- #1

Jason03

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http://i14.photobucket.com/albums/a322/guitaristx/lk.jpg

Heres my work so far...

http://i14.photobucket.com/albums/a322/guitaristx/Scan10026.jpg

Im checking to see if I am headed in the correct direction...

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- Thread starter Jason03
- Start date

- #1

Jason03

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http://i14.photobucket.com/albums/a322/guitaristx/lk.jpg

Heres my work so far...

http://i14.photobucket.com/albums/a322/guitaristx/Scan10026.jpg

Im checking to see if I am headed in the correct direction...

- #2

tiny-tim

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I'm finding it rather difficult to read your work.

I don't see anywhere "let the maximum additional deformation of the spring be x inches" …

- #3

Durato

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- #4

Jason03

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so the fact that the spring is compressed 6 in. is not takin into consideration in my equations?

- #5

Durato

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Sorry, didn't catch that's what you're doing (I was looking for an x in the equation to solve for the displacement). But this is really on the wrong side of the equation. It should be on the left side since initially it is displace this amount.

Here's how I would attempt it. First, write your left hand side equation with initial total energy, including spring, kinetic, and gravity potential. Then, on the left hand side, place your final total energy, which should include spring and gravity potential (no kinetic since its stopped). Take 'X' to be your TOTAL (not change) in compression of spring. Find the height change of the box when it goes 'X' distance down the slope. This will be used to find on the right hand side of equation the new gravitational potential energy. Set it all up, and solve equation!

Here's how I would attempt it. First, write your left hand side equation with initial total energy, including spring, kinetic, and gravity potential. Then, on the left hand side, place your final total energy, which should include spring and gravity potential (no kinetic since its stopped). Take 'X' to be your TOTAL (not change) in compression of spring. Find the height change of the box when it goes 'X' distance down the slope. This will be used to find on the right hand side of equation the new gravitational potential energy. Set it all up, and solve equation!

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- #6

tiny-tim

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so the fact that the spring is compressed 6 in. is not takin into consideration in my equations?

The 6 in is used to calculate the

You need to find the

- #7

Jason03

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[tex] \frac{1}{2}MV^2 + \frac{1}{2}KX^2 + mgh = \frac{1}{2}KX^2 [/tex]

the left side of the equ would be the block at the top of the incline...should there be gravitationl potential on the right side as someone had said earlier?...that doesn't make sense to me...

- #8

tiny-tim

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[tex] \frac{1}{2}MV^2 + \frac{1}{2}KX^2 + mgh = \frac{1}{2}KX^2 [/tex]

the left side of the equ would be the block at the top of the incline...should there be gravitationl potential on the right side as someone had said earlier?...that doesn't make sense to me...

Hi Jason03!

(btw, don't say "someone" … this is a

Yes, that's basically correct … you've followed the proecdure that

Except it shouldn't be [tex] \frac{1}{2}kx^2 [/tex], it should be kx, shouldn't it?

The gravitational potential can be on the left, on the right, or both, depending on where you put your "zero level" of gravitational potential.

Before you go any further, *draw a fresh diagram,* with every number or letter that you're going to use marked clearly … if you try to set up an equation like this in your head without a complete diagram, you'll almost certainly make a mistake.

- #9

Jason03

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I made my "zero level" for gravitational potential at ground level...and it is kx^2

so I came up with this

[tex] \frac{1}{2}(1500\frac{lb}{ft})(.5ft)^2 + \frac{1}{2}(6.2slug)(8\frac{ft}{s}) + (6.2 slug)(32.2\frac{ft}{s^2})(25Sin20) = \frac{1}{2}(1500\frac{lb}{ft})(x)^2[/tex]

so if I solve for x would that be giving me the total compression in x ...or just the compression due to the block hitting it...

so I came up with this

[tex] \frac{1}{2}(1500\frac{lb}{ft})(.5ft)^2 + \frac{1}{2}(6.2slug)(8\frac{ft}{s}) + (6.2 slug)(32.2\frac{ft}{s^2})(25Sin20) = \frac{1}{2}(1500\frac{lb}{ft})(x)^2[/tex]

so if I solve for x would that be giving me the total compression in x ...or just the compression due to the block hitting it...

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- #10

tiny-tim

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...and it is kx^2

Yes … sorry … got confused.

[tex] \frac{1}{2}(1500\frac{lb}{ft})(.5ft)^2 + \frac{1}{2}(6.2slug)(8\frac{ft}{s}) + (6.2 slug)(32.2\frac{ft}{s^2})(25Sin20) = \frac{1}{2}(1500\frac{lb}{ft})(x)^2[/tex]

The question says "bringing the package to rest", so you should assume that the package stays in contact with the spring. That means you have to include x in your h.

so if I solve for x would that be giving me the total compression in x ...or just the compression due to the block hitting it...

x, as

Apart from that (and apart from the fact that I don't really understand the units!), the equation looks ok.

- #11

Jason03

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didnt I include x in my h by the term (25 sin 20)?...

h = xsin25 = 25sin20 = 8.5 feet

h = xsin25 = 25sin20 = 8.5 feet

- #12

tiny-tim

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h = xsin25 = 25sin20 = 8.5 feet

sin25?

x (or x + 0.5) is the compression of the spring. 25 is the original distance of the package.

If the package compresses the spring by x (or x ± 0.5), then that has to be added to the 25 before finding h.

- #13

Jason03

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ohhh i see what your saying...so the angle is 20 and the x traveled is 25 so:

[tex]h = (25+x)sin 20[/tex]

so would that be

[tex] h = 7.5 + xsin20[/tex]

[tex]h = (25+x)sin 20[/tex]

so would that be

[tex] h = 7.5 + xsin20[/tex]

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- #14

tiny-tim

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[tex]h = (25+x)sin 20[/tex]

so would that be

[tex] h = 7.5 + xsin20[/tex]

Yes, except it's [tex]h = 8.5 + xsin20[/tex].

And with that definition of x, you now have to use 1/2k(x+h)^2.

- #15

Durato

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6.2*(32.2)*(25*sin(20)+x*sin(20))

where x = the extra compression of the spring along the slope

This change on the left hand side you have already stated (h = (25+x)*sin(20))

On the right hand side, spring potential energy must become .5*K*(.5+x)^2, because .5 is

the initial compression and x is the extra compression of the spring along the slope.

Oh, and I noticed that in the equation, on the left hand side, you need to square (I think?) the 8 ft/s.

(btw, how do you format equations into neat looking text?)

- #16

Jason03

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and here's the link to how to use LATEX

http://www.mathhelpforum.com/math-help/latex-help/

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