# How does energy remain constant when the NET work done is zero?

If I were to go from X to Y and then back to X, the NET work done by me would be zero. But, how will the energy remain constant? Really confused about energy and work here. Energy is the ability to do work, so if the net work is zero, the energy should remain the same. But, don't we utilize energy in moving from one point to another and then back?

• Do we utilize energy moving from one point to another and back? Certainly we would if dissipation was present (e.g. friction, air resistance, waste heat from converting chemical energy to mechanical energy), but then a nonzero amount of net work is done by dissipative forces in that case. Have you considered a situation where there's no dissipation at all? Commented Sep 9, 2020 at 11:18
• Ah, yes, let's assume an ideal case in which there's no friction! Still, won't I be utilizing energy to move forward and back? Commented Sep 9, 2020 at 11:31
• What does it mean to "utilize" energy? Where does that energy go? Remember, it can't be created or destroyed. Commented Sep 9, 2020 at 12:46
• This is another example of textbooks/teachers not giving enough focus on the imprtance of properly defining the system. See my answer here: physics.stackexchange.com/a/735981/123502 Commented yesterday
• Related: Why does holding something up cost energy while no work is being done? and links therein. Commented yesterday

If I were to go from X to Y and then back to X, the NET work done by me would be zero.

You need to be careful here. Forces, not objects, do work. Additionaly, the total work done on you is equal to your change in kinetic energy. Therefore, a more precise thing to say would be

If I go from X to Y and back to X such that I start and end at rest at X then the net work done on me would be zero.$$^*$$

If you are talking about walking, then typically accelerations are caused by friction between the ground and your feet. So ultimately if you have started and stopped at rest then friction has done no net work on you. This makes sense. As you speed up, friction points in the direction of your motion, which is positive work being done. As you slow down the opposite is true. Combining these gives no net work.$$^{**}$$

The whole "energy is the ability to do work" can be a good qualitative picture, but it's usually not as ambiguous when you instead focus on the more precise definitions and relations.

$$^*$$ We don't really need to return to the starting point here to get at the heart of the discussion, but I have left it in since it's your scenario.

$$^{**}$$ This is actually not entirely correct, since, if you aren't slipping, the point of contact between your feet and the ground doesn't move as you take a step. Therefore, friction isn't actually doing any work here. However, if one is more careful with breaking up the various body parts, moving pieces, forces, etc. then one arrives at the same conclusions anyway. I'm choosing a less precise description in order to have a clearer answer.

Ah yes, I think this question can only be answered by the second law of thermodynamics. The second law of thermodynamics states implicitly makes a statement that you can never have a 100% efficient cyclic process. If you first take the body from point A to point B , then you'd have to do some more work to suddenly reverse the process back into the original state.

So, in reality, it is not actually possible to loop back the body back from point A to point B, there will definitely be ton of losses such as friction losses, air resistance losses etc. The amount of loss you can only figure out by measuring the initial and final energy.

You may find this stack exchange interesting, even though the answer talks about thermodynamics, it applies equally well to the example you have involving forces.

But, don't we utilize energy in moving from one point to another and then back?

Actually no. You don't even need to go back. All you have to do is start from $$A$$ at rest and end at $$B$$ at rest.

In the absence of dissipation (like air resistance finally transformed into "heat" raising the temperature of air), you don't consume energy at all. You just convert it into motion (called kinetic energy) when you accelerate and get it back when you slow down.

Have a look at power indicator on an eletric car. When you slow down, the car is capable of converting some of the kinetic energy back into electricity and store it in the battery. If things were perfect, it could recover 100% of the energy spent for acceleration when you deccelerate.

Of course, things are FAR from perfect in real life because of many physical processes like friction, air resistance, heat lost in chemical reactions...

But, don't we utilize energy in moving from one point to another and then back?

This actually depends on the circumstances. If by "we" you mean a human body moving from one point to another, then yes we use energy in both directions. Living bodies are notoriously difficult to use as an example of the basic laws of physics because there's so much going on. In my opinion, that is one of the greatest challenges of learning introductory physics. While the rules apply obviously to simple machines like springs and levers and whatnot, it's often far less intuitive to see what happens inside a living body, with all of the complicated sensory feedback loops and chemical reactions involved. And, of course, we always have the most experience with... well.. our own body.

However, in other circumstances, it's easier. Consider a wind up toy car with a spring. You push on the front of the car to move it backwards, winding up the spring. Then you let the car push on you, letting the unwinding of the spring move the car forward. You control the force of your hand such that the car ends up motionless at its starting point (don't let it go free, as one usually does with a windup car). In this circumstance, you will find that the car does exactly the same amount of work to your hand on the way back as your hand did to the car while winding the spring. If the wheels didnt' slip, it should be natural to see that the spring should be in the exact same state of tension as it was when you started. The net potential energy added/removed from the car is 0.

If the sum was not zero, then the car would need to have different energy than it started. Either it would need to have kinetic energy (which it doesn't, because we structured the problem to make sure it is at rest at the end), or it would need potential energy (stored in the spring). But intuitively, we see the spring should unwind to exactly the same state (since it is geared to the wheels, and we structured the experiment not to engage the clutch that lets the wheels spin independently from the spring).

If I understand your question, you need to distinguish between MECHANICAL work and METABOLIC work. Our bodies are using energy all the time just to stay alive through breathing, cellular respiration, heart pumping, etc. Experimental biologists measure this via the amount of oxygen consumed, and maybe other ways.

To physicists and engineers, work is defined in different ways: $$\int F dS$$, $$\int P dV$$, $$- \nabla U$$ etc depending on the context.

Traveling on level ground does not require any work in theory. Many physics problems involving rolling disks and the like assume no dissiptation. But we are all aware that real wheels have small dissipative forces, and cars require energy in order to travel on flat roads or even to sit idling in traffic. The metabolic work will always be greater than the mechanical work done. Akin to the efficiency calculation of an engine or other thermodynamic processes.

So, yeah, in the real world, moving from place to place takes energy unless the process is reversible. Like a mass on an ideal spring.