Is it more work to put more (apparent) effort to get the same outcome?

I was taking my dogs for a walk yesterday evening when this question occurred to me.

The two dogs were pretty enthusiastic about the walk and wanted to run on ahead, so the leads were taut and they were sort of gently pulling me along. But they're quite small so I could easily control the speed at which we walked. At this stage they were presumably using more energy than that required to simply walk at that same speed given the resistance I was providing.

Suddenly, a cat spotted us and ran away. The dogs, upon seeing this, tried to give chase, running and pulling the leads with all their might. But I held onto the leads and continued walking at the same speed as before. So the result was two dogs frantically trying to pull and run but not going any faster.

So my question is: Are the dogs doing any more work (or using any more energy) when trying to chase the cat than when they were gently tugging along? Intuitively I think yes, (especially given the gasping and panting coming out of both of them) but I can't figure out why: they are moving at the same speed and pulling the same weight (i.e. me) as before....

• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Nov 10 '19 at 15:42
• Work with regards to what? If you are talking about general energy output, the answer is yes. But if you are talking about work in the classical sense, it gets much more complicated. – CuriousOne Nov 14 '19 at 16:21
• I have created a new question replacing the dog and person with a car and a sled. physics.stackexchange.com/questions/514029 – James Nov 15 '19 at 14:08
• I'll bet you never expected your question to generate such controversy. Yes your dogs do more work as you intuition tells you. But I think the reason you "can't figure out why" is because you probably view the extra work they do as being "unproductive". You are walking at constant speed regardless of how much work your dogs are doing. My point is their work is unproductive. Certainly your dogs must feel it unproductive since they get no closer to the cat! – Bob D Nov 15 '19 at 16:43

The dog does more work when pulling harder.

Work is force times distance. Consider the following two walks.

1. Walk 100 meters while pulling with 10 Newtons of force.

2. Walk 100 meters while pulling with 100 Newtons of force.

In the second walk (10,000 Nm), the dog does 10x the work of the first walk (1,000 Nm).

• Comments are not for extended discussion; this conversation has been moved to chat. – tpg2114 Nov 15 '19 at 12:24

The dogs may be expending more energy but they are not doing any more physics work which is net force times displacement. Similarly you would not be doing physics work if you are holding a heavy object without moving it, though it would certainly require physical effort on your part. The same if you pushed hard against a wall that doesn’t move.

Physical effort does not necessarily equal physics work. The energy expended is internal. Richard Feynman in his physics lectures explains it this way:

The fact that we have to generate effort to hold up a weight is simply due to to the design of striated muscle. What happens is when a nerve impulse reaches a muscle fiber, the fiber gives a little twitch and then relaxes, so that when we hold something up , enormous volleys of nerve impulses are coming in to the muscle, large numbers of twitches are maintaining the weight, while other fibers relax. When we hold a heavy weight we get tired, begin to shake, ...because the muscle is tired and not reacting fast enough.

UPDATE:

You may be following the discussions that @James and I have been having. In his answer he used an example of a dog exerting more force over the same distance and thereby doing more work. But my point is the dog can exert the same force on you and do less work. This is because it is you, and not the dog, that actually controls the amount of work the dog does.

Let's take his example but instead of varying the force the dog applies to you, you vary the distance you walk while the dog applies the same amount of force. You do this by pulling harder on the leash by planting your feet more firmly against the ground to keep the dog from pulling you along. In effect, you are increasing the static friction force between you and the ground to oppose to the force of the dog and slow you down.

1. Walk 100 meters with the dog exerting a force of 100 Newtons on you (work by dog =100,000 N.m).

2. Walk 50 meters with the dog exerting the same force of 100 Newtons on you (work by dog =50,000 N.m).

3. Walk 0 meters (stand still) with the dog exerting the same force of 100 Newtons on you (work by dog = 0 N.m)

Clearly the dog is making the same physical effort trying to pull you and will tire equally for each case, yet the work the dog does is not the same. In the third case no work at all is done by the dog.

Conclusion: The same physical effort by the dog does not necessarily equal the same physics work.

Hope this helps

• But in this case, the dogs ARE moving. – James Nov 8 '19 at 15:24
• @James I didn’t say they weren’t moving. Per the OP they were “trying to pull and run but not going any faster”. In other words their increased force did not result in acceleration. That means their extra effort did not perform physics work. That’s the point of my answer – Bob D Nov 8 '19 at 15:48
• But physics work is force times distance. Acceleration isn't required. If the force is increased then work is increased. – James Nov 8 '19 at 15:51
• Think of it this way... which is more work: 1) going for a walk around the block, or 2) going for a walk around the block while dragging a box behind you? – James Nov 8 '19 at 16:09
• I think the NET work done on the OP is zero because the dog's extra work is countered by the OP's extra (opposing) work to avoid being accelerated. Instead of the OP, imagine that the leash is attached to a box being dragged by the dogs. Do you think the dogs would be doing more work then? In the case of dragging a box, the dog's work would go into frictional heat. Do you think the work done by the dogs should depend on what's attached to the other end of the leash, or solely on the tensile force in the leash? – James Nov 8 '19 at 16:41

The dogs are definitely doing more work when chasing the cat over a given distance.

The distance is the same but the force is larger : your motion is not accelerated, which means that the net force on you is zero. The dogs are pulling you through their leash, and your feet are resisting this force by friction, except for an extremely small resultant to accelerate and slow you down between one step an another (your center of mass does not move in a completely smooth way) but they are certainly not "pulling your weight". Your weight is a vertical force, and it is balanced by the reaction of the ground under your feet (if you are distracted and step where the is no ground, over a manhole, for instance, you fall into it) Again since the dogs don't really accelerate, the net sum of the forces on them is essentially zero. The pull back the leash exerts on them must exactly balance the effort by their legs. So the harder you pull back, the harder they pull on their legs. This is the work they are doing, moving (not accelerating) against the pull of their leash. So at the same speed, if they pull harder, the do more work.

The discussion about work/power is moot. They produce more power, walking at the same speed, so at the end of a given time they end up having done more work, when covering a given distance.

The dog is doing more work becuse work = (force)(distance).So your dogs pull harder on you (more force) but don't move any faster so force is more while distance remains the same.

If the dogs pull harder they do more work. Trivially: work is force times distance.

The wording of your question leaves a small crack of doubt as to whether they are actually pulling harder after they see the cat: you include "(apparent)" in the title, and in the question say "the result was two dogs frantically trying to pull" [emphasis added]. Turns out, a lot can fit through that small crack. But all of it boils down to this: if the only thing that increases post-cat is the level of excitement of the dogs--that is, the force they exert on the leash remains the same--then there is no change in the rate at which they do work.

• I see your point, but at the beginning "they were sort of gently pulling me along". After seeing the cat they were "pulling the leads with all their might". – James Nov 15 '19 at 17:43
• @James, I totally agree with you. I'm fairly confident that OP means that force increases. Hence my one-sentence answer. The rest is a caveat in case we're both mistaken about the intended meaning. – Ben51 Nov 15 '19 at 17:47

Work in physics has a specific meaning. Force times distance over time.

Imagine that your leash is connected to a 100 pound weight which is on ice skates, that travels on ice. You however have an ice-free concrete sidewalk to walk on. Imagine that you pull with a constant force. The weight starts to move slowly. As you add force it speeds up, and then it keeps speeding up while you pull it faster with the same force, until you are running as fast as you can and you can't keep giving it more force. The whole distance you travel during that time counts in the work you've done. (Not even considering the work you did moving your own body.)

Now imagine that the weight does not have ice skates but has a big surface, and instead of ice it's traveling over sandpaper. It takes a big effort to get it to move at all, and once it's moving it takes a big effort to keep it moving. When you run out of breath and have to stop, the work is counted by the distance you took it.

In both cases, the rest of your effort went into friction. You produced heat. One way it was only a little bit of heat, and the other way it was a lot. The heat you produce doesn't count as work.

It's the same with the dog leash. When you use effort to oppose each other, you produce heat in your muscles (as always when you push or pull), and when you stretch the leash, that produces a little heat in the leash, and the extra force on your shoes produces a little heat from static friction in the shoes and the sidewalk, and so on. No extra distance traveled means no extra work.

Force times distance.

This simple idea applies to big mechanical systems. It's easy to mis-use it for other things. For example, imagine you have a Van de Graf generator that puts out sparks at a million volts. You expose one side of a capacitor to it, so electrons move into the capacitor. It takes force to move electrons in, more force than it would take if electrons were allowed to move out of the other side. The electrons move, say one inch. Force times distance. If the capacitor lead was a foot long, would it do 12 times as much work?

• You say that work is "Force times distance over time". That is the definition of power, not work. I might be mistaken, but are you saying the dogs do no extra work when they pull harder? Work is force times distance, so why is there not more work when the force is higher over a given distance? – James Nov 10 '19 at 14:34
• Force is mass times acceleration. If you are pulling harder but not actually speeding up, then it isn't force. Power is the rate that work is done, work is the integral. – J Thomas Nov 10 '19 at 18:42
• Consider a dog dragging a box attached to its leash at a constant speed. Are you saying that there is no force in the leash because there is no acceleration. I think the dog is doing work equal to the leash force times the distance dragged. That work is being converted into frictional heat. In this scenario, the net force on the dog/box system is zero. The dog is still doing work though. Correct? – James Nov 10 '19 at 19:49
• I have to admit that I may be wrong. But it looks to me like you can figure out how much force went into accelerating the box over time by how much it moved. Force that went into making heat doesn't count as work done. But here is another way to possibly think about it. When there are opposing forces, maybe one does positive work and the other does negative work. You could calculate the distance traveled in both directions if there was no opposition, and consider them both work, and then the actual distance gives you the net work. I think either approach could give a workable answer. – J Thomas Nov 11 '19 at 1:58
• "Force that went into making heat doesn't count as work done." That's the idea that is confusing me. That would mean that I do no work if riding a stationary bike. It would also mean that falling water in a hydro-electric plant does no work if the generated electricity is used to heat a building. I'm obviously no expert. Thanks for the discussion. – James Nov 11 '19 at 4:03

My previous answer with the example of the dog exerting the same force but doing less and less work I believe proved that physical effort does not always equal work as defined by Physics. That is how I interpreted, perhaps incorrectly, what your question was really about.

But while it is also true, as @James points out, that if the dogs exert a greater force acting through the same distance the dogs perform more work, I think you may be questioning whether or not that additional work or effort is, in effect, productive. If that was also part of your question, then I believe the answer to that is no.

When the dogs do more positive work (force times displacement in the same direction as the force), you, the walker, are also at the same time doing an equal amount of negative work (force in a direction opposite to the direction of the displacement), in order that there is no change in displacement for the same time, and therefore net work done remains zero.

So one can ask, does the additional work done by the dogs, which obviously requires more physical effort on the part of the dogs, improve the "outcome", that is, is their additional physical effort productive. From the dogs' perspective, has it gotten them any closer to the cats! The answer to that is no.

So my question is: Are the dogs doing any more work (or using any more energy) when trying to chase the cat than when they were gently tugging along? Intuitively I think yes, (especially given the gasping and panting coming out of both of them) but I can't figure out why: they are moving at the same speed and pulling the same weight (i.e. me) as before....

So the answer to the beginning part of your question is, as @James pointed out, yes the dogs are doing more work. But I would point out that they would be gasping and panting just as much if they exerted the same force for the same duration, but over a shorter distance if you further restrained them from pulling you along, thereby having them perform less physics work with the same effort, as shown in the examples of my original answer. But if the doubts expressed at the end of the question has to do with whether or not the additional effort on the part of the dogs is productive, I believe the answer to that, in the context of the above, is no.

Hope this helps.

@Jame's answer is a more simplistic one to approach the problem. However, I agree with him that the dogs will be doing more work compared to if they don't pull the leash. But I want to discuss a case here of by "how much", it is not directly linear after all.

Again, if we review, work is a dot product of two vectors. As a general rule, in your example, the amount of work is actually equal to $$Fscos(\theta)$$

$$F$$ = amount of force the dogs pull on the leash

$$s$$ = total distance traveled and

$$\theta$$ = the angle between the leash and the dog.

This tells us that the force component that actually gives a factor to the amount of work is the component parallel to the displacement or parallel to your dog.

So a case here is if your dogs are not of the same height. Both of them, though exerts same amount of force (*theoretically) and traveled the same displacement, won't be doing equal work. Because the angle where they pull their own leashes will be different - greater angle for the smaller one, and smaller angle for the bigger one.

This is a perennial question on this site, and always leads to confusion over the distinction between "work" and "effort". To avoid confusion, we need to be explicit about the many terms in the energy.

Energy contributions

Crudely, the energy of the dog can be written as the sum of four terms, $$E = U_{\text{pot}} + K_{\text{KE}} + U_{\text{chem}} + K_{\text{thermal}}.$$ The first two are macroscopic, and the last two are microscopic. They are:

• $$U_{\text{pot}}$$ represents the macroscopic potential energy due to external fields, such as $$M g h_{\text{CM}}$$ for gravitational potential energy
• $$K_{\text{KE}}$$ represents their macroscopic kinetic energy, i.e. the kinetic energy of all the parts big enough for you to see, such as their overall center of mass motion, the motion of their limbs, etc.
• $$U_{\text{chem}}$$ represents the potential energy stored in the chemical bonds of the molecules they can metabolize; it is what is released by the mitochondria, powerhouse of the cell.
• $$K_{\text{thermal}}$$ represents the thermal energy, the microscopic kinetic energy of individual molecules jiggling around in thermal motion.

For simplicity, we're going to ignore any heat transfer with the environment. Then the work-energy theorem (which is now equivalent the first law of thermodynamics) only tells you about the total change in energy, $$\Delta E = W = \int F \, dx.$$ Therefore, in isolation it tells us almost nothing about how each of the four individual terms change. The amount of "effort" roughly follows $$\Delta U_{\text{chem}}$$, but an understanding of how it changes in various situations requires a detailed treatment of metabolism and biomechanics.

Dog on a leash

For the example in this problem, we have $$\Delta U_{\text{pot}} = \Delta K_{\text{KE}} = 0$$ so if the dogs apply a force $$F$$ over a distance $$x$$, $$-Fx = \Delta U_{\text{chem}} + \Delta K_{\text{thermal}}.$$ At this point it would be tempting to set $$\Delta K_{\text{thermal}} = 0$$, so conclude that $$-Fx = \Delta U_{\text{chem}}$$ so the "effort spent" by the dog is proportional to $$Fx$$. While that might be true for an idealized machine, it isn't remotely true for biological systems. In general, the chemical energy spent is much more than $$Fx$$, and is not necessarily even linear in $$F$$ or $$x$$. The extra spent energy goes into thermal energy, which is why the dogs will start panting and a human would start sweating. Explicitly, the energy conservation equation above might actually look like $$(-100 \text{ J}) = (-1000 \text{ J}) + (900 \text{ J}).$$

It's not even obvious how $$\Delta U_{\text{chem}}$$ depends on $$F$$ and $$x$$. For example, as $$F$$ increases, the dog might have to pull in a less biomechanically efficient way, causing $$\Delta U_{\text{chem}}$$ to increase faster than linearly in $$F$$. Or, for very high $$x$$, the dog could get tired and switch to a less efficient metabolic pathway, causing $$\Delta U_{\text{chem}}$$ to increase faster than linearly in $$x$$. Or, even if you kept both $$F$$ and $$x$$ the same, the chemical energy spent might vary if the dog decided to pull a different way.

Person squatting a barbell

An even more extreme example of this is a person squatting a heavy barbell. For simplicity, let's assume they do it slowly, and that the barbell is much heavier than they are, so that we can neglect the person's gravitational potential energy. Then for the system containing just the person, $$\Delta U_{\text{pot}} \approx 0, \quad \Delta K_{\text{KE}} \approx 0.$$ If the barbell has mass $$M$$ and its range of motion is $$d$$, then in the first half of the exercise, the barbell does work $$Mgd$$ on the person,
$$Mgd = \Delta U_{\text{chem}} + \Delta K_{\text{thermal}}.$$ In the second half of the exercise, they raise the weight back up, so $$-Mgd = \Delta U_{\text{chem}} + \Delta K_{\text{thermal}}.$$ For an idealized machine, we would have $$\Delta K_{\text{thermal}} = 0$$, so $$Mgd = \Delta U_{\text{chem}}, \quad -Mgd = \Delta U_{\text{chem}}$$ for the two halves of the exercise. That is, a machine can gain energy from lowering a weight. Then it pays it back by raising it back up, leaving it in the same state as before.

None of this is remotely true for a biological system. In fact, even the sign of $$\Delta U_{\text{chem}}$$ is different! It takes effort for biological muscles to steadily lower something, so $$\Delta U_{\text{chem}}$$ is negative for both halves.

In fact, it's even worse: you will in general feel more tired, in the long run, due to the lowering portion than the raising portion. The reason is that the lowering portion stretches your muscle fibers, which causes more of them to break. Again, none of this contradicts energy conservation. The point is that energy conservation alone doesn't tell us the answer; we have to actually understand the biology.

they are moving at the same speed and pulling the same weight (i.e. me) as before

It's not that simple.You will notice that while keeping your dog steady you are leaning back. Degree of leaning back depends on the type of dog - more massive dog / it's power - bigger leaning will be. This game is a bit similar to Tug of War. Who wins ? The ones who have more strength :

Now, let's get to the Physics of this game :

Dog forces are colored in red and your forces are colored in green. Several things happens when dog tries to pull you more aggressively and you lean back because of it :

• Dog induces torque $$\tau_{dog} = T_{\perp} \cdot H_{_{COM}}$$
here $$T_{\perp}$$ - tension force perpendicular component to your body
$$H_{_{COM}}$$ - your height until Center Of Mass
• You induce torgue $$\tau_{man} = W_{\perp} \cdot H_{_{COM}}$$ which is caused by perpendicular component of your weight
your torgue direction is with an opposite sign than that of dog's
• your static friction force experienced by your foots is : $$F_s = \mu_s\left(|W_{\parallel} + T_{\parallel}|\right)$$ That's why your dog can't do anything about it - by leaning back it's easier for you to hold-on dog steady, because you convert part of it's tension force against him itself - to your static friction force, which it needs to overcome to be able to move. But hence, greater pull will induce even more bigger static force.

So my question is: Are the dogs doing any more work (or using any more energy) when trying to chase the cat than when they were gently tugging along? Intuitively I think yes, (especially given the gasping and panting coming out of both of them)

Since work is strictly the product of force times displacement, or distance, in the direction of the force, yes the dogs are doing more work.

but I can't figure out why: they are moving at the same speed and pulling the same weight (i.e. me) as before...

In order for your speed to increase because of the force the dog exerts on you, there needs to be a net horizontal force on you so that you accelerate. When the dogs apply a greater force on the leash you are also applying an equal greater force on the leash in the opposite direction in order to restrain the dogs. Refer to the picture below (only one dog shown).

The external horizontal forces acting on you are the horizontal component of the tension in the leash, and the opposing static friction force between your feet and the ground. As long as maximum static friction force is not exceeded and you do not slip the forces will be equal and you will not accelerate (change speed) due to the additional force the dog exerts on you.

The same external horizontal forces are acting on the dogs and they do not accelerate (change speed). If the dog increases its force on the leash the static friction force between its paws and the ground also increases so the net force on the dog remains zero. However, being that your dogs are "quite small", with a weight probably an order of magnitude less than yours, and assuming the coefficients of static friction are of the same order of magnitude, it is likely the dogs will cause themselves to slip before their force could become great enough to cause you to slip.

So while the dogs are indeed doing more work, I would say that the additional work they do is not "productive" since it does not result in a change in speed. They are expending more energy internally but the additional effort certainly doesn't get them any closer to the cat!

Hope this helps.