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I have problems fully understanding the concept of work, so please forgive me if this is simple. If I take a box of mass $ m $, and lift it a distance $ d $ vertically, why is the work I have done equal to $ gmd $, where $ g $ is the force gravity exerts on the box? I understand that work is equal to force times distance--so I'm not asking about the definition of work--but if I exert an upward force equal in magnitude to gravity's, won't the box remain motionless, i.e., net zero force, in which case the velocity is constant, and displacement and work done will be equal to zero?

Edit: To be clear, what I'm asking is not a duplicate of "Why does holding something up cost energy while no work is being done?", because I'm not asking about work done on an object with zero displacement, nor is it a duplicate of "What exactly is F in W=∫baFdx?", because I'm not asking about the distinction between the work done by an individual force and net force.

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marked as duplicate by Emilio Pisanty, David Z Nov 10 '13 at 4:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
You forgot to divide by time. –  Optionparty Nov 9 '13 at 22:46
    
Work depends on distance and force, not time, though, right? –  BlueBomber Nov 9 '13 at 23:53
    
Comment to the question (v1) Note that the definition of work depends on which force is considered. –  Qmechanic Nov 10 '13 at 1:29
    
Qmechanic, I'm specifically focused on the work I do on the box in this question. –  BlueBomber Nov 10 '13 at 2:37
    
It is completely unclear, then, what your question actually is. If you're asking whether the work is zero for a stationary object, the answer is yes. Otherwise, you need to spend your time making it clear what your question is instead of what it is not. –  Emilio Pisanty Nov 10 '13 at 12:55

2 Answers 2

You ask

if I exert an upward force equal in magnitude to gravity's, won't the box remain motionless, i.e., net zero force, in which case the displacement and work done will be equal to zero?

The box will not necessarily remain motionless, but it is true that the net work will be zero. If the force you exert is equal and opposite to that of gravity, then Newton's Second Law tells us that the acceleration of the box is zero. This still allows for the box to be moving at a constant velocity from the initial height to the final height.

To address the following question more generally (even for the case when the force you exert is not necessarily equal to that of gravity)

why is the work I have done equal to $gmd$

tt helps to know the so-called Work-Energy Theorem. The theorem says that the work done by the net force on an object equals the change in its kinetic energy; \begin{align} W_\mathrm{net} = \Delta K \end{align} Now suppose that the box is at rest at some point $a$ and that you move it to some other point $b$ at rest. Then the change in kinetic energy will be zero $\Delta K=0$. On the other hand, the work done by the net force is the sum of the work done by you, and the work done by gravity; $W_\mathrm{net} = W_\mathrm{you} + W_\mathrm{gravity}$. Combining these facts with the Work-Energy theorem gives \begin{align} W_\mathrm{you} = -W_\mathrm{gravity} \end{align} as desired.

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Thank you for your reply. Without reaching for any theorems (we are talking fundamentals that should be explicable without them, so it is like putting the cart before the horse, it feels), how can one say how much work I did on the box? You know its initial and final positions and velocities, but you don't know how fast I lifted it at each point in time, etc. I just don't see how the textbooks give a plain answer of "mgd" for this. If net acceleration of an object is zero, and it started at zero m/s, the answer cannot be mgd, or else no work is done, right? What am I missing? –  BlueBomber Nov 10 '13 at 2:42
    
@BlueBomber If you exert a force $mg$ then sure the acceleration will be zero, but as I said in the answer, it's velocity can still be any nonzero constant, so you will simply be pulling the box up at a constant velocity, and the work you do will be the force you exert, namely $mg$, time the distance traveled, namely $d$. It doesn't matter what velocity the box has during its trip; if you exert a force $mg$ over that distance along the whole path, the work done will be the same for every velocity. –  joshphysics Nov 11 '13 at 4:26

Yes, the net work on the box will be zero. The same holds if the box were sitting on a table. The table counters the force of gravity on the box. However, your muscles are not static. The muscles do not lock into position; they require a continual feed of chemicals to keep the static force on the box. That is why you are still doing work when holding the box: not on the box, but in keeping your muscles contracted.

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Thanks for the comment, but to be clear, I'm not asking about the work done on an object with zero displacement; that question has already been answered here. –  BlueBomber Nov 10 '13 at 0:02

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