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At the first-year physics level -- absolutely a normal force can do work. If an object moves parallel (or antiparallel) to a normal force, then the normal force has done work.
But how can that be? A block slides to the left on a flat surface, say. The normal force is upward, movement is left, so the motion is not parallel to the normal force. No work is done by Fn.
That's correct.
Even for a block sliding up or down an incline, the normal force does no work. The normal force by definition is perpendicular to the surface, and the block slides along the surface; no component of the normal force is parallel to the motion.
That's correct, too.
So how in the sam hill can a normal force possibly do any work?
What if the surface itself is moving?
Consider a person standing in an elevator. The normal force is the force of the elevator floor on the person. As the elevator moves upward, so does the person... so the normal force is parallel to the person's motion, doing work. If the elevator moves downward, the normal force is antiparallel to the motion, and so does negative work.
Now I leave you with this... what if the elevator is slowing down as it moves upward, such that the normal force is less than the man's weight. Does the normal force do positive work, negative work, or zero work? I'll answer in a few days in the comment section.
You tried to trick us with that last question. Whether an individual force does positive or negative work only depends on direction of motion. Since the normal force and the direction of motion both point up, we get a positive work done by the normal force. However, since the force of gravity is doing a larger negative work, the net work is negative and net work is what equals a change in kinetic energy.
ReplyDeleteAh, well... There are no "trick" questions in physics, just *tricky* questions. Michael, you nailed it. It's a common student misconception to conflate motion with acceleration when calculating the sign of work, or to conflate net work with work done by a specific force.
ReplyDeleteGCJ
hi Greg, i m a high school student and was having a major doubt in this, plz help
Deletemy question is that we are considering a normal force here.... so what i could get from this discussion was that normal to the surface and not the direction of motion ?! that's why the answer is that yes normal force can do work. But if the force is normal to the direction of motion then also can it do some work ??
Bcz we were taught that if the angle between the force and direction of motion is 90, then work done is zero
PLZZ HELP
SARISHA
Sarisha, a "normal force" by definition means the force of a surface on an object, acting perpendicular to the surface. The normal force does not necessarily act perpendicular to motion - that's the whole point of this post! :-) I see your confusion... hope this helps!
DeleteAlso, doesn't the normal force do work on an object that bounces into and then rebounds off a surface. Initially negative work to bring it to rest then positive work during rebound to give it kinetic energy again?
ReplyDeleteYes to that, too... It's difficult to describe clearly the distance through which the force acts. I've always considered the displacement of the ball's center of mass during the collision, when the ball compresses; this model is consistent with your observation, James, and is sufficient for introductory physics. This is a very confusing question when investigated at a deeper level, because the point of contact doesn't move relative to the surface... or does it?
ReplyDeleteI've been curious about a situation involving normal force doing work for a while, maybe you can help me straighten it out:
ReplyDeletewe know if a block slides down a frictionless ramp, the normal force does not do any work since it is perpendicular to the motion of the block.
And, as you pointed out, if the surface moves, then normal force might do work. For example: if a block slides down a frictionless ramp and the ramp itself is on a frictionless horizontal surface, then the normal force from the block on the ramp would do some work. (a component of that force must be parallel to the motion of the ramp, causing it to slide backwards as the block falls).
The issue I'm having is in understanding whether normal force would be conservative or non-conservative.
I think it must be nonconservative since the path the block takes determines how much work is done: if the block slides down, then some work is done. If the block slides down, then back up, then back down again, more work would be done. Plus, we know the mantra "the only conservative forces in AP Physics 1 are gravity and spring force"
Now, if it's a nonconservative force, then I don't think energy would be conserved in this case which sounds odd. Here's my reasoning:
If we take the system to include the block, the ramp, and the earth, the forces acting in and on the system would be as follows:
- gravity on the block (does work but is conservative)
- gravity on the earth (doesn't really do any work since the earth doesn't really move at all, it's conservative anyway)
- gravity on the ramp (does no work since the ramp does not move vertically)
- normal force from the ramp on the block (no work since it's perpendicular to the block's motion)
- normal force from the block on the ramp (yes work is done and I think it's nonconservative)
- normal force from the horizontal surface on the ramp (does no work since the ramp does not move vertically)
This gives us a single, nonconservative force doing work in the system. We know that work done by nonconservative forces changes the mechanical energy of the system so energy can't be conserved.
I've seen numerous examples that claim energy WOULD be conserved in cases like these and that makes sense from a logical standpoint. But, based on our definitions for work and energy, I don't see how that can be true. Am I missing another nonconservative force? Would the normal force be considered conservative here? It seems more like an elastic collision than anything else.
I know we can use conservation of momentum to analyze situations like this but the normal force has me thrown for a loop. Any ideas?
"what if the elevator is slowing down as it moves upward, such that the normal force is less than the man's weight. Does the normal force do positive work, negative work, or zero work?"
ReplyDeleteWhat's the answer, sir?
Oh, I just saw Troy's comment now, 1.3 years later. Sorry, Troy. The normal force is generally not conservative, because we can't define a potential energy for it. However, if we think of it as a "flex force" due to the springy-ness of the surface molecules -- see a fall 2016 post - then it might be partly conservative. That's why I say the exact details of the ball bouncing off the floor problem are way beyond first year physics.
ReplyDeleteAs for anonymous, I'm happy to answer... first, tell me your thought. It's okay to be wrong, of course, but I want to have context for my response. :-)
Thanks!
greg
I think the normal force does positive work because the force is directed upwards(positive) and also the displacement is in the upwards direction.
ReplyDeleteI have a confusion that does the man standing in the elevator also do some work on the elevator?
What I think is... when the person is standing on the elevator, he pushes the floor of the elevator due to his weight and there is also some displacement so there must be some work done by the man.
For the elevator to move up, it has to apply force more than the weight of the man so here the force exerted by the man(his weight) acts as opposing force like the force of friction in horizontal motion. And as the frictional force does work, the force exerted by the man(due to his weight)
should also do some work(negative work).
When I Google it or ask it on Quora, I don't get satisfactory answers. I hope I will be able to clear my doubts here.:)
And please correct me, if I am wrong somewhere.
Thanks!
Rakshit
(no more anonymous)
Hey, Rakshit... I think you're making this way more complicated than it is. You had the complete explanation in the first sentence.
ReplyDeleteDoes the man do work on the elevator? Sure... the force of the man on the elevator is down, the elevator's displacement is up, so the work done by the man is negative.
The big deal, though, is that the original question asked only about the work done by the normal force -- that is, the work done by the elevator on the man. Everything else is superfluous, and interferes with understanding.
(The only real MISunderstanding here is that force isn't necessary for motion. The elevator can be moving upward regardless of the value of the normal force. If the normal force is equal to the man's weight, the elevator must be moving with constant speed. If the normal force is greater than the man's weight, the elevator slows down. And, if the normal force is less than the man's weight, the man moves up but slows down.
An object which touches a horizontal surface experiences a normal force from that surface. Can that normal force on an object ever do work?
ReplyDeleteI hope you don't mind me joining this rather late. I used to fly aeroplanes and intellectual difficulty arises there also. My problem and, it seems that of others, is this; in the "constant velocity" ascending elevator the normal force exactly equals the "contact force" which is exactly opposite. There is no "net force" because the vector quantities exactly cancel. Even if you allow that the elevator does "positive" work and gravity does "negative" work the two also appear to exactly cancel. How is any net work accomplished? (It obviously is because the system ends up with more potential energy but I can't see that a net force has been moved through a distance, which was always my understanding of the definition of work. I have ignored the acceleration/deceleration which also cancel in a friction-less system). Interestingly (I think!) in the aeroplane example there is an analogous situation in flight at constant velocity; there thrust and drag are equal and opposite and similarly cancel but despite movement of the system there is no change in energy.
ReplyDeleteHamish Griffiths
Hamish, great question! I believe the issue lies with "double-counting" the force of gravity. The general formula for the conservation of energy is:
DeletePE+KE+Wnc=PE+KE
In words that would be: potential energy initial + kinetic energy initial + work done by non-conservative forces equals potential energy final + kinetic energy final
So, where do you consider gravity's energy? As Work =F*d=mg*h? Or as PE=mgh?
If the earth IS NOT part of the system, then there are two non-conservative forces: gravity and normal. Since they point in opposite directions, they cancel (or since one adds energy and the other removes energy, they cancel). Since the earth is not part of the system, then no PE can be stored. The equation boils down to 0=0.
If the earth IS part of the system, then there is one conservative force (gravity from the earth) and one non-conservative force (normal force from elevator). The normal force does positive work as it lifts the person up, and on the other side of the equation, positive PE is gained since the person has been lifted off the ground. The equation boils down to N*d=mgh (with N=mg due to equilibrium and the distance moved is h, so mg*h=mgh).
I hope that makes sense!
Thanks,
-Michael
This comment has been removed by the author.
ReplyDeleteEdited for reasons of previous stupidity!
ReplyDeleteHi Michael. Thank you for your explanation but I'm still struggling; I must be missing something but please indulge me! Surely in your first system, gravity and normal forces are "conservative" and cancel, not as you describe. In your second, gravity is conservative and the elevator force (generated by the motor) is non-conservative. By this I understand that the elevator motor must perform more work than the increase in energy of the system because some is lost as heat, noise, vibration etc. I don’t understand what you mean by “gravity’s energy”. At the end of the ascent the increase in energy, mgh Joules, is the same work performed, F*d, also Joules (less losses in the second case) but I can’t see how F is derived. It still seems like no net work has been done at the top of the elevator journey, but it apparently has. I can't quite picture why just because one force is conservative and the other is not they can't be calculated in the ordinary way for vectors, and therefore sometimes cancelled. In this case I would find it hard to draw a vector diagram showing a force which has moved. Would you be able to do that?
(1) From a force point of view, the two forces (gravity and normal) are equal and opposite and cancel out. When forces cancel this means there is no acceleration - which matches our scenario.
Delete(2) From a Work point of view, the two forces (gravity and normal) are doing the same amount of work, but Normal is doing positive work, and gravity is doing negative work. When Works cancel, there is no change in kinetic energy - which matches our scenario.
(3) From a Potential Energy point of view, as the elevator rises, it gains gravitational potential energy. Also, Normal force is doing positive work. We see that all of the positive work that the Normal force does is stored as gravitational potential energy; so, the elevator's KE hasn't changed, but it has gained potential energy - which matches our scenario.
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The key point is how gravity is considered with respect to energy. It can EITHER be considered as a conservative force (i.e., the work it does is CONSERVED as potential energy as in explanation 3 above), or a non-conservative force (i.e., the work it does is NOT conserved as potential energy as in explanation 2 above). In explanation 2, we don't consider PE related to gravity, and in explanation 3, we don't consider the work related to gravity.
I hope that helps!
I'm coming into this late, but want to point out that you have to be very careful how you define your system. I come via a Modeling perspective (for the last 14 of 22 yrs teaching HS physics), and found system diagrams and a system boundary to help clarify things. These are just ways of keeping track of a) all of the objects which are interacting in a particular situation, b) which ones are included in our system, and c) what forces act ON the system (i.e. cross the system boundary).
ReplyDeletePart of the confusion in Hamish's example is that he hasn't clearly defined his system. You made this point earlier in the thread. If the system is defined as a passenger on an elevator, the only relevant forces are those exerted ON the passenger: Fn on the passenger by the elevator; and gravity, Fg on the passenger by the Earth. Only forces exerted directly on the system can affect its motion and do work on it. Forces exerted BY the system are not relevant, nor is the motor. Trying to consider all of these at once muddles the problem.
For example, since contact forces (everything except gravity & magnetism for our purposes) can only be exerted between objects which are actually in contact, the motor can't do work on the passenger. The motor exerts a tension force Ft on the cable, which exerts a tension force on the elevator, which exerts a normal force on the passenger. If our system is the passenger, the only one of these that counts is Fn. Fn and Fg are thus the only forces acting on the passenger. In order to discuss issues such as whether the two tension forces are equal or not (depends on whether the elevator is accelerating) we would have to redefine our system to be the elevator and/or the cable.
hello from 2019, doesn't there have to be a third force to slow it down?
ReplyDeletehello from 2020,this was very helpful thank you
ReplyDelete