Calculating Wasted Energy: The Efficiency of Force Applied at an Angle

[bobie]

Edit: Suppose you are pulling a weight along a track at an angle (in the picture 45°).
horse pulling a cart on a track

If the object is dislocated by a distance D let's assume that the work done/energy tranmitted to the object is $E_{45}$ = K J. If you had pulled along the track, would the distance covered by the object be $D_0$ = K/cos45° and $E_0$ = K/cos45 = 1.41 * K? if this is correct, can we conclude that the energy wasted/ calories burned doing no mechanical work is 41 %?.

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

I know 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 that by 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 wasted?**
If this logic is valid, the same logic has been applied to the horse.

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

 

[xAxis]

Weight is proportional to mass, which doesn't directly determines the force you are pushing object with. The fact that you pushed object weighing 100N (cc 11kg) for 1 meter doesn't mean that work done is 100 j.

 

[Chestermiller]

I assume the track is an incline, and it rises vertically a distance of 1 meter, so that the distance you push the weight up the incline is 1.41 meters. But the force you need to exert to push the weight up the incline is only 70.7 N. So the work is still 100 joules.

Chet