# Why is work done not equal to force times time?

I am having an issue with the definition of work. I know there are several questions already posted which seem the same as this one but the issue is that all of them either haven't been answered or they have been answered in a tricky way which indirectly fools us. That's the reason I am posting this question again. I did research work on this and to prove this I will post this in the end. Also, I am going to write the question a bit more clearly.

Work done on an object is the amount of energy changed in the system of an object. And energy is a fundamental thing in physics. Now work is defined to depend directly on displacement and not time. So the issue is I will explain according to an example.

## Example

Suppose I set up a machine that constantly applies $$1N$$ of force and during this process it spents $$1Joule/sec$$ on any object. Now I made this machine to put its effort on a block over a displacement of 1 meter. This means that the total work done by the machine is 1 Joule. Now the issue comes when we think of the second meter displacement. What will happen is that by the time block enters the second meter, it already has a velocity which means that it is going to take less time to cover the second meter than the first-meter. This means that the energy spent by the machine in the second meter is less than first but still it has done the same amount of work in the second meter as force is the same and displacement is also 1 meter.

This means that the work done must depend on time rather than displacement. What's the issue here ? I can think of several examples like this where the same contradiction takes place. So prove that Work done depends on displacement instead of time. Also please don't use any formula of energy to prove the work done formula because I believe it's a way of fooling people. Thanks and appreciation to anyone who tries this.

## Proof that I did find answers but they didn't satisfy me (You can skip this)

• “please don't use any formula of energy to prove the work done formula because I believe it's a way of fooling people” I really dislike when a questioner pre-rejects answers. Why shouldn’t the answer to a question about work use a formula about energy. It is an absurd restriction. And how would a mathematical derivation be “fooling” people.
– Dale
Oct 11, 2020 at 11:40
• @Dale Why I am not asking answers which use energy formulas is because energy itself is an inspired concept from work. Please take your downvote back. Let me explain my saying again. When I am proving a thing which has proved the other thing then the first thing will always come true. Example: My name is Ritanshu Implies that people will call me Ritanshu. Now if you say that people call you Ritanshu hence your name is Ritanshu, that's wrong because my name is not because people call me Ritanshu but because I told people my name is Ritanshu. You get that? Oct 11, 2020 at 13:47
• @Dale You weren't able to understand what I meant. I said that if statement B is inspired by statement A then using statement B we can easily prove the existence of statement of A. In that naming context, the second statement "people calling me Ritanshu" is because of the first statement " My name is Ritanshu" and now if you say that because people call me Ritanshu hence my name is Ritanshu, yes surely you can prove that but that would be wrong way to prove that because statement A inspired B and B is not independent Oct 11, 2020 at 17:40
• instead of arguing with me why don’t you take note of the many downvotes on your question and many upvotes in my comment and fix your question. I am by far not the only person who dislikes this aspect of your question. If you are concerned about circular reasoning then state that the question. The circularity you allude to is not a result of using a specific formula, but from mixing the thermodynamic and mechanical definitions (each of which is non-circular on its own, but mixed together they look circular)
– Dale
Oct 12, 2020 at 11:06
• Your machine is a perpetual motion machine Jan 15 at 19:04

Why is work done not equal to force times time?

You have definitions backwards. It's not like we said "Ah yes, 'work' is important, what should its definition be?" The reason work is defined is because it is useful to explain physical phenomena. In other words, the quantity $$\int\mathbf F\cdot\text d\mathbf x$$ is useful, so we linked it to a term we call "work".

If you think there should be other useful quantities, then that's fine.$$^*$$ But saying "work really should be fill in the blank" just doesn't make any sense.

So prove that Work done depends on displacement instead of time

Work has an exact definition: the integral given earlier that depends on displacement. So this proof you are demanding is nonsensical. It's like asking someone to prove that the word "red" represents a color.

$$^*$$ If you don't already know, what you propose $$\int\mathbf F\,\text dt$$ is actually the change in momentum of a particle if $$\mathbf F$$ is the net force acting on the particle. This has the name "impulse".

• Excellent answer. No offense to Ankit, but the fact that the other answer was chosen by the OP is somewhat disturbing. It suggests to me that OP still does not recognize the core issue, and was instead lulled into complacency by technical examples. Jan 15 at 18:57
• OPs question is nearly: "why do we define things?" The answer, of course, is because of the utility of a definition; its compactness. Jul 11 at 4:53

You cannot set up a machine that "uses 1 Joules/s to exert a force of 1N on any object", precisely because it would lead to the sort of contradictory statements about work and energy spent with objects moving at velocities relative to the machine that you observe.

You can either set up a machine that exerts a constant force of 1N and uses variable energy to do so depending on the work it needs to do, or you can set up a machine that consumes a constant amount of power and exerts a variable force with it.

• So if the power is consistent then the force would vary due to change in what factor? Oct 11, 2020 at 9:55
• @user921307 Due to the change in displacement! For example the machine would need to move itself in addition to exerting a force on the object to keep in contact with the moving object, so if the power is constant it has less and less available to invest into exerting the force as it has to invest more into accelerating itself just to keep up. The details depends on how exactly this machine is constructed and how it is exerting the force. I suggest you try to actually think about how you would actually build a real-world machine like this. Oct 11, 2020 at 10:01
• what if the object itself is the machine? The object sucks something from one direction and throws towards the other. Or it can very well be a machine that pulls a rope that can be attached to a wall. There are several methods that I can think of which are like this. Oct 11, 2020 at 10:08
• @user921307 if your suck-throw machine uses a constant power then the output force will vary as the machine struggles to push on the thrown material. If your rope pull machine produces a constant force then the power will vary as the machine works hard to keep up with the rope.
– Dale
Oct 11, 2020 at 12:44

Look I don't know if this answer satisfies your condition or not but I am going to convince you that work should be force times displacement and not force times time.

Okay let us assume that $$Work = Force × time$$.

I can prove the above method wrong using two examples.

Example 1 : Now imagine an electron moving in the horizontal direction enters a region of uniform magnetic field whose direction is into the plane of your screen. So it experiences a force perpendicular to its velocity and starts an uniform circular motion as shown in figure below

Now from our definition of work (as force times time) , the electron should gain energy since it is experiencing a force for some period of time . So it's kinetic energy and hence speed should be increasing but experimental measurements show that the speed of an electron in a region of perpendicular uniform magnetic field remains the same i.e it follows a uniform circular motion.

Example 2 : This one is based on the fact that energy has no direction i.e it is a scalar physical quantity.

Now from your definition of work (i.e $$W = F × t$$) , you see that there is a vector physical in the above relation i.e the $$F$$ . And of course time is scalar. So a vector times a scalar will finally give you a vector physical quantity . So , work is a vector physical quantity from this relation.

Wait what !!!!

It is absolutely clear that our assumption of work equals force times time leads to contradictions with experimental measurements and physical understanding. So we must change our assumption .

Now we have two things that can set with the above property of constant speed of an electron . And in both the possibilities the work done on the electron by that magnetic force would be zero.

1. $$W = F\cdot s = Fs \cos \theta$$

$$OR$$

1. $$W = F \cdot V = Fv \cos \theta$$

First let us assume that the second possibility of work done is correct. So , the work done on our assumed electron will be zero (since force is perpendicular to the velocity at every instant) and thus no change in kinetic energy.

Okay this assumption looks good. Now assume a particle projected upward and is under the influence of gravity only. So , it is experiencing a downward force and thus from our definition of work , we can notice that the total work done by gravity would be negative since $$\cos \theta = \cos 180°$$. Now if we want to define power (rate of work done), it will be

$$P = F\cdot a = Fa \cos \alpha$$

Now we know that the acceleration term in the above equation is the acceleration due to gravity and so $$F$$ and $$a$$ both are in the same direction ($$\cos \alpha = \cos 0° = 1$$) and thus power will be positive !!!!.

How is it possible that total work done is negative while total power is positive?

This completely signifies that we made a wrong assumption .

So we are now left with just one option and this one sets good with all experimental measurements. So,

$$W = F \cdot S$$

Note : If you are new to electrons in a magnetic field then you can replace the electron and imagine a ball tied with a rope . So in this case , the ball will speed up even if you don't apply a tangential force to it. Also in the first example , it is assumed that work equals change in kinetic energy . If you don't accept it, check out the second example. It will be more helpful and convincing than the first one.

Hope it helps ☺️.

• @user921307 and will continue until I get the answer. wasn't the answer fruitful ? If you still have some doubt you can ask me. Oct 11, 2020 at 18:00
• Yes it was, You meant there is no kinetic energy change in the system and I note this down in my diary but the thing is that the objects momentum changes and that to me seems like some work has been done there. As I have already stated, kinetic energy is inspired idea of work and introducing it before proving work mean something fault Oct 11, 2020 at 18:03
• @user921307 why do you think that for changing momentum you need to do some work ? Please elaborate. BTW Force is itself sufficient to change momentum. Oct 11, 2020 at 18:06
• But the thing is that force should come from somewhere and if something applies a force and there is change in momentum then it is losing energy, isn't it? In your example, the magnetic field isn't coming out of anywhere, It must have an energy source Oct 11, 2020 at 18:07
• Oct 11, 2020 at 18:08

This means that the total work done by the machine is 1 Joule.

We are not just talking about the work done by the machine. We are talking about the work done by the machine on the block. Therefore, it doesn't matter whether that machine spends 1 J/s. That energy amount isn't necessarily all converted into work done on the block.

The work done by the machine on the block will always be 1 Joule because it is being pushed with a force of 1 N over a displacement of 1 m, regardless of how much energy that is spent on producing that 1 N force.

it already has a velocity which means that it is going to take less time to cover the second meter than the first-meter. This means that the energy spent by the machine in the second meter is less than first

No, it does not mean that. You previously mentioned that the machine pushes with 1 N. That is regardless of the block's initial velocity. If the block is already moving, then your machine's force still applies 1 N.

That 1 N causes an acceleration. This acceleration increases the velocity (kinetic energy is added). It is that velocity increase that matters, and not the start velocity. The work you put in is not energy that make the block move, it is energy that makes the block accelerate. If your machine didn't touch the block at all over that second metre, then the block would still move through that second metre - but it wouldn't experience any energy gain. If you apply a force on this block, then it will experience an energy gain, because you will increase its velocity.

That energy gain that your machine provides happens when you apply a huge force. If your force isn't huge then it must be upheld over a longer displacement before the same energy gain has been achieved. Therefore force and displacement are the relevant factors. It doesn't matter how much time it takes - if you push with a huge force for a long time on a wall, then no velocity increase happens. No kinetic energy is gained. Because there is no displacement over which that speed increase can take place.

• Then where would the other energy used up by the machine go? You would say that it goes into heat or something like that but assume that we have a perfect machine. Then where does it go? Oct 11, 2020 at 10:29
• @user921307 Then the force would be larger than just the 1 N. If you only want that 1 N of force and you also have a "perfect" no-waste, no-loss theoretical (and impossible) machine, then the machine would have to stop expending energy at some point. It wouldn't make sense that it forever and ever would be spending 1 J/s Oct 11, 2020 at 10:42

I recently came across the same question.., and thought of an answer that convinced me. I would like to share that example, in addition to many of the answers above.

Let’s start with energy, since the concepts of work and energy were developed simultaneously.

When a body at rest starts moving, we knew it had gained SOMETHING, and we called That SOMETHING, as ENERGY.

Suppose your throw a ball, you apply a force on the ball for a time interval t.And, according to Newton’s third law of motion, a force of equal magnitude is applied BY the BALL ON your HAND.

If Energy were force * time, then the ball would lose an equal amount of energy as it gains, since force is being applied on it, and it is applying a force on the other body for equal durations of time. So according to this, there is no energy gain, and nor is there any work done.

But, according to the way we started: we knew that the body was gaining SOMETHING, and we called this energy.

Since there is no net change in the energy of the system if we define it as force * time, we don’t define energy or work that way. The same holds for work also, since energy change in the system is due to work done on the system.

• for you statement Since there is no net change in the energy of the system if we define it as force * time, we don’t define energy or work that way , I don't really agree. Since defining it as force times displacement would also give the same result that there is no net change in energy of the system. Apr 14 at 17:20

The idea is something like this: suppose you have a block and you push it with a constant force, F, for time T. Now in that duration, the block travels a certain displacement, S.

Of course, you can calculate the change in kinetic energy of the block by finding the initial and final velocities of the block. However, an alternative method will be to multiply the force, F by the displacement of the block, S. The 2 methods give the same numerical result. Can you show why?

About the example that you gave, yes, the object does cover the second meter in a shorter duration. Assuming the machine exerts a constant force, the change in velocity, V, of the block will be lesser in the second meter. However, the thing here is that kinetic energy is not proportional to the velocity but to the square of velocity. In other words, the smaller V in the second meter is made up for by the fact that the object enters the second meter with some velocity.

Here's another way to put it (which I think makes some sense, but may not be correct, so do correct me if it's wrong): say you accelerate your object horizontally by shooting out a stream of particles at it from the origin. Let's assume that the particles do not stick to the object so the mass of the object does not change (ie. they bounce of the object elastically) . What does the acceleration of the object depend on? It depends on the relative speed between the gas particles and the object. This means that as the velocity of the rocket increases, the speed that the gas particles leave the origin in the lab frame must increase. Obviously, this means that the you must shoot the gas from the origin at a higher speed and expend more energy per unit time. So even though the object covers the second meter in shorter time, you have to spend more energy per unit time shooting the gas particles out at a higher speed, and it turns out that the energy you expend in both intervals are the same.

• physics.stackexchange.com/questions/450402/… Here's a similar question that I had asked before, maybe it can be of some help? Oct 11, 2020 at 10:09
• I know that kinetic energy has a formula and work done and change in kinetic energy in some special contexts will be same but as I said in the question it all formula game. Let me show you $W = Fs = mas = m \frac{v^2}{2}$ Oct 11, 2020 at 10:24

If you push the block with a constant force (one Newton, in your example) and the surface under the block will give no resistance, then the block moves with a constant acceleration.
I'm, not sure what's the importance of your machine in the question. For the second meter, the machine pushes on the block with a force of one Newton too, but it needs to push for less time because the block already has an initial velocity (acquired from the acceleration during the first meter). The same applies to the third meter, etc.
So the kinetic energy given by the machine to the block in the second meter is (obviously) bigger ($$E_{kin}=\frac 1 2 m v^2$$), relative to the machine, than the energy given to the block in the second meter. And this is done in less time than the energy given in the first meter (this is also the case in gravitational acceleration, with $$9,8 m/s^2$$ as the acceleration, though the potential energy, in this case, is reduced in sync with the increase of kinetic energy).

So your machine will have to give more energy to the block in the second meter but in less time. If your machine uses $$1J/s$$ this isn't compatible with giving more energy to the block in less time.

You state $$W=Fs=mas=ma\frac 1 2 at^2=\frac 1 2 a^2 t^2=\frac 1 2 m v^2$$. So the work done is exactly the increase in kinetic energy.
But what if you push the block on a surface with friction, in such a way that $$a=0$$? Then $$W=Fs=Cs$$ because the block isn't accelerated and all work has conversed into heat.

If you define $$W$$ as $$W=Ft=mat$$, what does this represent? Well, it represents momentum. And that is useful, but not for defining $$W$$. Momentum is momentum, $$W$$ is $$Fs$$ (or the integral form).

Of course, you can start from your definition of $$W$$ and call it the work done after which you call the normal definition "kinetic energy" (if no friction is present), but that's a "wrong-way-round-reasoning".

I think its dependance on displacement is due to the fact that their is some momentum in the mass you are exerting work on and this moment will give you higher displacement as the velocity fades.

In other words if the 1 N block uses 1 jule to move one meter it takes 1 joule from the machine and each extra meter it takes another jule and time doesn't appear in this relation.

probably if the equation was made without integration you may find time in both sides that cancels out as it appears it doesn't have anything to do with work.

In a simpler definition : Work is how much energy does it cost to change the position of an object from point A to B.

If you made this machine and calculated power you will see an increase in power consumption linearly equal to the displacement.