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Is the NET work done on a system, by Normal Force always ZERO?

Hmm, it's interesting! I'm sure about the static friction, it's net work is zero on a system. But, is the same true for normal force also?

I tried to think of a contradiction, and here's what I've got. Consider a partially elastic collision between a ball and a fixed wall. The ball rebounds in the opposite direction, and there seems no other force working on the ball except the normal from the wall. And, since the wall is fixed, the work by Normal from the ball to the wall is zero.

The normal from the wall, does work on the ball, but the reaction normal from the ball to the wall dosen't. Hence net work done $\ne 0$.

These are my thoughts. Please elaborate and feel free to point out any errors.

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    $\begingroup$ So you are saying the contradiction is that the normal force from the wall to the ball produces a work, but the normal force from the ball to the wall doesn't produce work? $\endgroup$ Commented Dec 20, 2021 at 11:26
  • $\begingroup$ Also both Normal forces and static forces can do work IIRC, an example of the first would be standing in an elevator (where motion is vertical, just like the normal force) and for second should be standing in a bus (where the motion is horizontal, just like the frictional forces) $\endgroup$ Commented Dec 20, 2021 at 11:28
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    $\begingroup$ Alright, thanks. Actually, what I meant by NET work is, the work done by the force we're interested in, in the whole system where the force acts. Am I being vague… For example, consider the static friction example. The static friction does positve work on the man, but it does an equal and opposite work on the bus too, making the NET work zero. I've actually read in "Halliday and Resnick, Vol.1" that the NET work done by static friction is always Zero. I was just thinking, is it the same for a normal force? $\endgroup$
    – King
    Commented Dec 21, 2021 at 1:20

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You are right that work is done on the ball, the force $F$, acting to the right, moves through the distance $-d$ and does work according to 'Work done = Force x distance' of $Fd$.

It reduces the Kinetic Energy of the ball from $10$J (for example) to zero.

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But, by Newton's 3rd law, there will be the same force acting to the left, on the wall.

It also moves through the same distance and causes a gain in potential energy in the wall, as it bends slightly. The energy stored in the wall is $Fd$ and the total work done by the pair of normal forces is $Fd-Fd=0$J.

So, if the 'system is just the ball, there is a net work done by the normal force, in the above example. However if you include both of the forces in the 'system', the net work done has to be zero due to Newton's 3rd Law.

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  • $\begingroup$ Thanks a lot. But what if the wall is perfectly rigid and the ball is compressible? $\endgroup$
    – King
    Commented Dec 21, 2021 at 6:46
  • $\begingroup$ @King good question, but in reality there has to be some compression so that the K.E. becomes stored P.E. at the moment when the ball is stopped, that then becomes K.E. again (due to motion in the opposite direction) later on. $\endgroup$ Commented Dec 21, 2021 at 8:09
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Is the NET work done on a system, by Normal Force always ZERO?

Hmm, it's interesting! I'm sure about the static friction, it's net work is zero on a system. But, is the same true for normal force also?

The term "net work" usually refers to the dot product of the net force and the displacement of the center of mass. Since it refers to the net force it does not make sense to ask about the net work done by an individual force. Net work is done by the net force.

However, if you simply mean the work done by the normal force, then there are indeed many instances where the normal force does work. The easiest example is in an elevator. The normal force on the bottom of the feet of the passengers does work and changes both the KE and the PE of the passengers. Static friction can also do work. For example, for a box in the bed of an accelerating truck, the static friction force acting on the box does work on the box thereby increasing the box's KE.

The key is that for the normal force or the static friction force to do work on an object the surface must be moving. There are many examples of moving surfaces, so such examples abound.

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  • $\begingroup$ By net work done by normal or by static friction the OP meant the net work done by all normal forces in the system, action and reaction pairs included, which would indeed be zero in both the examples given by you. $\endgroup$ Commented Sep 29, 2022 at 1:21

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