress and equal that energy when it comes to a stop at distance x in the formula W= 1/2 kx²

 

[SammyS]

d=2.8m was given. I'm using it for the distance to find work down by the falling elevator before it touches the spring. when the spring is fully compressed its work should be the same but opposite.

Adding to what Nathanael posted ...

You have found the work done by gravity and friction in falling prior to contact with the spring. How much work do gravity and friction do over the distance, x, when the spring is being compressed?

(I would have thought that you would have these spring problems licked by now.)

 

[J-dizzal]

Adding to what Nathanael posted ...

You have found the work done by gravity and friction in falling prior to contact with the spring. How much work do gravity and friction do over the distance, x, when the spring is being compressed?

(I would have thought that you would have these spring problems licked by now.)

these problems are licking me.
when the spring is being compressed it does a total of -48160J? because its oppose of the KE

edit; or would it be w=17200x where x is the distance the spring is compressed.

 

[Nathanael]

when the spring is being compressed it does a total of -48160J? because its oppose of the KE

48160J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).
Your mistake is basically the same as your mistake in post #4 of your previous problem.

 

[SammyS]

these problems are licking me.
when the spring is being compressed it does a total of -48160J? because its oppose of the KE

edit; or would it be w=17200x where x is the distance the spring is compressed.

BINGO!

It's the 17200x which you have been neglecting and Nathanael has been hinting at strongly.

 

[J-dizzal]

48160J is the work done on the elevator before the spring is compressed... But as the spring compresses, gravity and friction still act on the elevator (doing additional work).
Your mistake is basically the same as your mistake in post #4 of your previous problem.

.30M N/M = 300000NM?

 

[Nathanael]

.30M N/M = 300000NM?

Yes, 0.3 Mega N/m is 300,000 N/m

 

[J-dizzal]

BINGO!

It's the 17200x which you have been neglecting and Nathanael has been hinting at strongly.

but then i get 2.8m again
48160 +17200x = W, solve for x its 2.8m

 

[Nathanael]

but then i get 2.8m again
48160 +17200x = W, solve for x its 2.8m

By "W" you mean the work done by the spring, right? You should get a quadratic equation.

 

[J-dizzal]

By "W" you mean the work done by the spring, right? You should get a quadratic equation.

W=1/2 k (48160+17200x)x I am really confused.

 

[SammyS]

W=1/2 k (48160+17200x)x I am really confused.

I agree. You're really confused.

Read the last line of Post #11 .

 

[J-dizzal]

I agree. You're really confused.

Read the last line of Post #11 .

was that the correct equation to you?

 

[J-dizzal]

was that the correct equation to you?

so W=1/2 k x² I am not sure how 17200x fits into that

 

[Nathanael]

The spring's potential energy is 0.5kx² and (since the final speed is zero) this is equal to the work done by gravity&friction, which you've found is equal to 17200N times the distance the elevator falls.

Everything in the above sentence makes complete sense right?

So the million dollar question is, how far does the elevator fall? Hint: the answer is not 2.8m

 

[J-dizzal]

The spring's potential energy is 0.5kx² and (since the final speed is zero) this is equal to the work done by gravity&friction, which you've found is equal to 17200N times the distance the elevator falls.

Everything in the above sentence makes complete sense right?

So the million dollar question is, how far does the elevator fall? Hint: the answer is not 2.8m

why 17200N what is the N?

 

[J-dizzal]

why 17200N what is the N?

17200N=1/2 kx² Sorry I am not getting it.

 

[Nathanael]

why 17200N what is the N?

The N just means Newtons. I included it because the equations don't actually have to use SI units to get the correct answer, and the number 17200 is meaningless without a unit. You should really get in the habit of not plugging in numbers until you're finished with the problem (or at least almost finished, because sometimes it gets messy). So instead of writing 17200, I would write (mg-F_f).

 

[J-dizzal]

The N just means Newtons. I included it because the equations don't actually have to use SI units to get the correct answer, and the number 17200 is meaningless without a unit. You should really get in the habit of not plugging in numbers until you're finished with the problem (or at least almost finished, because sometimes it gets messy). So instead of writing 17200, I would write (mg-F_f).

mg-F_f = 1/2 kx²

 

[Nathanael]

mg-F_f = 1/2 kx²

Does this equation mean something to you?

The units are not even the same on both sides. Force is never equal to energy.

Look at the bigger picture and take it slow.

 

[J-dizzal]

Does this equation mean something to you?

The units are not even the same on both sides. Force is never equal to energy.

Look at the bigger picture and take it slow.

(17200N)x=1/2 300000N/m
x=8.721m not sure what I am doing wrong.

edit. i can't put an x² on the right side because then i cannot solve for x

 

[Nathanael]

(17200N)x=1/2 300000N/m
x=8.721m not sure what I am doing wrong.

The units are not even right again.

edit. i can't put an x² on the right side because then i cannot solve for x

First, yes you could still solve for x
Second, you can't just leave out an important part of an equation because it makes it harder to solve!
We are making no progress. Let us start from square one...
Please explain to me your approach to this problem. How are you going to find where the velocity is zero?

 

[J-dizzal]

if the energy of the elevator hits the spring with 48160J how far does the spring compress? 17200x =48160?

 

[J-dizzal]

The units are not even right again.First, yes you could still solve for x
Second, you can't just leave out an important part of an equation because it makes it harder to solve!
We are making no progress. Let us start from square one...
Please explain to me your approach to this problem. How are you going to find where the velocity is zero?

velocity =0 when the spring distance is maximum, or when the work done by the spring = 48160 +17200x where x is the distance of spring compression.

 

[Nathanael]

velocity =0 ... when the work done by the spring = 48160 +17200x where x is the distance of spring compression.

Yes, exactly. Now what is the work done by the spring?

 

[J-dizzal]

Yes, exactly. Now what is the work done by the spring?

48160J +17200x = Wspring

 

[Nathanael]

48160J +17200x = Wspring

Yes but do you know an expression for the work done by a spring being compressed?

 

[J-dizzal]

Yes but do you know an expression for the work done by a spring being compressed?

.5kx²
but when i put them together i get .3211 = x(x-17200) can't go any further.

 

[Nathanael]

.5kx²
but when i put them together i get .3211 = x(x-17200) can't go any further.

Well yes, it's a quadratic equation. You need to use the quadratic formula.
(The trick to solving these equations is called "completing the square," but if you "complete the square" for the general expression ax²+bx+c=0 then you will arrive at the quadratic equation, so you can just use that.)

 

[J-dizzal]

My homework is past due (midnight). i really need to understand this though.

 

[J-dizzal]

Well yes, it's a quadratic equation. You need to use the quadratic formula.
(The trick to solving these equations is called "completing the square," but if you "complete the square" for the general expression ax²+bx+c=0 then you will arrive at the quadratic equation, so you can just use that.)

im getting -8534 and -8665.6 using the quardratic eq. i forgot completing the squares

 

[Nathanael]

im getting -8534 and -8665.6 using the quardratic eq. i forgot completing the squares

Double check yourself, I get 0.6269m

 

[Nathanael]

My homework is past due (midnight). i really need to understand this though.

The work done by the spring is equal to the work done by gravity minus work done by friction. This comes from the work-energy theorem which says the change in kinetic energy (in this case zero) is equal to the net work done.

The work done by the spring is 0.5kx²
The work done by gravity is mgD (where D is the distance fallen)
The work done by friction is F_fD (where D is the distance fallen)

So the equation is 0.5kx²=(mg-F_f)D

The important thing to realize is that D=(2.8+x)

Your mistake was to say D=2.8

 

[J-dizzal]

The work done by the spring is equal to the work done by gravity minus work done by friction. This comes from the work-energy theorem which says the change in kinetic energy (in this case zero) is equal to the net work done.

The work done by the spring is 0.5kx²
The work done by gravity is mgD (where D is the distance fallen)
The work done by friction is F_fD (where D is the distance fallen)

So the equation is 0.5kx²=(mg-F_f)D

The important thing to realize is that D=(2.8+x)

Your mistake was to say D=2.8

Ok thanks, yea I am having trouble tonight.
my eq is 0=x² -17200x - 0.3211 does this look right? i keep getting huge values for x.

 

[Nathanael]

Ok thanks, yea I am having trouble tonight.
my eq is 0=x² -17200x - 0.3211 does this look right? i keep getting huge values for x.

It should be 0=150,000x²-17,200x-48,160

I assume you got your equation by dividing by 150,000? But you forgot to divide the 17200 by 150000.

 

[J-dizzal]

It should be 0=150,000x²-17,200x-48,160

I assume you got your equation by dividing by 150,000? But you forgot to divide the 17200 by 150000.

ok i see now. I am so out of it i can't even do algebra lol. so then part c would be 48160-17200(.6269) J is the energy of the spring pushing and then subtract 48160+friction?

edit. maybe Fs would be a better equation to start with.

 

[Nathanael]

48160-17200(.6269) J is the energy of the spring pushing

It should be a plus sign not a minus. Or you could use 150000(0.6269)^2

For this part you are going to have a change in spring energy, a change in gravitational energy, and the total change in energy will be the work done by friction. The kinetic energy will be zero again.

Maybe you should get some sleep before doing this part

Think about part D as well, that part is more interesting. It will be more satisfying to figure it out yourself so give it some time before asking questions.

 

[J-dizzal]

It should be a plus sign not a minus. Or you could use 150000(0.6269)^2

For this part you are going to have a change in spring energy, a change in gravitational energy, and the total change in energy will be the work done by friction. The kinetic energy will be zero again.

Maybe you should get some sleep before doing this part

Think about part D as well, that part is more interesting. It will be more satisfying to figure it out yourself so give it some time before asking questions.

Ok thanks for all the help again Nathanael. I'll have a new series of problems tomorrow but hopefully will have time to get back at this problem.

 

[SammyS]

Ok thanks for all the help again Nathanael. I'll have a new series of problems tomorrow but hopefully will have time to get back at this problem.

Hopefully Nathanael will be well rested to prepare for this.