Conservation of energy when slowing an object down 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?
 A: 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.
A: 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 $$F d = \frac{1}{2} m v^2.$$
