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Let's say I push a block 5 meters with a force of 10 newtons. It can then be said that I did 50 joules of work on the block, since $$W = F*d = 10 * 5$$

However, when I push on the same block which is now up against a wall, the block does not move, and therefore, according to the equations, I do no work. However, I would like to ask if I truly do no work on the block, or if I actually do transfer 50 joules to the block, but the wall does work in the exact opposite direction, therefore we can either say that work on the block is (50 + -50) J or (0 + 0) J.

Which is technically correct?

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    $\begingroup$ "work" does require movement. Your intuition is in the right direction though, the wall does aplly a force on the block, countering your push. But as neither of these two forces causes movement, the work done is zero. Note also that unlike force, work is directionless. If it was as you speculate, that sum of yours would have been 50 + 50, not 50 + (-50). $\endgroup$
    – PcMan
    May 26, 2021 at 7:54
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    $\begingroup$ @PcMan work can be positive or negative $\endgroup$
    – Bob D
    May 26, 2021 at 8:53
  • $\begingroup$ You and the wall are exerting equal but opposite forces, and are thus changing the block's momentum equally but oppositely (a net change of zero). However, neither you nor the wall is doing any work on the block. Positive/negative work is equivalent to the gain/loss of kinetic energy (see work–energy theorem). $\endgroup$
    – Jivan Pal
    May 26, 2021 at 18:20
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7 Answers 7

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In your second example no work is done by you or the wall on the block because there is no displacement of the block. That is not to say you didn't expend any energy pushing on the wall. But the work you did is internal physiological work, and not physics work. Richard Feynman explained it this way in his physics lectures:

The fact that we have to generate effort to hold up a weight is simply due to to the design of striated muscle. What happens is when a nerve impulse reaches a muscle fiber, the fiber gives a little twitch and then relaxes, so that when we hold something up , enormous volleys of nerve impulses are coming in to the muscle, large numbers of twitches are maintaining the weight, while other fibers relax. When we hold a heavy weight we get tired, begin to shake, ...because the muscle is tired and not reacting fast enough.

That said, work can be positive or negative. Work is positive if the direction fo the force is the same as the direction of the displacement of the object. Positive work transfers energy to the object. In your first example you did positive work of 50 Joules on the block.

Negative work occurs if the direction of the force is opposite to the displacement of the object. Negative work takes energy away from the object. If in your first example you pushed the block on a floor with friction, the floor does negative friction work equal to the kinetic friction force times 10 meters. Friction work takes the energy from the motion of the block and dissipates it as heat.

If the block started at rest and ended at rest at 10 meters the negative friction work equals your positive work for a net work of zero. Per the work-energy theorem the net work done on an object equals its change in kinetic energy.

Hope this helps.

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  • $\begingroup$ And to expand on the muscle explanation... it's my understanding that even when your arm (or whatever) doesn't appear to be shaking, it actually is, just on a scale too small to see. So in actuality you are doing tiny bits of positive and negative work repeatedly which is turning into heat. $\endgroup$
    – Michael
    May 27, 2021 at 16:11
  • $\begingroup$ @Michael Yes, initially you don't appear to be shaking. Its when fatigue sets in, as explained by Feynman. $\endgroup$
    – Bob D
    May 27, 2021 at 16:14
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However, I would like to ask if I truly do no work on the block, or if I actually do transfer 50 joules to the block, but the wall does work in the exact opposite direction, therefore we can either say that work on the block is 50 + (-50), or 0 +0

$0 + 0$ correct as pointed out by other answers. However, I would like to add that the $ 50 + (-50) $ would have been the case when (suppose ) you push the block for $5 m$ with $10 N$ of force that moves the block constant velocity and when friction is present. In this case the work done by you is equal and opposite to the work done by friction and there is no change in the object's kinetic energy.

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  • $\begingroup$ The distinction being that surfaces with friction don't completely prevent movement, unlike walls. (ok walls can be broken if you apply really high force) $\endgroup$
    – smci
    May 26, 2021 at 20:41
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You are doing exactly no work.

Think of a table. With its normal force it can hold up an apple forever. Does it constantly spend energy on doing that? No. Otherwise it would eventually "run out", which doesn't happen.

No energy is transferred between objects if no displacement or thermal interaction happens.

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  • $\begingroup$ But after trying to push the block through the wall, I am fatigued. $\endgroup$
    – Filippo
    May 26, 2021 at 10:22
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    $\begingroup$ @Filippo True, but that is not because you have supplied any work to the wall. That is just because you are using an inefficient human body to produce the force you are exerting, and the human body spends energy just to produce such a force. (It even spends energy while not producing the force). The fatigue you feel is due to energy being spent (muscles contracting, elongating, tensing, your circulation and adrenaline increasing etc.) within the body, not due to any work being done on the wall. $\endgroup$
    – Steeven
    May 26, 2021 at 10:36
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    $\begingroup$ @Filippo Biomechanics makes things a lot more complicated - it's often much simpler to replace yourself with an inanimate object that does the same thing. Instead of you pushing on the block, just lean a heavy brick against it that provides the same horizontal force. The problem is identical in terms of forces, but you don't need to feed or fuel either the brick or the wall. When trying to figure out if work is being done, you can just examine the apparatus and ask, "do I need to power this thing?" For a brick leaning against a wall, the answer is "no". $\endgroup$ May 26, 2021 at 17:24
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When you talk about the block being pushed against the wall at that moment only, there is already no displacement present of the block if ( which is 5m you have provided). This 5m is possible only when it moved before touching the block.Therefore ,W=0 for the block being pushed just near or touching the wall.

Even if let us say you had already pushed the block and after 5m , the wall stopped the block. Then , we calculate work done at the reference point. For example , after 5m before touching wall . W=50J but after touching the wall , W=0 , not (50J-50J) because at that reference , there is no displacement of the block.Remember, our system is block.

Points: In terms of work done to understand it more better.

  1. F = (m*a)

where mass of block is present but there is no acceleration of the block due to the wall pushing it back. Wall doesn’t let the block create acceleration for itself.

  1. Displacement

Since there is no displacement either of the block nor the wall. It doesn’t matter if wall displaces or not because our system is block. So , we must only consider whatever happens to the block. Therefore , displacement = 0 as well.

Hence , W = F(0) * S(0).

Do let me know if you have any difficulty.

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Actually, for a real box, you will do some work when you compress the box. To hold it compressed you will have to contract your muscle cells continuously, to keep the force steady. They contract, expand, contract, expand, etc. Holding the box compressed (or just pushing on an incompressible box) is easier when you, say, place a strong ventilator in front of it. That way you don't get tired by the work done while continuously making your muscles contract.
In the same way, keeping a book on constant height requires a constant force. This can be given by a static table or by your stationary hand. Keeping your hand stationary does require work. Again you have to make a part of your muscles contract continuously. your hand is in a dynamic equilibrium instead of a static one like the table.

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No. This isn't correct. The definition of work is a technical definition in physics. In physics, the work done is defined as $$W=\vec F \cdot \vec d$$ This definition is useful in applying the conservation of energy and work-energy theorem. When you push on the block up against the wall, you apply force to both the block and the wall but you do no work. Of course your body needs energy to supply the force.

Hope this helps!

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All the above answers are referring to classical mechanics of work which is defined as $W=F*d$, i.e. as linear momentum in a closed system in units of poundes*inches, but since there is no motion work equals zero. By Newton's conservation laws energy is conserved.

If we look at system setup as an open system (Ludwig von Bertanlaffy's Open Systems Theory, the external energy (muscles, $E= F* dv/dt$ - as equivalent to Newton Law of Motion, $m*a = \mathrm{lb}*\mathrm{s}$, is pushing the (brick against the wall and we consume energy $E$. In open systems we have case of energy conversion by interaction between systems (external energy (muscles) , brick (mass $1$) and the wall (mass $2$)), as the energy is returned back to environment in a form of pressure potential energy, $F/ A$ ($\mathrm{lb}/\mathrm{in}^2$ of contact area). Here instead of converting energy to motion in form of work ($W=F*d$, time independent), we convert external energy to work ($E= m * dv/dt$ time dependent (lb sec). In gravity environment, since brick and wall are rigid, based on Emmy Noether principle of Systems Symmetry, momentum is conserved, but energy is not conserved. If brick or wal wall were elastic or plastic, energy would be converted in brick or wall deformation relative to Newtonian global coordinates.In free space without any external constraints on systems, force from muscles would set brick and wall masses in linear motion relative to each other following Newton Law of motion.

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  • $\begingroup$ The parentheses here are confusing because they do not all match each other. $\endgroup$ Jun 2, 2021 at 14:25

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