- #26

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17200N=1/2 kxwhy 17200N what is the N?

^{2}Sorry im not getting it.

- Thread starter J-dizzal
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- #26

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17200N=1/2 kxwhy 17200N what is the N?

- #27

Nathanael

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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-Fwhy 17200N what is the N?

- #28

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mg-FThe 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}).

- #29

Nathanael

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Does this equation mean something to you?mg-F_{f}= 1/2 kx^{2}

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.

- #30

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(17200N)x=1/2 300000N/mDoes 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.

x=8.721m not sure what im doing wrong.

edit. i cant put an x

- #31

Nathanael

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The units are not even right again.(17200N)x=1/2 300000N/m

x=8.721m not sure what im doing wrong.

First, yes you could still solve for xedit. i cant put an x^{2}on the right side because then i cannot 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?

- #32

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- #33

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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.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?

- #34

Nathanael

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Yes, exactly. Now what is the work done by the spring?velocity =0 ... when the work done by the spring = 48160 +17200x where x is the distance of spring compression.

- #35

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48160J +17200x = WspringYes, exactly. Now what is the work done by the spring?

- #36

Nathanael

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Yes but do you know an expression for the work done by a spring being compressed?48160J +17200x = Wspring

- #37

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.5kxYes but do you know an expression for the work done by a spring being compressed?

but when i put them together i get .3211 = x(x-17200) cant go any further.

- #38

Nathanael

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Well yes, it's a quadratic equation. You need to use the quadratic formula..5kx^{2}

but when i put them together i get .3211 = x(x-17200) cant go any further.

(The trick to solving these equations is called "completing the square," but if you "complete the square" for the general expression ax

- #39

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

- #40

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im getting -8534 and -8665.6 using the quardratic eq. i forgot completing the squaresWell 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 ax^{2}+bx+c=0 then you will arrive at the quadratic equation, so you can just use that.)

- #41

Nathanael

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Double check yourself, I get 0.6269mim getting -8534 and -8665.6 using the quardratic eq. i forgot completing the squares

- #42

Nathanael

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

The work done by the spring is 0.5kx

The work done by gravity is mgD (where D is the distance fallen)

The work done by friction is F

So the equation is 0.5kx

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

Your mistake was to say D=2.8

- #43

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Ok thanks, yea im having trouble tonight.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 0.5kx^{2}

The work done by gravity is mgD (where D is the distance fallen)

The work done by friction is F_{f}D (where D is the distance fallen)

So the equation is 0.5kx^{2}=(mg-F_{f})D

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

Your mistake was to say D=2.8

my eq is 0=x

- #44

Nathanael

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It should be 0=150,000xOk thanks, yea im having trouble tonight.

my eq is 0=x^{2}-17200x - 0.3211 does this look right? i keep getting huge values for x.

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

- #45

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ok i see now. im so out of it i cant 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?It should be 0=150,000x^{2}-17,200x-48,160

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

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

- #46

Nathanael

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It should be a plus sign not a minus. Or you could use 150000(0.6269)^248160-17200(.6269) J is the energy of the spring pushing

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.

- #47

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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.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.

- #48

SammyS

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Hopefully Nathanael will be well rested to prepare for this.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.

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