Conservation of Energy: Why are Normal Forces Zero?

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SUMMARY

The work done by normal forces is zero due to their perpendicular relationship with displacement. The equation for work, W = Fd cos(θ), indicates that when θ is 90 degrees, cos(90°) equals zero, resulting in zero work. This is further supported by the definition of work as dW = F · dr, where the normal force does not contribute to displacement in the direction of the force. Therefore, the normal force's direction remains unchanged, and the displacement does not align with the force, confirming that the work done is indeed zero.

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  • Understanding of basic physics concepts, particularly forces and work.
  • Familiarity with vector mathematics and dot products.
  • Knowledge of angular motion and its effects on forces.
  • Basic grasp of energy conservation principles in physics.
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  • Study the relationship between normal forces and displacement in various physical scenarios.
  • Learn about vector components and their role in calculating work done by forces.
  • Explore the principles of angular velocity and its impact on force direction.
  • Investigate the conservation of energy in different mechanical systems.
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t2r
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Hello everyone,
someone could explain me please, why the work of the normals forces are 0 ?



He used with conservation energy equations.
How should I refer to the displacement point ?

Thx everyone !
 
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Because the work done by a force is ##W= Fd\cos\theta##. The normal force is perpendicular to the displacement by definition in which case ##\theta =90^o## and ##\cos(90^o)=0##.
 
Well, the work is, by definition ##dW=\vec{F}\cdot d\vec{r}##, and a normal force is, again by definition a force which fulfils ##\vec{F}\cdot d\vec{r}=0##. So the work is zero almost by definition again.
 
The direction of the Normal forces is changed, and also the Δx is not downwards because the angular velocity.

Am I right ?
 

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t2r said:
The direction of the Normal forces is changed, and also the Δx is not downwards because the angular velocity.

Am I right ?
Nop, the direction of the Normal forces is the same before and after, and also the ##\Delta x## is downwards in one case and horizontal in the other. From where you can see that ##\vec{F}\cdot d\vec{r}=0## as we said before.
 
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