Falling elevator onto a spring.

In summary: J 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... you are forgetting to account for the work done by gravity and friction as the spring compresses.In summary, the elevator cab with a weight of 2000 kg and a constant frictional force of 2.4 kN experiences a snapped cable while at rest on the first floor, where the cab bottom is 2.8 m above a spring with a spring constant of
  • #36
J-dizzal said:
48160J +17200x = Wspring
Yes but do you know an expression for the work done by a spring being compressed?
 
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  • #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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