Why is there no work for someone walking on flat plane

In summary, the conversation discusses the definition of work, which is the product of force and displacement vectors. It is represented as |F|*|d|cosθ, where θ is the angle between the two vectors. When θ = 90, cos90=0 and the work done is 0. This is because work is only done in the direction of the force. The conversation also explains the difference between cos and sin in the vector dot product and how it relates to finding the work done. The example of a person walking is used to demonstrate the concept of work and how it is dissipated. However, the human body is not an efficient system to use for this concept.
  • #36
haruspex said:
It's not the shape, more the rigidity.
Real world objects are not completely rigid. A tyre deforms as it rotates, producing a flat area in contact with the road. This constant deformation takes work. But rolling resistance is not limited to that. It might be mostly that, but the term is a catch-all encompassing other losses as such axle friction.
I was referring to, rolling friction being illuminated, but still the force to accelerate the mass translationally and rotationally and then to decelerate it, would be needed to calculate the work done over time. (power).
 
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  • #37
zanick said:
I was referring to, rolling friction being illuminated,
I guess you mean eliminated.
zanick said:
but still the force to accelerate the mass translationally and rotationally and then to decelerate it, would be needed to calculate the work done over time. (power).
You've lost me.
 
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