I Why Substitute Force Magnitude in Spring Work Calculation?

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The discussion centers on the substitution of force magnitude in the work calculation for springs, specifically using Hooke's Law (F = -kx). Participants debate why the book suggests substituting the magnitude of force (kx) instead of the restoring force (-kx), leading to confusion about the signs in the integral for work done (W = ∫(-kx dx)). Clarification is provided that in one-dimensional scenarios, the signed force accounts for direction, eliminating the need for cosine. The conversation emphasizes the importance of understanding the physics behind the equations rather than just memorizing them, highlighting the distinction between the vector force and its magnitude. Overall, the discussion underscores the complexities of force direction and work calculations in spring mechanics.
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i read through work done by a spring force derivation and have a simple question about the substitution.
i'm copying from the book...
Hookes Law - F = -kx
W = Fdcos∅
since ∅ is 180°, W = -Fd = -Fx
W = ∫(-Fxdx)
now the book says, from Hookes Law equation "the force magnitude F is kx. Thus, substitution leads to W = ∫(-kxdx)"
why are they saying to substitute the magnitude of the force and not the restoring force of (-kx) resulting in a positive formula in the integral with the two negatives?
 
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The actual formula for work done is <br /> W = \int \mathbf{F} \cdot d\mathbf{x}. In one dimension this is <br /> W = \int F\,dx where F is the signed force, -kx, yielding <br /> W = -\int kx\,dx.
 
pasmith said:
The actual formula for work done is <br /> W = \int \mathbf{F} \cdot d\mathbf{x}. In one dimension this is <br /> W = \int F\,dx where F is the signed force, -kx, yielding <br /> W = -\int kx\,dx.
this formula is missing the cosine of the angle in one dimension
 
dainceptionman_02 said:
this formula is missing the cosine of the angle in one dimension
No it's not - that's what the signed force is. In 1d it's either parallel to dx (+ve sign) or anti-parallel (-ve sign). There are no other options.
 
but Halliday still wrote it in the way that i showed in the original post...
 
One must know what work you are talking about. The work done by the spring will be the negative of the work done on the spring. If one understands the Physics, the sign is clear. This is why understanding the physics is always better than memorizing the equation.
 
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dainceptionman_02 said:
but Halliday still wrote it in the way that i showed in the original post...
Then he isn't working in 1d. That's fine; as is working in 1d with a signed force and no cosine.

The problem here is keeping straight which force you are talking about, which way it's pointing, and which body you're doing work on. The spring exerts a force ##F## in the ##-x## direction on the mass, so the spring does work ##-Fx## on the mass (that's the mass' kinetic energy decreasing). The mass exerts a force ##F## in the ##+x## direction in the spring so the mass does work ##Fx## on the spring (that's the spring's potential energy increasing).
 
A vector in 1D is just a signed number!!
 
I think I can say something usefull here, the problem in the book is probably with the overload of the symbol F: It is used to mean both the vector force and the magnitude of the force. When the book says ##W=Fdx\cos\theta## this F is the magnitude of the force.

As it is well know the work (infinitesimal) is the dot product ##dW=\vec{F}\cdot d\vec{x}=|\vec{F}||d\vec{x}|\cos\theta##.

I mean when you see that ##\cos\theta## in the expression for the work you know that the dot product is expanded so the other symbols must be the magnitude of the vectors.
 
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