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If a stone is launched upward, of which is subject to gravity and air resistance, which of the following will have a greater kinetic energy? The stone at a point on its way up, or the stone at the same point on its way down?

I know $KE=1/2mv^2$, and $PE=mgy$.

If I had to guess I would say the kinetic energy is equal up and down, but I'm not sure. Could someone explain this concept to me?

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The kinetic energy of the stone will not be equal at the same point on the way up or the way down, due to the presence of friction from the air. To show this, it is enough to compare the speed of the stone just after it leaves your hand (on the way up) with the speed just before it lands back in your hand (on the way down).

First consider the case without air resistance. You launch the stone with speed $u$ directly upwards. The initial energy you give the stone is $$ KE = \frac{1}{2}mu^2. $$ The stone goes up until it reaches its maximum height, where all of its initial kinetic energy is converted into potential energy. Then the stone comes back down, and by the time it reaches your hand again all of this potential energy has been converted back to kinetic energy. Let's call the final speed of the stone $v$. The final and initial energies must be the same. Since all the initial kinetic energy has been converted back to kinetic energy, the initial and final kinetic energies are the same, or in maths: $$ \frac{1}{2} m u^2 = \frac{1}{2} mv^2 $$.

Now add air resistance to the picture. As the stone moves through the air, some of its kinetic energy is used up and converted into heat energy, raising the temperature of the stone and the air slightly. The initial and final energies of the entire system (stone and air) still have to be the same, so we now have \begin{eqnarray} \text{Initial energy} &=& \text{Final energy} \\ \frac{1}{2}mu^2 &=& \frac{1}{2} m v^2 + (\text{heat energy from friction}). \end{eqnarray} Rearranging this equation, you get $$\text{Final kinetic energy} = \frac{1}{2}mv^2 = \frac{1}{2}mu^2 - (\text{heat energy from friction}),$$ which is smaller than the initial kinetic energy. A similar argument shows that the kinetic energy is less on the way down than the way up for any point on the trajectory.

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Let us first examine the ideal case, where we ignore air resistance.

Without Taking Air Resistance Into Account:

When the stone is traveling upwards, it will go through a certain point A, and will be travelling with a specific velocity. The only force acting on this stone is the force of gravity, and this only depends on the mass of the object. So for the whole process, the stone will go up through point A, reach the top of it's trajectory, and come back down through point A traveling at the same speed it went up with.

Since it is traveling at the same velocity both on the way up and on the way down, the kinetic energy is the same. Now let's look at what happens when we consider air resistance.

Taking Air Resistance Into Account:

In this situation, the stone will go through the point A at a certain velocity, and the only forces acting on it will be the force of gravity and the force of air resistance. When the stone is traveling upwards, the force of gravity is going to be removing kinetic energy from the stone, and the force of air resistance will be as well.

On the way down, the force of gravity will be restoring kinetic energy to the stone, but the force of air resistance will still be acting to slow the stone. Therefore, the stone will have less kinetic energy on the way down than it had on the way up, solely due to the influence of air resistance.

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ha, you got there first...just! good answer :) – Mark Mitchison Oct 28 '12 at 22:49
Man I thought I had that one for sure! – Michael Leonard Oct 28 '12 at 23:44

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