Understanding Work Calculations: Clarifying the Role of Forces and Intuition

In summary: But this is exactly the problem. I stick to the definition and I don't philosophise but I keep reading on PF and elsewhere that people do get confused. I am certain that this confusion because, for instance:The floor is doing negative work on the box via friction....the erroneous belief that friction is the source of energy transferred.Friction does not provide energy, it only transfers it.The box is not doing any work on the static floor, indicating that all the energy taken from the box by the negative work is dissipated by friction.The difference in Energy is described using the term Efficiency. The question of 'who dunnit?' need not apply. Perhaps this is the problem; people have to look for an agency instead
  • #1
brawler234
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A box is moving at a constant velocity. 250N of applied force is acting on a box while 250N of friction is acting on it on the opposite direction. If the box travels 4 meters in the same direction as the applied force, how much work is done?

I know that work = force * displacement.

Is the force in the equation supposed to be the net force (which would make work 0J) or just the applied force in the same direction of motion (which would make work 1000J)?

[Moderator's note: Moved from a technical forum and thus no template.]
 
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  • #2
brawler234 said:
Is the force in the equation supposed to be the net force (which would make work 0J) or just the applied force in the same direction of motion (which would make work 1000J)?
The question should state which work is to be computed.
 
  • #3
A.T. said:
The question should state which work is to be computed.

Hmm... so there is work done by the box pusher as well as work done by the friction of the floor?
 
  • #4
brawler234 said:
Hmm... so there is work done by the box pusher as well as work done by the friction of the floor?
Yes, and there is the net work done on the box, which the sum of them.
 
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  • #5
This is the standard situation for practical moving objects, like cars, boats and sledges.
The work done by the pusher is 250 x 4 = 1000 J. This energy is then transferred to heat resulting from friction.
In any case where the vehicle is not accelerating, the work done is entirely expended in heat.
You don't have to consider the force of reaction in this calculation.
 
  • #6
A.T. said:
Yes, and there is the net work done on the box, which the sum of them.
Work done 'by' and 'on' is always problematical. I have difficulty about work done 'by' friction because I feel that friction is not a source of Energy. It's an intermediate Force in an Energy Transfer process and, when Work is done via a friction mechanism, there is Energy input and Energy output. The distance moved by an Effort Force may be the same as the distance moved by a Load Force but the two forces will not be the same so there is Energy deficit.
The difference in Energy is described using the term Efficiency. The question of 'who dunnit?' need not apply. Perhaps this is the problem; people have to look for an agency instead of just describing a process.
 
  • #7
sophiecentaur said:
I have difficulty about work done 'by' friction because I feel that friction is not a source of Energy. It's an intermediate Force in an Energy Transfer process
Why is this a problem? Work is defined as a transfer of energy so since friction transfers energy it does work. There is no implication that friction is the source of the energy transferred.
 
  • #8
sophiecentaur said:
Work done 'by' and 'on' is always problematical. I have difficulty about work done 'by' friction because I feel that friction is not a source of Energy. It's an intermediate Force in an Energy Transfer process and, when Work is done via a friction mechanism, there is Energy input and Energy output. The distance moved by an Effort Force may be the same as the distance moved by a Load Force but the two forces will not be the same so there is Energy deficit.
The difference in Energy is described using the term Efficiency. The question of 'who dunnit?' need not apply. Perhaps this is the problem; people have to look for an agency instead of just describing a process.
There is nothing problematical if you stick to the definition, and don't philosophize so much.

The floor is doing negative work on the box via friction. The box is not doing any work on the static floor, indicating that all the energy taken from the box by the negative work is dissipated by friction.
 
  • #9
Dale said:
There is no implication that friction is the source of the energy transferred.
That's where we must disagree. You have a loss mechanism that is supplying energy. Don't you see how that isn't straightforward.

A.T. said:
There is nothing problematical if you stick to the definition, and don't philosophize so much.
But this is exactly the problem. I stick to the definition and I don't philosophise but I keep reading on PF and elsewhere that people do get confused. I am certain that this confusion because, for instance:
A.T. said:
The floor is doing negative work on the box via friction.
. . . which is of course quite correct. But there is a dissonance there which represents a hurdle, not, of course when you 'get it'; I am talking about people who don't get it.

I guess all I would really want is for the definition of Work to be used straight and the direction of energy flow noted, when necessary, without the by and on. Anyone who isn't aware of the problems that 'people' tend to have with the implied agency involved should just read the confusion that's found in many posts about the way friction works. It is a mechanism that both supplies energy and dissipates energy at the same time. Isn't that a potential source of confusion?

Walking and driving a car are both topics that appear regularly and the by and on question is always in there somewhere and gets in the way. That's basically my evidence that people need help with this.
 
  • #10
sophiecentaur said:
I guess all I would really want is for the definition of Work to be used straight and the direction of energy flow noted, when necessary, without the by and on.
The definition of work involves a force applied over a distance. To apply that definition you have to specify which force (exerted by what on what) is meant.
 
  • #11
sophiecentaur said:
That's where we must disagree. You have a loss mechanism that is supplying energy. Don't you see how that isn't straightforward.
I do see that it isn’t straightforward but if you use the correct terms then it helps. For instance here you say “a loss mechanism that is supplying energy”. I would say “a loss mechanism that is transferring energy”. Do you see how that helps? “Supplying” sounds like it is a source, but “transferring” makes no such implication.

Anyone who has purchased something using PayPal should be able to understand the idea of a mechanism that transfers a quantity with some loss without being a source.
 
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  • #12
[I deleted a comment I made in this space. Let me reword it a little, and I'll repost.]
 
  • #13
[This is nearly identical to my original wording. Only very minor changes were made.]

Work and kinetic energy are ultimately relative*; they depend on the frame of reference (FoR).

*(Although this applies to Einstein's relativlity too, let's just stick to Galilean relativity for this problem.)

In this problem, assuming a FoR stationary with the ground, the dot product between the applied force and the block's displacement is positive, meaning the work is being done by the applied force, and that work is done on the force of friction.

But one could just as easily define a new FoR to be the same velocity as the block's velocity. In that case, no net work is being done on or by the block, because the block's displacement is zero (the block isn't moving). But the ground is moving. So here, since the ground's displacement is in the same direction as the force of friction (on the block), the work is done by the force of friction. And the work is ultimately done on the applied force (since the applied force is ultimately in the opposite direction as the ground's movement).

So it doesn't really matter what is, and what is not, considered an "energy source," but rather the important part is what one chooses as one's FoR. Once a FoR is chosen, one can determine "by" and "on" by the following:
  • If the dot product between the force and displacement is positive, the force does positive work; the work is done by the force. (If it helps to remember this, just remember, "if work was easy, they wouldn't call it 'work.'")
  • If the dot product between the force and displacement is negative, the force does negative work; the work is done on the force.
 
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  • #14
Dale said:
I would say “a loss mechanism that is transferring energy”.
Yes. Personally, I like to use the word "transfer" but it is common for people to use or imply "supplying" - which is an 'agency' word. I guess it's something to do with the causal chain idea and a word like "supplying" implies a causal direction.

I just read post #3 again and it makes my point totally.
brawler234 said:
Hmm... so there is work done by the box pusher as well as work done by the friction of the floor?
That post implies to me that @brawler234 is uneasy about the notion of two sources of the same energy. But that's not always a problem. When you pull a string, you do work on it and the string does work on the toy; there's no explicit loss in the system and that's inherently different from the use of friction when walking across mud because (choice of FOR again) although we 'know' the energy chain direction, it is not blindingly obvious that you have to include the vectors in order to know that happens with the energy. (Not an obvious thing for non-Physicists to do)
Don't get me wrong. I don't have a problem with this particular stuff but I am constantly recognising where other people seem to. And the reason they have the problem is that "friction" is a loaded word in English. Friction slows things down and it has no earthly business speeding things up. That idea looms all the time.
 
  • #15
sophiecentaur said:
Friction slows things down and it has no earthly business speeding things up. That idea looms all the time.
But that's a misconception about friction, not about work.
 
  • #16
sophiecentaur said:
Friction slows things down and it has no earthly business speeding things up.
It all depends on your chosen frame of reference (FoR). Think about what happens when you put a "stationary" block on a "moving" conveyor belt.
 
  • #17
collinsmark said:
It all depends on your chosen frame of reference (FoR). Think about what happens when you put a "stationary" block on a "moving" conveyor belt.
The English Langauge doesn't use frames of reference and nor does peoples' intuition. That sentence of mine explains why people have such problems with friction; it doesn't represent what I think personally. If all you need to do is use the phrase FoR to get things straight, how is it that work and friction are such popular topics of confused questions on PF?
Just stop anyone in the street and ask them what friction does. That is my point. It's not Physics, it's language.
 
  • #18
A.T. said:
But that's a misconception about friction, not about work.
Yes. That's true in the strict sense and I am by no means trying to argue against the Physics. My opinion is that it gives people more problems because of the language that is used.
Of course, anyone who can be dispassionate about the topic and go along with the formal approach won't have a problem. But, as I wrote above, there are many posts which indicated that there is a problem and it would be a help if the wording were modified in order to avoid the misdirection.
 
  • #19
sophiecentaur said:
Just stop anyone in the street and ask them what friction does.
There are many words in physics which also appear in common language, leading to naive misinterpretations: force, energy etc.
sophiecentaur said:
... it would be a help if the wording were modified in order to avoid the misdirection.
Good luck convincing physicists to rename all that.
 
  • #20
A.T. said:
rename all that.
Straw man, I'm afraid. All I want is to leave out the On and By adverbs. Isn't Science all about change?
 
  • #21
sophiecentaur said:
All I want is to leave out the On and By adverbs.
See post #10.
 
  • #22
A.T. said:
which force (exerted by what on what)
You already brought up Newton 3 so the same magnitude force is involved. The direction defines the flow of energy. Why is by and on involved necessarily - although it could be handy for convenience. All your statements on this topic assume that people have the same level of appreciation of the situation that you do. My point is that they don't. Perhaps they 'should' ?

Moreover, if you take a snapshot of just part of a power chain and an arbitrary FoR, the by and on is unknowable. Is the car clutch being a driver or a brake? Which direction does on or by apply to?
 
  • #23
sophiecentaur said:
Which direction does on or by apply to?
The definition of work involves a force. If that force is exerted by A on B, then the work calculated using that force is the work done by A on B.
 
  • #24
A.T. said:
The definition of work involves a force. If that force is exerted by A on B, then the work calculated using that force is the work done by A on B.
It seems to me that the energy transferred by kinetic friction could be usefully divided into three piles:

1. The work done by friction on B. i.e. the frictional force of A on B multiplied by the incremental motion of B's contact surface in your chosen lab frame. This will often be negative and could be phrased as "mechanical energy drained from B by friction".

2. The work done by friction on A. i.e. the frictional force of B on A multiplied by the incremental motion of A's contact surface in your chosen lab frame. This will often be negative as well and could be phrased as "mechanical energy drained from A by friction".

3. The dissipated energy. i.e. the frictional force of either on the other multiplied by the relative motion of the surfaces. Pick any sign convention you like so that this comes out non-negative.

If you add up the three, the result should come to zero. Energy is conserved.
 
  • #25
jbriggs444 said:
If you add up the three, the result should come to zero. Energy is conserved.
Right, so you can compute 3. directly from 1. and 2. without the need to pick a sign.
 
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  • #26
A.T. said:
There is nothing problematical if you stick to the definition, and don't philosophize so much.
A.T. said:
The definition of work involves a force applied over a distance. To apply that definition you have to specify which force (exerted by what on what) is meant.
A.T. said:
But that's a misconception about friction, not about work.
A.T. said:
The definition of work involves a force. If that force is exerted by A on B, then the work calculated using that force is the work done by A on B.
A.T. said:
Right, so you can compute 3. directly from 1. and 2. without the need to pick a sign.
All those statements are fine and I wouldn't disagree with them but they only work in a world where no one has a problem with their intuition. Of course we can do the Maths and come out with an answer. Many of us have no conscious problems with applying strict logic to every situation we encounter. But the fact is that an awful lot of people who post about this topic do have problems with it. Is there really any point in telling them that they just shouldn't? Isn't one of the aims of PF to clear up this sort of thing, rather than just re-stating what is 'obvious' to the cognoscenti?
The OP has asked a question in terms of someone who doesn't 'just get' the problem and would probably appreciate more than just a reiteration of the textbook line.
 

1. What is the formula for calculating work done?

The formula for calculating work done is W = F * d * cos(theta), where W represents work, F is the applied force, d is the displacement, and theta is the angle between the force and the displacement.

2. How is work done related to energy?

Work done is directly related to energy, as it is the transfer of energy from one system to another. When work is done on an object, energy is transferred to that object, and when work is done by an object, energy is transferred away from that object.

3. Can the work done be negative?

Yes, the work done can be negative. This occurs when the applied force and the displacement are in opposite directions, resulting in a negative value for work. This indicates that energy is being transferred away from the system.

4. What is the unit of measurement for work done?

The unit of measurement for work done is joules (J). This is equivalent to a force of 1 newton (N) acting over a distance of 1 meter (m).

5. How can work done be calculated for a non-constant force?

If the applied force is not constant, work done can be calculated by breaking the displacement into small intervals and calculating the work done for each interval using the formula W = F * d * cos(theta). The total work done will be the sum of the work done for each interval.

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