Assuming that I am holding a ball in midair, it has potential energy and 0 kinetic energy. Next, I move the ball in a downwards direction, it has kinetic energy but then I stop the ball in midair. Where did the kinetic energy get transferred to or what other energy did the energy get converted into?
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$\begingroup$ Do you move the ball or let it fall freely and catch? $\endgroup$– ACBCommented Jul 19, 2021 at 5:50
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$\begingroup$ @ACB move the ball $\endgroup$– Newton HofsteiderCommented Jul 19, 2021 at 5:51
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3 Answers
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First imagine moving the ball in horizontal direction using your hand.
In this case, your hand generates and passes kinetic energy to the ball during the motion. And when you stop it, the hand takes back the kinetic energy it provided and converts it into heat.
Now, if the direction is vertically downwards, the ball does lose potential energy. Where does it go? It results in slightly less effort that you feel in moving the ball down (imagine a very heavy ball), than in moving it any other direction.
In this case, depending upon how fast you moved your hand, during the journey, your hand either continuously provided extra kinetic energy to the ball (than it would have gained by losing potential energy) or absorbed some of the kinetic energy of the ball and converted it into heat. In any case, in the end, your hand took any leftover kinetic energy from the ball and converted it to heat.
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In the real world, if you moved the ball from the higher position to a lower position, the "lost" potential energy ended up as heat in your arm muscle.
By way of:
As the ball starts being lowered (assume you lower it at freefall speed for simplicity), the potential energy becomes kinetic energy of downward motion.
Your hand then applies an excess force upwards to the ball, to decelerate it to a stop.
To achieve this upwards force of your hand, your arm muscles must contract.
To achieve the contraction of your muscle, your arm needed to consume some fuel(ATP, derived from glucose and oxygen).
The end result, without considering efficiency of each step, is that the slight reduction of potential energy of the ball now manifests as a very slight amount of heat in your arm.
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When the ball was at a higher location, say $h$, if you let it move down to a lower height $h’$ by free-falling, so that it has a final speed $v$ when you catch it, then it’s kinetic energy has been turned into potential energy so that $$\frac{1}{2}mv^2=mg(h-h’)$$
But moving it with your hands, means that the lost potential energy moving it from $h\rightarrow h'$ has been transformed into the heat energy of your muscles contracting (as they must contract to move the object and prevent it from falling) and other biological and biochemical processes that are exothermic.
