- #26

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..and I don't know what else to use to show that work is a scalar.

Now this is going in circles, it's a scalar because it's defined to be a scalar. [gives up]

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- Thread starter Forrest T
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- #26

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..and I don't know what else to use to show that work is a scalar.

Now this is going in circles, it's a scalar because it's defined to be a scalar. [gives up]

- #27

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The theorem is particularly simple to prove for a constant force acting in the direction of motion along a straight line. For more complex cases, however, it can be claimed that very concept of work is defined in such a way that the work-energy theorem remains valid.

Why insist on understanding work without any reference to energy? Energy is the more fundamental quantity.

- #28

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1) Dot product of two vectors is scalar, thus work is scalar.

2) Work is not vector; you could not write it as Wx+Wy+Wz because IT ISN'T VECTOR.

3) Yes...is that clear

- #29

jtbell

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In the 1700s, there was a lot of confusion about whether the "proper" quantity that combines mass and velocity contains v (momentum) or v^2 (kinetic energy). Finally physicists realized that we need both of them, for different purposes.

We define work the way we do, so as to have a conserved quantity via the work-energy theorem:

[tex]\int{\vec F \cdot d \vec s} = W = \Delta K = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2[/tex]

When we integrate force over time (instead of over distance) we get the change in momentum, via the impuse-momentum theorem:

[tex]\int{\vec F dt} = \vec I = \Delta \vec p = m \vec p_f - m \vec p_i[/tex]

- #30

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@JeffKoch

I understand what you're saying, but...

InThe Feynman Lectures on Physics, Volume I, an exploration of torque produces a motivation for the definition of the vector (cross) product. I think there is a similiar procedure with work, but I don't knowwhywork is defined as a scalar. The work-energy theorem in one dimension is not useful for this, and I don't know what else to use to show that work is a scalar.

So you asking why work is scalar and why dot product is a scalar? Hard to resolve this without falling back to the definition of energy and that energy is a scalar (therefore it doesnt matter if we have kinetic energy on the x,y or z axis we just add em to find the total energy). If work wasnt a scalar how it could be a vector what

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