What is the work done by a person on a box during horizontal displacement?

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The discussion centers on the work done by a person on a box during horizontal displacement, questioning the standard physics answer that states no work is done in this scenario. The consensus is that while lifting the box vertically involves work, the horizontal movement does not contribute to work because the force exerted is perpendicular to the displacement. Some participants acknowledge that, in reality, minor forces may be applied to initiate or stop the box's movement, but these are considered negligible for the purpose of the problem. The conversation highlights the distinction between theoretical physics and practical considerations in real-world scenarios. Ultimately, the conclusion remains that, under ideal conditions, no work is done during horizontal displacement.
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I think this post is appropriate here, because it is a conceptual problem, and I am questioning the standard answer.
From the halliday Quick quiz 7.2:

A person lifts a heavy box of mass m a vertical distance h and then walks horizontally a distance d while holding the box, as shown in Figure 7.5. Determine (a) the work he does on the box and (b) the work done on the box by the force of gravity.

http://dl.getdropbox.com/u/698994/ilustra%20forums/person_lift_box.jpg​
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the Halliday's answers:
(a) Assuming the person lifts with a force of magnitude mg, the weight of the box, the work he does during the vertical displacement is mgh because the force is in the direction of the displacement. The work he does during the horizontal displacement is zero because now the force he exerts on the box is perpendicular to the displacement. The net work he does is mgh + 0 = mgh.
(b) The work done by the gravitational force on the box as the box is displaced vertically is -mgh because the direction of this force is opposite the direction of the displacement. The work done by the gravitational force is zero during the horizontal displacement because now
the direction of this force is perpendicular to the direction of the displacement. The net work done by the gravitational force - mgh + 0 = -mgh. The total work done on the box is mgh - mgh =  0

Could someone discuss the "The work he does during the horizontal displacement is zero because now the force he exerts on the box is perpendicular to the displacement" part? I think the person did work during this step.
 
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vinirn said:
Could someone discuss the "The work he does during the horizontal displacement is zero because now the force he exerts on the box is perpendicular to the displacement" part? I think the person did work during this step.
Why do you think he did work? In what direction does he exert a force on the box? In what direction does the box move?

(In reality, he must have accelerated the box a tiny bit to start it moving, then deaccelerated it to stop once he got where he wanted. Ignore these minor complications.)
 
Doc Al said:
(In reality, he must have accelerated the box a tiny bit to start it moving, then deaccelerated it to stop once he got where he wanted. Ignore these minor complications.)

These minor complications are what I was referring to.
 
vinirn said:
These minor complications are what I was referring to.
Well, that's good thinking on your part. :smile:

You can always imagine that the person moves horizontally so slowly that the amount of work he must do to get the box moving can be arbitrarily small. And once he's moving, the horizontal force exerted on the box is zero, assuming he's moving at a constant speed.
 

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