Falling elevator onto a spring.

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The discussion revolves around a physics problem involving a falling elevator cab and a spring. Participants are trying to calculate the speed of the cab just before it hits the spring, the maximum compression of the spring, and the subsequent bounce height of the cab. Key points include the importance of considering both gravitational and frictional forces during the spring's compression, as well as the need to apply conservation of energy principles correctly. The conversation highlights confusion around the equations used and the relationships between work, energy, and spring compression. Ultimately, the participants aim to derive the correct distance the elevator compresses the spring while ensuring all forces are accounted for.
  • #31
J-dizzal said:
(17200N)x=1/2 300000N/m
x=8.721m not sure what I am doing wrong.
The units are not even right again.

J-dizzal said:
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?
 
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  • #32
if the energy of the elevator hits the spring with 48160J how far does the spring compress? 17200x =48160?
 
  • #33
Nathanael said:
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.
 
  • #34
J-dizzal said:
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?
 
  • #35
Nathanael said:
Yes, exactly. Now what is the work done by the spring?
48160J +17200x = Wspring
 
  • #36
J-dizzal said:
48160J +17200x = Wspring
Yes but do you know an expression for the work done by a spring being compressed?
 
  • #37
Nathanael said:
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.
 
  • #38
J-dizzal said:
.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.)
 
  • #39
My homework is past due (midnight). i really need to understand this though.
 
  • #40
Nathanael said:
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
 
  • #41
J-dizzal said:
im getting -8534 and -8665.6 using the quardratic eq. i forgot completing the squares
Double check yourself, I get 0.6269m
 
  • #42
J-dizzal said:
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 FfD (where D is the distance fallen)

So the equation is 0.5kx2=(mg-Ff)D

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

Your mistake was to say D=2.8
 
  • #43
Nathanael said:
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 FfD (where D is the distance fallen)

So the equation is 0.5kx2=(mg-Ff)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.
 
  • #44
J-dizzal said:
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.
 
  • #45
Nathanael said:
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.
 
  • #46
J-dizzal said:
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 :smile:

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.
 
  • #47
Nathanael said:
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 :smile:

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.
 
  • #48
J-dizzal said:
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. :smile:
 
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