Where Does Potential Energy Go When Descending Slowly?

[misogynisticfeminist]

Yea hi, I have a little question here regarding COE.

If I climb up a very high tower or a very high place, using $$E_p=mgh$$, I would have a very high potential energy. But say, if I climb down the tower using a flight of stairs very slowly, I would actually have a very low average velocity. Because velocity= displacement/time and my time taken is very huge.

But using $$E_{k}= \frac {1}{2}mv^2$$, the kinetic energy I gain is very small. So where is this energy "siphoned" off to?

 

[HallsofIvy]

Everytime you take a step down, your body absorbs the energy into its muscles. Eventually, the energy is turned into heat.

 

[Chi Meson]

A little hint from a HS teacher:

If the questions is "where did the energy go?"

The answer is ALMOST always "heat and sound."

 

[misogynisticfeminist]

Ok, to hallsofivy, does that mean that if I use a very low velocity, i would gain a lot of heat energy? So, if my velocity is infinitely low, my heat energy in my body is infinitely high? I don't know any physiology, is it correct here?

and to chi Meson: assuming that the whole step is smooth and in a vacuum.

 

[ZapperZ]

Ok, to hallsofivy, does that mean that if I use a very low velocity, i would gain a lot of heat energy? So, if my velocity is infinitely low, my heat energy in my body is infinitely high? I don't know any physiology, is it correct here?

and to chi Meson: assuming that the whole step is smooth and in a vacuum.

The problem here is that you are seeing only two forms of energy possible in your scenario, and that is just incorrect.

If I drop you from a height, and if we ignore all over insignificant forces, then ALL of your PE is converted to KE. These are the ONLY two possible types of energy relevant in this case. However, when you are walking down a flight of stairs, it is wrong to only consider these two because now, there are OTHER forms of energy involved, as has been mentioned. The fact that you are not sliding down the stairs at the same rate as you would if you were free-falling should already indicate that there are other external forces acting besides just gravity. At the simplest level of analysis, these external forces do work and take up some of the initial PE.

So if you finally realize of other means of energy transfer besides just PE -> KE, then this question has been answered.

Zz.

 

[saksham]

A small thought- Undergrad at Bremen

Ok, to hallsofivy, does that mean that if I use a very low velocity, i would gain a lot of heat energy? So, if my velocity is infinitely low, my heat energy in my body is infinitely high? I don't know any physiology, is it correct here?

and to chi Meson: assuming that the whole step is smooth and in a vacuum.

The law of conservation of energy as you mentioned will hold only when you have a free falling body when it starts to fall with PE=mgh and KE=0 and when it just reaches the ground, it has PE=0 and KE=0.5*m*(v)^2.
When you are climbing down the stairs as you have mentioned above, you break the whole thing into series of steps you take down to climb down to the bottom.
When you take the first step down, you let yourself fall to the step in the staircase just below where you are standing now and you drop some PE and gain KE, that is why you can decend a step. Then, if in vacuum, you body (joints especially) absorbs the "mechanical shock" and then brings you to rest. Then this process goes on until you are at the bottom.
Looking back at it, you have converted the entire PE you had into small sets of KE while decending the stairs and then you converted those energies into heat energy in the parts of your body that experienced the "mechanical shock" when your feet made impact with the steps while you landed.
Overall, the energy was conserved.

This is what I feel is correct. But maybe there is some other answer to explaining how the energy is conserved, but I am sure that Energy is conserved in your situation where subatomic particles are not involved and we don't get deep into the level of quantum mechanics.

 

[HallsofIvy]

Ok, to hallsofivy, does that mean that if I use a very low velocity, i would gain a lot of heat energy? So, if my velocity is infinitely low, my heat energy in my body is infinitely high? I don't know any physiology, is it correct here?

and to chi Meson: assuming that the whole step is smooth and in a vacuum.

If you are motionless at the bottom, no matter how quickly you got there, you have, in some way dissipated the potential energy. You can't "absorb" more than your original potential energy so it certainly couldn't be greater than that. You could say that your heat energy would be "infinitely close" to your original potential energy. It's not a matter of how fast you come down, its a matter of your speed AT the bottom. (Actually, a small amount of energy will be lost to the steps in the walk down. If you jump out a window, a large amount of energy would be lost to the ground that stopped you. I would count the energy that went into breaking your bones as "absorbed by your body!)

 

[reilly]

The reason you have a high potential at the top of the tower is that you've done a lot of work while climbing the tower -- against the g force -- to the tune of mgh. Now, when you climb down, the situation is reversed -- the g force is still there, so the magnitude of the work to go down is mgh. After a complete cycle, up then down, the total work done against gravity is zero -- as it must be for any conservative force. The energy to climb and descend comes from your body -- Krebs Cycle and all that.

(This question is sometimes used in PhD oral exams. It's really a question about the properties of a conservative force.)

Regards,
Reilly Atkinson

 

[Chi Meson]

Nothing I can add to that!