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Work energy incline with spring

  • Thread starter Jason03
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  • #2
tiny-tim
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Hi Jason03! :smile:

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
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On the right hand side of the energy conservation equation, you're assuming the object is going to be at a height of 0 (in other words, you need to include the final gravity potential energy on the right hand side, if I'm not mistaken). In addition, you're solving for displacement of spring, displacement isn't already found (on the right hand side you already plugged in 'x'?)
 
  • #4
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so the fact that the spring is compressed 6 in. is not takin into consideration in my equations?
 
  • #5
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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!
 
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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 initial energy of the spring.

You need to find the maximal energy of the spring, and then calculate x from that. :smile:
 
  • #7
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this problem is confusing me, would this work for setting up my equ...?


[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 doesnt make sense to me...
 
  • #8
tiny-tim
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this problem is confusing me, would this work for setting up my equ...?


[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 doesnt make sense to me...
Hi Jason03! :smile:

(btw, don't say "someone" … this is a friendly forum! :smile:)

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

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
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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...
 
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  • #10
tiny-tim
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Hi Jason03! :smile:
....and it is kx^2
Yes … sorry … got confused. :redface:
[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 you have written the equation, is the total compression (so the problem starts with x = .5ft).

Apart from that (and apart from the fact that I don't really understand the units!), the equation looks ok. :smile:
 
  • #11
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didnt I include x in my h by the term (25 sin 20)?........

h = xsin25 = 25sin20 = 8.5 feet
 
  • #12
tiny-tim
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h = xsin25 = 25sin20 = 8.5 feet
sin25? :confused:

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. :smile:
 
  • #13
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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]
 
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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. :smile:
 
  • #15
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The 'h' that you're defining is the vertical distance from the original position and the new compressed position. So, since you have no gravitational potential energy on the right hand side of the equation, this infers that h=0 (relative to 'zero' level). So , zero level must be at the end position of the spring compression. Therefore, on the left hand side of equation, potential energy must become

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?)
 

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