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Suppose you are pulling a weight along a track at an angle (in the picture 45°).

If the object is dislocated by a distance $D_{45}$ let's assume that the mechanical work done on/energy transmitted to the object is: $$ W_{45} = (\vec{F}\cdot\vec{d} = Fd\cos_{45}) = K J$$, which in the picture is called: working (real) power.

If you had pulled along the track, presumably the distance covered by the object would be $D_0$ = $D_{45} /cos_{45}$ and the work done would be $$W_0 (= E_0) = (\vec{F}\cdot\vec{d} = Fd\cos_{0}) = K/cos_{45} = K * 1.41 J$$, which in the image is called: total (apparent) power

If this is correct, can we legitimately suppose that the same amount of energy/calories were burned in both cases, that is: $E_0 = E_{45} = 1.41 * K$, but, since the work done is less, that is: ($W_{45} < W_0 = 0.7* W_0$) can we therefore conclude that the energy wasted in the first case (calories burned doing no mechanical work) is 41%? That energy was wasted trying to dislocate the track or derail the cart: non working (reactive) power

enter image description here

(If this is not correct, how can we calculate the energy wasted when we apply a force at an angle greater than 0°?)

update

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.

I knew that no mechanical work in excess is done, that is because of the peculiarity of the definition of work. I tried to dodge this ostacle speaking of calories burned: I thought we can desume the exact value by, so to speak, reverse engineering.

The logic is this: we have instruments (I am referring to instruments, so we avoid inefficiency of human muscles) to produce and measure force. If we measure the force something exerts on a blocked object, then remove the block and measure the oucome of that force, can't we deduce that the same force has been applied before and the same amount of energy/calories has been burned withoud doing any work? If this logic is valid, the same logic has been applied to the horse.

A biologist can tell yus the percentage of calories burned in excess with regard to the effective force applied/ transmitted: the inefficiency of the human machine is around 80%. WE burn 4 - 5 times more energy than we put to avail. Efficiency may vary from 18% to 26%. In the case above, energy actually burned would therefore be 41 * 4.5 = ca. 185%

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Let's zoom in on the horse to look at the forces being applied and the distance moved:

Horse

The force on the train is $F\cos\theta$ so 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, and the horse also moves a distance $d$. But remember that work is given by:

$$ W_{train} = \vec{F}\cdot\vec{d} $$

where the force $\vec{F}$ and the distance $\vec{d}$ are vectors and the $\cdot$ is the dot product of the two vectors. 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 the work done on the horse is:

$$ W_{horse} = Fd\cos(\pi-\theta) = -Fd\cos\theta = -W_{train} $$

The work done on the horse is equal and opposite to the work done on the train, or to express this in a more obvious way 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.

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  • $\begingroup$ What about the increased friction? $\endgroup$ – JonathanReez Apr 20 '15 at 14:59
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The key problem in your analysis is this mistaken assumption:

can we legitimately suppose that the same amount of energy/calories were burned in both cases

No, you cannot assume that. If you developed an efficient machine for pulling (instead of a horse which is very inefficient) then it would burn less energy in the "diagonal pull" case.

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  • $\begingroup$ No, you don't use the same power. Calculate it instead of assuming it. $\endgroup$ – DaleSpam Sep 22 '14 at 1:58

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