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Say I have an object sliding down a frictionless curved slide. Then conventional wisdom says that all gravitational potential energy gets converted to Kinetic energy, allowing us to conveniently find velocity.

Question: Since there is also a normal force (otherwise the object just falls through the slide!) this normal force also contributes to the work done. So it's not obvious that this conventional wisdom holds.

I have shown that if we decompose the gravity force into a normal and tangential component, and assume that the normal force cancels out the normal component (so the net force is the tangential component of gravity) then this conversion of energy indeed holds. But I was wondering why most authors never address this, and is there something I am missing.

Bonus question: The same question can be asked for pendulums and tension (playing the role of normal force).

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  • $\begingroup$ Please show your proof. $\endgroup$
    – velut luna
    Oct 25, 2017 at 11:07

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Normal force is a constraint force. The aim of this normal force in your example is to keep the object in contact with the slide. The normal force is always perpendicular to the displacement of the object and hence does no work.

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Since there is also a normal force [...] this normal force also contributes to the work done

To do work, the force must make something move. There mere fact that a force is present does not imply work being done (a table does no work on the apple it is holding up in spite of the constant normal force).

If a train moves in a rail and you jump out and push it from the side, then nothing happens. The train continues moving forwards and gains no sideways motion. You applied a force, but you did no work because you made no motion happen in the direction you were pushing.

Similarly in your example, the normal force is perpendicular to the motion down the incline. The object does not start lifting perpendicularly off from the incline, so this normal force causes no motion in this direction. It just balances out the other forces that a present along this perpendicular direction, in the same way that the table balances out the weight of the apple.

The apple could be moving over the table; maybe a sideways force does work on it to cause acceleration that creates this motion. But all that would make no difference for the normal force. It only cares about holding up the apple exactly as before - no work is done is there is no motion along the force direction, regardless of motion taking place along other directions. This is encompassed in the work formula:

$$W=\vec F \cdot \vec s$$

The dot product mathematically only takes parallel components into account. Are the vectors perpendicular, then it is zero.

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  • $\begingroup$ Thanks! My username probably indicates that the apple and train examples don't help, but N.ds=0 explains it all :) $\endgroup$ Oct 25, 2017 at 11:20

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