Falling elevator onto a spring.

[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.