# Homework Help: Work, Net work, Work done by friction

1. Jan 23, 2013

### dcasplen

1. The problem statement, all variables and given/known data
I am looking for some clarity on what should be general conceptual questions involving work, net work, and work done by friction. In General;

W[F]=the component of the applied force acting in the direction of motion X distance OR
Δenergy

W[friction]=μN and is opposite the direction of motion

W[net]= W[F] - W[f]

If the object is moving at a constant velocity, there is no net work (i.e. W[F]=W[f])

If there is net work there must be an acceleration

Finally, when work is done to overcome gravity, this work is just equal to the the force of gravity (mg for force and mgh for work). I shouldn't concern myself with initial acceleration when the object is lifted and deceleration as the object comes to rest because this is factored into an average acceleration? In a case such as this, the applied force does the work W=mgh. Does gravity do any work? or does gravity only do work when the object falls some h?

2. Relevant equations

W=F*d
W[f]=μN
W=ΔE

3. The attempt at a solution

Thanks in advance. I'm making sure i have these concepts straight. I've been seeing questions that give necessary parameters of a problem but state that the velocity is constant and ask for the net work. I feel like no calculations are necessary but would like to confirm that i do understand the terms work, net work, and work due to friction.

2. Jan 23, 2013

### Simon Bridge

You are confusing work with force.

Work done by an applied force is the component of the force in the direction of displacement, multiplied by the magnitude of the displacement. Mathematically, that's a dot product: W=F.x
You can also do it the other way.

Total work is change in energy... usually treated as the change in kinetic energy (change in total energy is zero) but it can be any energy relevant to the situation.
Work has dimensions (units) of energy.

f=μN is the force of friction, which is proportional to the component of the total applied force normal to the friction surface - work moving a distance x against friction is W = fx.

Work itself is a scalar: it does not have a direction.

What you have written says that you need less energy when there is friction in the system.
You need:

W[net]=W[F]+W[f]
But all you really want to say is that the energy needed is the amount to accelerate the stuff and the amount lost by some processes (i.e. friction).

There is no net force.
Consider - if you push an object at constant speed over a frictional surface... your applied force is the same as the firction, F=f, so W=fvΔt because vΔt=x.

See above.

More like: because you are only interested (this time) in what happens during the time that the forces are balanced.
Ah... who does the work?
You should be able to work it out from the above now.

Last edited: Jan 23, 2013
3. Jan 23, 2013

### tms

Work is a scalar, and has no components.

4. Jan 23, 2013

### Simon Bridge

Thanks tms - that's a typo.
Edited and removed.

5. Jan 24, 2013

### tms

I figured as much, but thought it might confuse the OP.

6. Jan 24, 2013

### szimmy

Some of what you said confused me Simon.

"What you have written says that you need less energy when there is friction in the system.
You need:

W[net]=W[F]+W[f]
But all you really want to say is that the energy needed is the amount to accelerate the stuff and the amount lost by some processes (i.e. friction).

There is no net force.
Consider - if you push an object at constant speed over a frictional surface... your applied force is the same as the firction, F=f, so W=fvΔt because vΔt=x."

So what you're saying is that if there is a box with a rope attached to it, and I pull on the box then the final velocity will be less than if there was friction causing a negative acceleration. That doesn't really make sense now does it?

Work is the net force applied over a distance.
So W = (Fa - Ff)Δx
So Wnet = Wapplied - Wfriction
So OP is right.

Also in your 'consider' situation, the work that you stated is the work that you did on the box, not the net work on the box.
Since the box is traveling at a constant velocity, there is no change in kinetic energy, and since work is the change in energy, that means there is no work done, which further proves my point that W = Wapplied - Wfriction because if it was Wapplied + Wfriction like you said earlier, then the box would not travel at a constant velocity because the change in kinetic energy would be equal to 2FvΔt.

EDIT: I see where you were going with your explanation, and yes work is a scalar, but it can have a direction. The reason is work is the scalar product of the magnitude of force and distance multiplied together, and that multiplied by the cosine of the angle between the direction of motion and component of force, so in the case of friction, the angle is 180, and as you know cosine 180 is -1, meaning the net work is in fact FaΔxcos0 + FfΔxcos180
Or FaΔx - FfΔx AKA Wapplied - Wfriction

Last edited: Jan 24, 2013
7. Jan 24, 2013

### tms

That is a contradiction; scalars do not have directions by definition.

8. Jan 24, 2013

### dcasplen

Thanks everyone for your input!

I think my confusion has more to do with the english langauge than the physics involved.

--So the net force is the difference between applied force and the force of friction.
--This net force leads to a net work.
--A net force→net work→leads to an increase in energy, i.e. there must be an acceleration
--conversely, constant velocity implies no net force, work

If the above statements are true then I am now good with that verbage!

Now;

If I lift a stationary book with mass (m) a distance h I have increased the energy of the book by mgh. This indicates that gravity does no work and the work I did is mgh. Am I correct in saying that the difference in how we look at the force of gravity compared to that of kinetic friction is because the force due to gravity is conservative vs. kinetic friction being non-conservative? It seems the only other way to look at this situation is that if we consider that the force of gravity does indeed do work, then I would have to do twice the work in order for the energy of the book to increase. That would give a net work of mgh.

Simply asked: Does gravity do work to a mass as the mass is lifted a distance h?

9. Jan 24, 2013

### dcasplen

I think the issue with work being a scalar is mathmaticlly solved by szimmy below;

This is the correct equation rather than the Wnet=W[a]-W[f] that I wrote. By using the forces the direction can be correctly accounted for.

10. Jan 24, 2013

### Simon Bridge

Work done accelerating an object is done by the net force... expressed as: $W=F_{net}\Delta x = (F-f)\Delta x$ ... that what you mean?
When you want to talk about work, if you say what the work is associated with, you'll have an easier time making yourself understood.

The work done by you applying force F would be $W=F\Delta x$ ... you have supplied that much energy into the motion, but some of it has been lost to friction.

11. Jan 24, 2013

### PhanthomJay

You have a few misconceptions here, except you are on the right track when you talk about differences between conservative forces (like gravity and springs, most notably) and non conservative forces (like friction or your applied force).
Simply answered, yes, negative work.(work has no direction, since it is a scalar, but it can still have either a positive or negative value).

Rather than go into a difficult to understand explanation of conservative and non conservative forces, you should familarize yourself with some essential work-energy equations.

One such equation stems from the Work-Energy theorem, namely
$W_{net} = \Delta KE$, which is sometimes written as
$W_{total} = \Delta KE$
where $W_{net}$ (or $W_{total}$ if you want to call it that) is the sum of the work done by non conservative forces and conservative forces, that is
$W_{net} = W_{nc} + W_c$.
So re-writing the Work_Energy equation,
$W_{nc} + W_c = \Delta KE$

Now let's look at your lifting the book example, assuming you lift it slowly at constant velocity. Since there is no velocity change , then there is no KE change, hence
$W_{nc} + W_c = 0$, where $W_{nc}$ is the work done by you and is equal to mgh (your force and displacement in the same direction), and where $W_c$ is the work done by gravity and is equal to -mgh (force of gravity and displacement in opposite directions). The net work is 0, but you do positive work and gravity does negative work of the same magnitude.

The other important energy equation stems from conservation of energy,
$W_{nc} = \Delta KE + \Delta PE$, so for the same book example being lifted at constant velocity, then
$W_{nc} = \Delta PE$ , implying the work done by you is mgh, same result.

Combining these 2 important equations, it should be evident that
$W_c = - \Delta PE$, or in your example, the work done by gravity is -mgh.

12. Feb 2, 2013

### Fizzik-01

This is exactly the question I searched out and found Physics Forums to solve. So I joined.
I searched. I read. I finally found. But I am still not clear. I am a career mechanical
engineer who has always enjoyed physics, and now am beginning to teach it in High School. I took 3 semesters of Physics for Engineers and Scientists and got A's at a well-respected
Institute. I believe I follow the explanation above, except for one or two minor niggles, probably of terminology.
Plus, one large gap of understanding:
I conceive of Work as Energy in transition.
As in, from PE to KE, or KE to heat (via friction), or my chemical potential energy into KE
or PE of another body, etc.
So to sum up what I see from dcasplen, I do work to raise the book. Let's say I raise the 2
kg book 1.5 meters, in the process doing +3 J work. I did the +3 J by virtue of my breakfast
Wheaties, which came into me by mouth, transformed via digestion, and left via transformation into my physical muscle motion.
The book now has +3 J more PE than at time=0. So far so good.

But some say gravity did -3 J work on the book. How much net energy change has the book experienced? I say that since it now has +3 J of PE, 0 change in KE, 0 change in thermal, chemical, nuclear, photonic, accoustic, electromagnetic, and every other kind of energy, known and unknown, then exactly +3 J of work must have been done on it.

However, if gravity DID do -3 J work on the book, what is not adding up?
I did +3 J work on the book. Gravity did -3 J on the book. But the book now has +3 J PE
compared to when I started. How does +3 - 3 = +3 ???
What am I missing? Gravity did not eat any Wheaties, did it? It has no volition by which to do work on a counter-moving object
Fizzik-01

13. Feb 2, 2013

### Simon Bridge

Welcome to PF;
You can get very messed up by the different ways people talk about work.
In your example 3J of Work got done - against gravity.
Probably other work got done too.

What happened is you lost 3J of energy to the gravitational field between the mass and the Earth. Since gravity gained the energy, it counts negative.

You understand pushing a car? If you were to push horizontally against a car but you and the car moved in the opposite direction of the direction you were pushing anyway: how much work do you do?

14. Feb 2, 2013

### Fizzik-01

Hi, and thanks, glad to be here.
Yes, it can be confusing. I hope you can help clear up the confusion, or bring in someone who can.

I have pushed many a car. I follow your point. To reiterate briefly:
1. If it does not move: no work.
2. If it moves the same direction I am pushing, I do work on it. Depending on the situation the car retains the energy of the work I did may stay with the car (as PE or KE, for example) or not (if the friction retarding the car exactly matched my work, and drained it away as heat or abrasion-into-wear-particles, or some such.)
3. If the car moves against me and my herculean efforts, I get run over. Or if I am luckier than that, I maintain my footing but give up ground. The car does work on me. Am I also "doing" (negative) work on the car? Either way, I am not sure it is clarifying things for me.

This car example may no be so illuminating, since some of the processes are non-conservative. Can we stick to the example where I am lifting the book in a conservative process against gravity? I am only pursuing this because I feel the need to be able to explain an assigned problem set to the AP students. I am feeling a bit stuck.

I do +3 J of work. The book gains +3 J of PE. Makes sense.
The assigned question asks, how much work does gravity do on the book as I lift it? The apparent answer is -3 J, as already discussed. Maybe I should be asking if this question even makes sense to ask? That is, having defined the book as the system, should the sum total of all the different "work" done on the system equal to the change to the Total Energy of the system? And if it does, what are all the different "works" to include? One must be missing from my original analysis, as the work does not add up.

A sudden thought: Maybe the work gravity does is NOT on the book, but rather on ME!
That would explain the loss by me of the 3 J and the gain by an outside entity (book + gravity system) of 3 J. Could this be it?

Or maybe a closely related example may help illuminate my confusion or question.
A pendulum has a total mechanical energy TE = PE + KE.
At the top of the swing, KE is zero so at that point TE = PE + 0 = +mgh (where h is the vertical component of the swing path).
Gravity does a certain amount of work on the pendulum as it swings down: mg(-h).
At the bottom of its swing, PE is zero and TE = 0 + KE (still = +mgh)
The work done by gravity resulted in NO new energy in the pendulum.
Is this because while gravity did work on the pendulum, the pendulum did "an equal and opposite amount of work on gravity" somehow? Or what?

Gravity's work does track the conversion of PE to KE during the downswing of the pendulum, even though no new energy is introduced.

There is a similar, mirror process on the upswing. Gravity does -mgh work, and the KE drains away as it is converted by the gravitational work process into PE.

But if work is being done on the pendulum (positive and negative) during the two portions of its swing, and the pendulum is neither gaining nor losing Total Energy, where does the "work" of gravity go to and come from?

Something is amiss.
I realize you are smart and probably a whiz at physics. Can you help me actually understand? I have been struggling with this one for weeks.
Thanks,
Fizzik-01

15. Feb 2, 2013

### PhanthomJay

Fizzik-01: Your question is an excellent one and perhaps not easily answered (I too have always struggled with it, being an engineer myself). If you have any faith at all in the work-energy theorem, which states that the work done by all forces (total work, or net work) acting on an object is equal to the change in KE of that object, and if we agree in the problem of lifting a book slowly at constant velocity that there is no change in its kinetic energy, then we must conclude that no net work is done on the object, and further, that since the work done by the non-conservative force (your applied force) is +mgh, then without doubt the work done by gravity must be -mgh. This certainly also stems from the fact that the work done by gravity is always and forevermore equal to the negative of the change in the objects gravitational potential energy, per my above listed equations, whether the object moves upways or downways or sideways or any whichway.
So your question appears to be, how in the world can the object gain PE of 3J if no work is done on it? To attempt to answer this, it may be worthwhile to look at the conservation of energy equation which in essence states that energy of a system is always a constant, that energy cannot be created or destroyed, only transformed into different types of energy, and which is often written as the work done by nonconservative forces acting on the object is equal to the change in the objects (PE +KE). But eliminating the concept of a 'work' term, the conservation of energy equation can instead be more aptly stated as
$\Delta KE + \Delta PE + \Delta E_{other}= 0$, where $\Delta E_{other}$ represents the change in the energy of the system due to work done by non-conservative forces (that is, $W_{nc} = - \Delta E_{other}$). So now when you look at this equation and apply it to the book example, where there is no KE change, the equation reduces to $\Delta PE + \Delta E_{other}= 0$, or in other words, the change in the objects Potential Energy is negated by the energy transferred into it from the work done by the non conservative force. In essence, then, we have $W_{nc} = - \Delta E_{other} = mgh$ .
I don't think this clears up anything, but it might be worth pondering.

16. Feb 2, 2013

### Fizzik-01

Thanks MockingJay,
You are right, nothing is cleared up yet, but you have given me some possibly productive lines of thought to pursue. It's too late tonight to pursue right now though... it will have to wait for later.
I sincerely appreciate the thoughtful answer you have taken the time to assemble!
THANKS ! ! !
Fizzik-01

17. Feb 2, 2013

### Simon Bridge

Similarly, you work is on gravity, not the book - moving the book around is just a means of keeping score.

It's helpful I think... though some people are uncomfortable with the physicality this gives to fields.

However - as PhantomJay points out, you need to view work in context of the conservation of energy.

18. Feb 3, 2013

### PhanthomJay

Some more food for thought:

The mechanical energy (KE + PE) of an object changes when work is done on it by non-conservative forces. The work done by gravity is already included in the PE term. Even though in the book example there is no net work being done, it is the work done by your applied non-conservative force that changes the books mechanical energy (in this case just a PE change).

The total energy of the earth-book-you system can never change, which includes the energy from the Breakfast of Champions Wheaties you had prior to exerting your force.

19. Feb 4, 2013

### Simon Bridge

It makes sense... work is done by something on something else, so you are always considering part of a system.

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