Work done on an object is equal to $$FD\cos(\mathrm{angle}).$$ So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.

However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.

But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?

Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction,

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    $\begingroup$ We already have quite a few questions on the distinction between physical work and muscular effort and how to reconcile the two. I think that physics.stackexchange.com/q/1984 is the leading example of the class, and you will find many other links therein (see the "Linked" sidebar). $\endgroup$ – dmckee --- ex-moderator kitten Aug 13 '19 at 20:05
  • $\begingroup$ Now that I look again at the answers to the linked question I have to say that many of them are quite poor. Nor do I find another question with a notable better pool of answers. $\endgroup$ – dmckee --- ex-moderator kitten Aug 13 '19 at 20:11
  • $\begingroup$ @dmckee I had started writing an answer before your first comment, and couldn't see any question that really addressed the specifics well (though TBH I didn't look through 4 pages of linked). If you think the answer should be removed, or posted on another Q instead, just let me know. $\endgroup$ – JMac Aug 13 '19 at 20:15
  • $\begingroup$ But my question is not really about physical work and muscular effort. It could be any force. Thanks anyway though $\endgroup$ – Jan F. S Aug 13 '19 at 20:42

Yes, when accounting for factors such as air resistance and biological processes, there is clearly energy used, and work being done somewhere.

If you're only considering a book moving at a constant velocity in the horizontal direction, you could say there is no net work done on the book.

If we analyze it further, you will find that if it moves with a constant velocity, there must be some other force acting on it to counteract the drag force. This is the force supplied by you when moving the book.

If this book is moving with constant velocity, this must mean that the net force acting on the book in the horizontal direction is $0$. If that is the case, the net work in the horizontal direction required to keep it moving at the horizontal velocity is zero. The work done by the person on the book is exactly countered by the work done by the air resistance on the book, and thus no net work.

Work is required for a person to move a book through the air at a constant velocity, because the work done by air resistance needs to be countered, but the net work on the book during that movement is none.

It's also often the case that it's assumed you can neglect air resistance. In that case, it would not require work to keep the book moving at a constant velocity. I suspect they may be making that assumption when they are discussing work.

You could also consider a situation where you are starting from a standstill, and ending at a standstill while covering some distance. You can see there is work done to accelerate and decelerate the book. From an energy/work perspective, the work required to slow down the book is exactly opposite of the work required to get it up to the moving speed. This also means that no net work is done on the book during the entire travel (though both the person and the air will have had net work done on them).

  • $\begingroup$ Thanks your answer pretty much reflects my own intuition. But one question though. Your last sentence. "though both the person and the air will have had net work done on them)." I don't follow this - wouldn't the same apply to the person as the book? $\endgroup$ – Jan F. S Aug 13 '19 at 21:51
  • $\begingroup$ @JanF.S Not for the air at least. That will be put into motion by the person passing through it. I guess you wouldn't call it "work" on the person, since friction with the ground would be countering the force from the book; but the person also does have a net energy change in the process, unlike the book. $\endgroup$ – JMac Aug 13 '19 at 22:03
  • $\begingroup$ What net change in energy would it be for the person? If he comes to a standstill then his kinetic energy would be zero again and potential energy is irrelevant. I assume you don't mean the biochemical changes in the muscles? $\endgroup$ – Jan F. S Aug 13 '19 at 22:07
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    $\begingroup$ @JanF.S Yeah, an internal energy to overcome the energy required to combat air resistance. The same would apply if it wasn't a person, but a device. There would be something providing energy so that it can overcome that. $\endgroup$ – JMac Aug 13 '19 at 22:10

Work-energy theorem would be quite helpful to you. $$dW_{net}=dK_{system}$$ As speed is constant,$$dK_{system}=0$$ $$dW_{hand}+dW_{resistance}=0$$ It doesn't imply that work done by your hand is zero but the net work done by your hand as well as resistive forces is zero.


You do work on the book. To carry the book in the horizontal direction you move with a force equal to Mass of the person plus mass of book times acceleration. F = Ma where M is the total mass. Hence work done is the force times the distance traveled. You have to keep applying a force to overcome atmospheric resistance and the friction of the ground.


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