Does an opposing force cause loss/ waste of energy?

I found this answer by John Rennie (in the question you see the train on a track):

when the train moves a distance $d$ the work done on the train is $Fd\cos\theta$. It's certainly true that the horse is exertiong a force $F$ that is greater than the force on the train,... The dot product is defined as: $\vec{F}\cdot\vec{d} = Fd\cos\phi$, where $F$ and $d$ are the magnitudes of the vectors and $\phi$ is the angle between the vectors. In our case the angle $\phi$ between the vectors is $\pi - \theta$, so $$W_{horse} = Fd\cos(\pi-\theta) = > -Fd\cos\theta = -W_{train}$$ the work done by the horse is equal to the work done on the train. So no mechanical energy is being wasted by pulling at an angle.

However this overlooks the fact that muscles are inefficient things and consume energy even when no mechanical work is being done. This does indeed mean the horse will use more energy to pull the train at an angle, but this is just down to way muscles work rather than fundamental physics. To calculate the extra energy the horse uses you'd have to go off and ask a biologist about muscle physiology.

in which the fact that "(...it's certainly true that the horse is exerting a force F that is greater than the force on the train)" F is greater than the net force $F>F\cos\theta =F_{net}$ is attributed to the way muscles work.

I am aware that mechanical work is the product $F_{net}\cdot d$, but, if we substitute the horse with a defined mechanical force, the result is the same and we cannot attribute the cause to muscles.

Also, we know how to use freebody diagrams and calculate net force subtracting opposing forces, why doesn't he consider the opposing force exerted by stiction (track: $F_{stiction} = F \sin\theta$) what in the original sketch is defined as nonworking reactive power) and say that the wasted energy (the force/work on the track) can be easily determined, even if this is not mechanical work?

Also, if a steel ball A ($J = k, Ke = E) hits another identical steel ball B it does work (on B), if it its a wall Wand this doesn't budge, it does no work on (on W), but why we cannot conclude that exactly same E was lost/wasted on the wall? • Hi. Is it possible that in the answer you have read, the answerer argues about the power used by the muscles? That means that in a muscle takes an amount of energy by some biological process, the work the muscle can give as output is lesser than the energy he has received. It's about efficiency. You never have 100% efficiency. But I'm not sure, I'm making an assumption here. – Constantine Black May 25 '15 at 20:25 • Yeah, the amount of energy consumed by muscles in performing some externally measurable amount of mechanical work is hard to calculate with any accuracy. Even if the muscles themselves are perfectly efficient, the heart and lungs are "burning" energy that is not reflected in the mechanical work. And try holding a 20 pound weigh with your arm out horizontally. Even though you hold the weight (almost) motionless, doing no "work", your body will quickly tell you that you're burning energy. – Hot Licks May 27 '15 at 0:05 1 Answer in which the fact that "(...it's certainly true that the horse is exerting a force F that is greater than the force on the train)" F is greater than the net force$F>F\cos{θ}=F_{net}$is attributed to the way muscles work. No, it's not. This is just a property of the dot product. The comment on muscle efficiency was merely an aside which, in hindsight, probably caused more confusion than it's worth. To be clear, the reason the force$F$exerted by the horse is greater than that on the train is because the force is acting at an angle. This is why the force on the train, which is constrained to move in one dimension along the tracks, is given by $$F_{net}=F\cos{\theta}<F.$$ But nothing is "lost" because, as John Rennie showed, the work done by the horse is equal to the work done on the train. There is, as you say, a force perpendicular to the tracks given by$F\sin{\theta}\$. However there is no work done because there is no motion in this direction. Hence, there can be no "wasted energy" as a result. Equivalently, we could say there is no net force in this direction, because the pulling force is balanced with a normal force from the tracks.

Similarly, a steel ball will do no work on a rigid wall because there is an equal and opposite (normal) force opposing that of the ball. The same cannot be said for a collision between freely-moving identical steel balls - it is easy to imagine the second ball gains at least some of the momentum of the first (i.e. there is some work done). I hope it is clear now why we cannot say the energy is the same in both cases.