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Spring Force Sign Question

  • #1
I am sorry if I am doing anything wrong. I'm new to this site and physics in general. Please feel free to correct me if I'm making some stupid mistake in formatting.

1. The first figure gives spring force Fx versus position x for the spring–block arrangement of the second figure. The scale is set by Fs = 160 N. We release the block at x = 12.0 cm. How much work does the spring do on the block when the block moves from xi =+9.0 cm to (a)x=+6.0 cm, (b)x=-6.0 cm, (c)x=-9.0 cm, and (d)x=-10.0 cm?


2. Homework Equations
I am probably missing an equation here, but here are the ones I used:

F = kx
Work = -.5k(Δx)^2​

3.
Okay, first I converted everything to meters, which just ended up moving the decimal two places left.

Second, I found k. (Force is Kg*m/(s*s), x is meters, so k is in Kg/(s*s), which is kinda a weird unit, but whatever)

F = kx
k = F/x
k = 160/.02
k = 8000​

Third, I plugged in the numbers for the work equation.

Work = -.5k Δx
-.5k = -.5*8000 = -4000
Δx = xf - xi
Work = -4000(xf - .09)^2​

Lastly, I plugged in the four ending points. (x's are in meters)

Axf = .06-----> Wa = -3.6 Joules
Bxf = -.06----> Wb = -90 Joules
Cxf = -.09----> Wc = -129.6 Joules
Dxf = -.1-----> Wd = -144.4 Joules​



Okay, as I was writing this up, I found out that I had made a stupid mistake in my calculations (I didn't square delta x in the work equations). However, I'm on my last shot on this online homework, and I want to get it RIGHT. So are these values correct? Or should they be positive? (I thought Joules were always positive, but I don't think I missed a minus anywhere...)
Thank you for your help!
 

Answers and Replies

  • #2
Doc Al
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F = kx
Work = -.5k(Δx)^2​
Careful! That's the work done by the spring when it is stretched a distance Δx from its unstretched position, not for an arbitrary Δx.

3.
Okay, first I converted everything to meters, which just ended up moving the decimal two places left.

Second, I found k. (Force is Kg*m/(s*s), x is meters, so k is in Kg/(s*s), which is kinda a weird unit, but whatever)
From F = ma, you should recognize Kg*m/(s*s) as equivalent to a Newton.

F = kx
k = F/x
k = 160/.02
k = 8000​
OK.

Third, I plugged in the numbers for the work equation.

Work = -.5k Δx
-.5k = -.5*8000 = -4000
Δx = xf - xi
Work = -4000(xf - .09)^2​
No. Using your equation, you want ΔW.

You'll need to redo your calculations.

(I thought Joules were always positive, but I don't think I missed a minus anywhere...)
Joule is a unit of work/energy. Work can be positive or negative.
 
  • #3
Thanks! Could you tell me why, exactly, I would be solving for the change in work? I know you mentioned this:
Careful! That's the work done by the spring when it is stretched a distance Δx from its unstretched position, not for an arbitrary Δx.
But how do I know that I'm not supposed to do what I did?


Additionally, how do you solve for the change in work? (I know that ΔW is Wfinal - Winitial, but what x's make up those works? Would it be that the Winitial Δx would be .12-.9? Or am I wrong, again?)
 
Last edited:
  • #4
Doc Al
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44,912
1,170
Thanks! Could you tell me why, exactly, I would be solving for the change in work? I know you mentioned this:


But how do I know that I'm not supposed to do what I did?


Additionally, how do you solve for the change in work? (I know that ΔW is Wfinal - Winitial, but what x's make up those works? Would it be that the Winitial Δx would be .12-.9? Or am I wrong, again?)
I think you'll have an easier time if you think in terms of the elastic potential energy stored in the spring. That is given by:
PE = .5kx2, where x is the displacement from equilibrium.

The work done by the spring in moving from one position to another will equal -ΔPE. (If the PE increases, that means the spring has done negative work.)

Note that Δ(y2) ≠ (Δy)2.
 
  • #5
...We haven't gone over potential energy yet; so I'm a bit clueless on what you just said there. Sorry about that.
 
  • #6
Doc Al
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...We haven't gone over potential energy yet; so I'm a bit clueless on what you just said there. Sorry about that.
No worries. In that case you'll just have to use your formula, but with the caveat that I explained in my earlier post.

For example, you'll first have to calculate the work done by the spring when stretched to +9.0 cm. Then compare that to the work done when stretched to +6.0 cm. And so on.

A most important tip is:
Note that Δ(y2) ≠ (Δy)2.
 
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  • #7
Ahh okay thank you very much!
 

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