Why Substitute Force Magnitude in Spring Work Calculation?

[dainceptionman_02]

TL;DR

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?

 

[pasmith]

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.

 

[dainceptionman_02]

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

 

[Ibix]

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.

 

[dainceptionman_02]

but Halliday still wrote it in the way that i showed in the original post...

 

[hutchphd]

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.

 

[Ibix]

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).

 

[hutchphd]

A vector in 1D is just a signed number!!

 

[Delta2]

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.