# Energy used to stop / slow an object

I'm trying to workout how much energy (if any), I use (imagining me as an efficient machine rather than a complex bio-mechanical human) when I lower or catch a weight. I understand that when I push it up, I have used energy to do work as I have the force of the object (it's mass and gravity) and distance. So I've got two similar questions...

If I am to stop a free-falling object, how much energy do I use? If I look at it from energy change I believe I need to use the same amount as the kinetic energy of the object at the moment before I catch?

If I slowly lower an object how much energy do I use? In an ideal world would the total be the same as in the free falling situation, assuming the height at which the object fell was the same.

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You can't answer your question by assuming an efficient machine, because the effect you're asking about is specifically a property of complex biomechanical systems. An efficient machine would actually gain energy by catching and lowering a falling weight. The fact that such an act is tiring for a human is exactly that we are not built like efficient machines. –  Colin K Sep 19 '12 at 2:18
@Colin K Thanks, that makes sense! –  user1529408 Sep 19 '12 at 10:48

This is an expansion of Colin K's comment (Colin, if you want to convert your comment to an answer I'll delete this).

When you catch a falling object the object does work on you because it exerts a force on you for the distance it takes you to slow the object. Suppose your arm was a spring. This isn't as far fetched as it sounds because making limbs act as springs is quite common in nature. Kangaroo's legs work this way. Anyhow, if your arm was a spring the object would compress your arm as you caught the weight. The work done by the object would be stored as elastic energy in your arm. Once you've caught the object you could let you arm spring back and propel the object up again, or you could drop the object and use the stored energy for something else. In both cases the total work done by you is zero.

The problem is that our arms aren't springs, and human muscles have to continuously use energy to maintain a force even when your muscle doesn't move so no mechanical work is done. The energy needed will depend on all sorts of factors that aren't related to Netwonian mechanics but instead to the details of your physiology. That means it's impossible to provide a sensible answer to your question.

The work involved in catching, raising and lowering weights is easily calculated by their change in kinetic and potential energy, but this will be only loosely related to the energy you expend in moving the weight.

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Ok that makes sense, thanks. Just to clarify when I am pushing the weight up I am using energy to do work (weight * distance)? –  user1529408 Sep 19 '12 at 10:47
Yes. If the object of mass $m$ rises a distance $h$ then the work you've done on the object is equal to the increase in it's potential energy i.e. $mgh$. The energy you expend to do this is a lot greater than $mgh$ because human muscles aren't particularly efficient. I think muscle efficiency is around 15-20%. –  John Rennie Sep 19 '12 at 14:33
Thanks, just wanting to make sure I was not completely incompetent. Also very interesting to learn about muscle efficiency is around 15-20%! –  user1529408 Sep 20 '12 at 13:58