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Alot of different books , authors and physicists define work differently in classical physics and mechanics.

For example , Halliday & Resnick FUNDAMENTALS OF PHYSICS defines Work as

Work is energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work, and energy transferred from the object is negative work.

Wikipedia also gives a similar definition,

In science, work is the energy transferred to or from an object via the application of force along a displacement

Link :https://en.wikipedia.org/wiki/Work_(physics)

While for example Feynman lectures , Hewitt and Walter lewins open lectures define work as its own formula Work is integral of Force with respect to displacement

While John r. Taylor's classical mechanics defines it as change in Kinetic energy

Which is more accurate and why it can't be directly said as transfer of energy ?

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    $\begingroup$ They are manifestly the same thing! $\endgroup$
    – naturallyInconsistent
    Commented Oct 1 at 12:17
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    $\begingroup$ Maby because there are so man different physicist writing books ;) $\endgroup$
    – Lukas Nullmeier
    Commented Oct 1 at 12:21
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    $\begingroup$ Taylor, in Classical Mechanics, does not define work as change in Kinetic energy.He shows that is related to KE but he defines is as dot product of force and dispalcement (in differential form), as most everybody. $\endgroup$
    – nasu
    Commented Oct 1 at 14:03
  • $\begingroup$ Interestingly, none of these definitions would consider the "work" done by an electrical circuit lighting a light or running a heater to be actually be work. I would define work as "energy usefully converted". I would include the caveat "usefully" because, for example, internal combustion engines produce more heat than motion - but motion is the goal. $\endgroup$
    – Grimm The Opiner
    Commented Oct 2 at 12:45

2 Answers 2

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I can appreciate your apparent frustration with the inconsistencies.

The Resnick statement "Work is energy transferred to or from an object by means of a force acting on the object" doesn’t include displacement. So as a definition of work it seems incomplete, unless you’ve left stuff out.

The Wikipedia article, which is similar to Feynman, is better, but it still lacks an important qualification, which is the displacement has to occur at the point where the force is applied. Static friction the ground applied to the drive wheel of a car causes it to accelerate without skidding, but the ground does no work (it doesn't transfer energy to the car). That's because there is no displacement of the point of application of the static friction force on the tire.

Taylor is incomplete (at least based on what you quoted). It sounds like the work energy theorem which is the net work done on an object equals its change in kinetic energy. I do work pushing a box at constant speed along a floor with kinetic friction even though there is no change in kinetic energy while I'm pushing it. That's because my work is not the net work done on the box. The net work is zero because my positive work equals the negative work done by kinetic friction for a net work of zero.

My view is the definition must include displacement of the point of application of the force.

Hope this helps.

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  • $\begingroup$ To be fair, you are not giving energy to the wall, you are expending it by yourself. $\endgroup$
    – Javier
    Commented Oct 1 at 13:04
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    $\begingroup$ @Javier I did say that. But the energy you expend is not physics (force x displacement) work. It is physiological work, as I explained in my answer here: physics.stackexchange.com/questions/594783/… $\endgroup$
    – Bob D
    Commented Oct 1 at 13:36
  • $\begingroup$ The Resnick statement seems to apply fine to the case of the motionless wall: No energy is transferred to it by means of a force, so no work is done on it. $\endgroup$
    – Chemomechanics
    Commented Oct 1 at 14:06
  • $\begingroup$ @Chemomechanics you’re right. I didn’t read the whole thing. I will update. But it still isn’t a definition of work as it doesn’t include displacement $\endgroup$
    – Bob D
    Commented Oct 1 at 14:36
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    $\begingroup$ Well, the very next sentence (of the chapter summary) reads "The work done on a particle by a constant force during displacement is $W=Fd\cos\theta=\vec{F}\cdot\vec{d}$ (work, constant force)". The premise of the original question is arguably incorrect, as @nasu has also pointed out. This is a case of incomplete reading and selective quoting. A single sentence in a textbook doesn't preclude fuller interpretation of a concept. $\endgroup$
    – Chemomechanics
    Commented Oct 1 at 15:26
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All the mentioned definitions are equivalent:

  • Your first two are stated in text,
  • your third one is stated mathematically,
  • and your fourth one is of experimental origin.

(I would, though, say that the first one is rather short and less precise since it leaves out any mention of displacement and thus can't be used alone.)

The precise mathematical definition of work is: $$W=\int \mathbf F\cdot \mathrm d\mathbf r,$$

and in words this indeed can be phrased as something like "transferred energy $W$ when a force $\mathbf F$ influences a displacement" ($\mathbf r$ is the displacement vector). Using words, such definitions become less precise, or at least not unique as there could be many equivalent formulations. This is the nature of language. The mathematical definition is a safer choice - that is our only precise "language".

Now, your fourth one is actually going back to the roots and is the origin for the other equivalent definitions. This one gives a practical means of defining work as follows: In a scenario where only the energy of motion (kinetic energy) changes, then the energy conservation law reduces to:

$$K_1+W=K_2.$$

The $W$ term represents energy that is added to the system and which causes the change in kinetic energy. This $W$ has then been given a name: Work. It turns out that this $W$ is added to the system due to forces that cause a displacement of the system - you can show that this is perfectly equivalent to the mathematical definition mentioned above.

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