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It is known that the mechanical energy of a system is conserved if the only forces acting on it are conservative. In class, however, my teacher pointed out that the mechanical energy of an object sliding down a frictionless slope is preserved, despite a non-conservative force (the normal force) acting on it. He said it was okay for non-conservative forces to act on it if were perpendicular to its displacement, i.e. did no work.

So what are the exact requirements for mechanical energy of a system to be conserved? Should only conservative forces be acting on it? Or are non-conservative forces allowed as long as they do no work?

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    $\begingroup$ "He said it was okay for non-conservative forces to act on it if were perpendicular to its displacement, i.e. did no work." That is the exact requirement. ;) $\endgroup$ – Chris Jan 27 '18 at 11:26
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The relation for mechanical energy conservation is a modification of work energy theorem when no work is done by non-conservative forces. Either these non-conservative forces should be absent or must not do any work for our modification to hold true.

The work energy theorem considers all the work by different forces in the left hand side of the equation and the net change in kinetic energy in the right hand side of the equation.

$$ W_{conservative}+W_{non-conservative}=\Delta KE$$

If conservative forces, like gravity, are involved and they do some work we can do a substitution with will give us the typical relation of mechanical energy conservation.

Change is potential energy is nothing but negative of work by conservative forces. When conservative forces, like gravity, do positive work the potential energy of the system decreases. Similarly, when they do negative work the potential energy of the system increases.

$$W_{conservative}=-\Delta U$$ In the absence of non-conservative forces, you can replace $W_{conservative}$ by $-\Delta U$ and make the work energy theorem more specific.

$W_{conservative}=\Delta KE$

$-\Delta U=\Delta KE$

And finally, you have reached the relation.

$\Delta KE + \Delta U=0$

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