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Work in physics is mathematically defined as force $F$ applied on an object multiplied by the displacement $d$ it covers in the direction of the force. In a system where, a restrictive force exists like friction due to contact between objects or in a field where work done changes into potential energy, displacement is finite.But according to Newtons second law of motion when a force is applied on an object, it produces acceleration in that object which causes a change in velocity in a "certain" time. Moreover when the force has been applied, according to Newtons 1st law of motion it should displace "infinitely" with a constant velocity which is changed due to the force applied. My question here is that what is the real definition of 'displacement' in work formula?

$$ W = F d \cosθ$$

Is the displacement infinite making work infinitely increasing or it is the displacement covered by that object while the force is being applied or when there is acceleration being produced?

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    $\begingroup$ Perhaps you should look at power (work per unit of time) to better understand work. For example, if you engine runs at a constant power to accelerate your car or spaceship (ignoring friction), then the work would be linearly increasing in time for as long as the engine is running. It also is always better to say "with no limit" instead of "infinitely", because the infinite is not a scientific concept (despite a widespread confusion). $\endgroup$
    – safesphere
    Commented May 9, 2018 at 19:46
  • $\begingroup$ More on the definition of work. $\endgroup$
    – Qmechanic ♦
    Commented May 9, 2018 at 20:08
  • $\begingroup$ I changed the title to be more clear about what the question is. So more other users can find it, use it, answer it - and give you reputation points. Change it again if my title is not good! $\endgroup$
    – Volker Siegel
    Commented May 10, 2018 at 12:12
  • $\begingroup$ You can push a wall for hours. Your boss tells you to move the wall. What work have you done? None, so you'll be fired. That's a joke, but check that making a force can be useless if you don't move the object. You're using your force to fight another force... but that's not changing the energy. $\endgroup$
    – FGSUZ
    Commented Jan 15, 2019 at 22:46

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The work occurs while the force is applied.

So the d in the formula is the distance the object moves while the work occurs.

Independent of that, the object, which was accelerated by the work, wil displace further afterwards.

There are two uses of the term "displacement" for different things in your question I think. Sorting that out may help clear up your confusion about it.

The "certain" time is the time during which work is done, because force is applied. After that, the object will move with it's new velocity, caused by the acceleration by the work. It will keep that velocity until a new force is applied, changing it's velocity again.

That new force could be friction, stopping a piece of wood pushed by the force, or gravitational attraction of a planet nearby an asteroid.

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  • $\begingroup$ You are walking downstairs. What work do your muscles do? Same force and displacement, but not nearly as much work as walking upstairs. $\endgroup$
    – safesphere
    Commented May 9, 2018 at 19:52
  • $\begingroup$ Is there some sort of proof of what you just said "The work occurs while the force is applied.". I mean can you prove it with an example or some geeks theorem :) $\endgroup$
    – Simab Asif
    Commented May 9, 2018 at 19:53
  • $\begingroup$ There is nothing to prove! It can not be proven! It may even be wrong, on the level of quantum mechanics. We - Newton in this case - have found our universe behave like this, and he wrote down the formula describing it. $\endgroup$
    – Volker Siegel
    Commented May 9, 2018 at 19:59
  • $\begingroup$ In math, you can prove things because math is about an abstract structure we have build ourselves and know completely. We can just say "if all that is true, this is true also, because a and b and c.". In physics, we do actually not even know whether Newtons laws are true in our whole universe. That it works here on earth, and probably our galaxy, does not mean it works everywhere. I would not expect that Newtons laws were meaningful during the first few seconds after the big bang. So it may not be applicable to the microwave background. $\endgroup$
    – Volker Siegel
    Commented May 9, 2018 at 20:08
  • $\begingroup$ And note that Newton not even knew that other galaxies exist, and what a galaxy is, to begin with. $\endgroup$
    – Volker Siegel
    Commented May 9, 2018 at 20:13
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Work in physics is mathematically defined as force F applied on an object multiplied by the displacement d it covers in the direction of the force.

Not really. The original definition, from the early 1800s, was applied to the amount of force needed to lift a bucket of water. Since this was occuring in gravity, the need for the term "during the application of the force" was not needed, that was assumed in the experimental setup - the bucket stopped when you stopped applying the force.

So then the definition isn't automatically useful when considering cases outside gravity, which is what is causing your confusion (and everyone else's, this is one of the most commonly asked questions here). So then you add the proviso that was inherent but unspoken in the original definition, "during the application of the force". You can see this in the Wiki article you linked to.

My suggestion: whenever you get confused by work, apply the work-energy equivalence. So if someone defines work as the change in energy and the result seems odd, reformulate the question as work being the force over distance. And if, as in this case, the force over distance seems to be odd, then consider it as a change in energy.

So in this case it seems odd that work isn't infinite when the displacement goes on forever. Ok, reformulate the same question in terms of energy. Was infinite energy applied to the object? No? The mystery instantly disappears.

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To give a short formula based answer, solving for $S$ from the equation: $v^2 = u^2 + 2as$ and substituting it into the equation: $W = fs$ gives $W = \Delta KE$. Which is true based on the work energy theorem. Consider that we used an equation that defines $S$ in a context of constant acceleration. So, we used the displacement that is covered during acceleration, not constant speed. Therefore, the $S$ in the definition of work is displacement during the period of application of force, which is the period of acceleration where the kinetics energy changes because of change in velocity. This brings the harmony between the basic definition of work done and the work energy theorem. In other words, the work energy theorem tells us exactly which displacement is there in the definition of work done. But we need some further explanation on what happens when there is resistive force like gravity or friction.

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  • $\begingroup$ Your 1st formula works only for uniform acceleration. $\endgroup$
    – FGSUZ
    Commented Jan 15, 2019 at 22:44
  • $\begingroup$ Yes, and it means the object upon which the force is exerted gains energy only during acceleration. $\endgroup$
    – Samuel Birhanu
    Commented Jan 20, 2019 at 20:42

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