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If it takes energy to slow an object down, and then the object also loses KE, then how is energy conserved? Don't you have a net loss of energy?

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    $\begingroup$ The energy of a system isn’t conserved when an external force does work on it. $\endgroup$ – G. Smith May 25 at 17:58
  • $\begingroup$ You could have asked a similar question when an object speeds up. The gain in energy is equal to the work done by the force applied to it. $\endgroup$ – G. Smith May 25 at 18:00
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    $\begingroup$ It doesn't take energy to slow an object down. What exactly do you mean by that? $\endgroup$ – Steeven May 25 at 18:04
  • $\begingroup$ Usually, you need to take into account something else like Newton's third law or friction. For example, in a frictionless environment when you slow down an object by pushing on it, you are pushed by the object, and then you are the one with kinetic energy now. So energy is merely transferred in that case. Newton's third law is actually very important to take into account in these cases. $\endgroup$ – SpiralRain May 25 at 18:05
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If it takes energy to slow an object down, and then the object also loses KE, then how is energy conserved? Don't you have a net loss of energy?

You never have a "loss" of energy. It is always conserved. It may just change its form. In order to slow the object down it takes negative work that takes the kinetic energy away from the object and does something with it.

That negative work could be dry friction work between surfaces in which case the loss of kinetic energy increases the temperature of the surfaces (their internal energy). Friction force opposes motion so the work is negative. Then the higher temperature surfaces can transfer heat to the lower temperature surroundings. Then it becomes the internal energy of the surroundings, etc.,etc.. If you follow all the energy transfers you realize the energy is never "lost" but simply morphs into different forms.

If you throw an object up in the air it slows down due to the force of gravity. Gravity does negative work (its force is also in the opposite direction to the motion). But in this case it takes the kinetic energy away from the object and gives it gravitational potential energy. When it starts falling down gravity does positive work on the object converting its gravitational potential energy into kinetic energy. If this is done in a vacuum, mechanical energy (kinetic plus potential) is conserved. If there is air drag, then once again some kinetic energy is lost due to air friction again eventually as heat and eventually becoming another form. But again, the energy is not "lost".

Hope this helps.

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Energy conservation means the total energy of the system, which includes the object and the person who does the slowing down.

If the object has mass $m$ and moves with speed $v$ the kinetic energy is $\frac{1}{2} m v^2$ so that the person must supply this amount of energy to stop the object. This work supplied by the person may be supplied in a variety of approaches. For example, if they apply a constant force $F$ then the object will come to rest over a distance $d$ where $d$ is determined from $$F d = \frac{1}{2} m v^2.$$

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    $\begingroup$ It is quite the opposite, the object is doing work until it stops. If it is stopped by a person, there comes the biological aspect. The body spends energy for absorbing energy as work is being done on the body. The object kinetic energy and body chemical energy are converted mostly to thermal energy. $\endgroup$ – Poutnik May 25 at 19:29

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