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Consider this:

  • 1 kg object vs 1000 kg object
  • No friction applies in my example and the machine I am using is 100% efficient.
  • Some machine using a fuel applies 1 N of force over 1 meter to each object which equates to 1 J of transferred energy.
  • It takes 1.41 s for the 1 kg object to reach 1 meter. It takes about 45 s for the 1000 kg object to reach the same distance.
  • Applying 1 N for 45 s surely must require more energy input to the machine than applying 1 N for only 1.41 s. Yet both objects acquire only 1 J (due to the kinetic energy giving more importance to velocity than mass) as the 1000 kg object gain only a fraction of the velocity that the 1 kg object does.

But seeing that 1 N of force throughout 45 s CLEARLY requires more joules than 1 N throughout 1.41 s, where does the rest of the joules go? I mean, energy is a conservative thing, right?

Where does my reasoning fail?

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    $\begingroup$ "CLEARLY" I would say clearly not. You use 1 joule, there is no "rest of the energy". Energy is not "conservative". $\endgroup$ – my2cts Jan 7 at 19:03
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It doesn't matter how long it takes for an object to move some distance due to an applied force. The definition of work is just $$W=\int\mathbf F\cdot\text d\mathbf x$$

This definition says nothing about how long it takes for the force to be applied over the displacement. So in a 1D example, a force of $1\ \rm N$ applied over a distance of $1\ \rm m$ corresponds to $1\ \rm J$ of work if it takes $1\ \rm s$ to complete or $1000$ years.

The flaw in your reasoning probably comes from you imagining yourself taking a long time to push a heavy object as compared to a lighter object. You might have to put in more effort, but the work you do is the same.

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  • $\begingroup$ Ok, BUT, what about the conservation of energy? Would you agree that my machine would have to burn more fuel in moving the 1000 kg object compared to the 1 kg object (yes/no)? $\endgroup$ – dsadsadsa Jan 7 at 17:34
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    $\begingroup$ @dsadsadsa Right, but that doesn't violate energy conservation. For example, if you were just holding a book up, you are doing no work on the book at all, yet you will start to feel fatigued.Try to hold a heavier book and you will fatigue faster. $\endgroup$ – Aaron Stevens Jan 7 at 17:38
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It is not unusual for forces to do little or no work. Look at swinging a ball on a string. You have to apply a force on the string to keep the ball going, but the force is perpendicular to the motion and does no work.

In this case you do get something for your extra effort, but it is not energy. When the pushing is done the kinetic energy $1/2\ m\ v^2$ is the same for both objects, but the heavier object has more momentum in the same ratio ~$45/1.4$ since $\Delta (m v)=F \Delta t$.

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