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It seems like a simple question but I was wondering where does the energy go when I place an object from a height on the floor.

Initially it's all stored as potential energy, and as I'm moving the object it has kinetic energy. But when I have placed it on the ground where does the energy go? I'm sure a bit is lost as heat through friction but where does the majority of the energy go?

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  • $\begingroup$ @Ollie113 thank you for your help $\endgroup$
    – bobsburger
    Apr 23, 2019 at 17:21
  • $\begingroup$ Muhammad. Please look at the following link in support of my answer to your question. courses.lumenlearning.com/physics/chapter/… $\endgroup$
    – Bob D
    Apr 23, 2019 at 20:12
  • $\begingroup$ Muhammad please clear something up. Are you placing the object on the floor or dropping it on the floor. It makes a difference. If you are bringing it to "rest" on the floor, that is, the object has zero velocity when touching the floor, then the kinetic energy is zero when it touches the floor. $\endgroup$
    – Bob D
    Apr 23, 2019 at 22:17
  • $\begingroup$ Sorry for the late response. As you mentioned I am "putting" it to rest on the floor and NOT dropping it. The change in energy when dropping is clearer as the bottle keeps bouncing and losing energy as sound, friction and such. $\endgroup$
    – bobsburger
    May 1, 2019 at 20:09
  • $\begingroup$ Thank you for the clarification. I am working on a revision to my answer that will, hopefully, better answer your question. $\endgroup$
    – Bob D
    May 1, 2019 at 21:41

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It depends, are you placing the object on the ground or are you dropping it? You said it yourself. As a mass falls towards the ground it loses its potential energy, which is transferred to kinetic energy. Think about what happens when you drop an object. If the object is a bouncy ball you can see that some energy is lost to heat/friction, but the rest is transferred to kinetic energy again.

What about if you drop a cannon ball on the ground? It would dent the floor right? What do you think happened to the cannon ball's kinetic energy in that scenario?

I'm sorry if my answer seems a little obscure but your question is marked "homework and exercises" so I'm trying to get you to think of the answer yourself rather than just giving it to you.

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  • $\begingroup$ If you're still struggling with the concept, think about dropping a cannon ball into a box filled with fine dust. You'd expect a cloud of dust to rise into the air right? Why is this? $\endgroup$
    – Ollie113
    Apr 23, 2019 at 16:36
  • $\begingroup$ The OP specifically says he "places" an object from a height on the floor. Placing it on the floor is not dropping the object on the floor. I will ask for clarification. $\endgroup$
    – Bob D
    Apr 23, 2019 at 22:15
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REVISED ANSWER:

The following is a revision to my answer in light of the following comment you gave me:

As you mentioned I am "putting" it to rest on the floor and NOT dropping it. The change in energy when dropping is clearer as the bottle keeps bouncing and losing energy as sound, friction and such

Thanks for the clarification. I will now focus on your following specific question, coming at it a little differently:

“where does the energy go when I place an object from a height on the floor”.

When you ask where the energy goes, there are two types of energy in this example: kinetic energy and gravitational potential energy. Let’s account for each.

Let’s consider first kinetic energy.

You take an object that is initially at rest (zero kinetic energy) at a certain height, say the height of the surface of a table above the floor. In order to bring it down the object needs to “get moving”. That means the object needs to initially acquire kinetic energy. In order for this to happen your upward force on the object has to be less than the downward force of gravity. There is then a net downward force and gravity does positive work on the object to give the object kinetic energy.

As the object approaches the ground you now need to exert an upward force greater than the downward force of gravity in order to decelerate the object (slow it down). You are doing negative work on the object (your upward force is in the opposite direction to the displacement of the object) taking kinetic energy away from the object and transferring it to yourself. You can also look at it as the object doing positive work on you transferring energy from itself to you.

The question of exactly how you absorb that kinetic energy is a physiological one that I am not qualified to answer. My guess is the work the object does on our muscles results in various forms of friction, raising tissue temperature and eventually being dissipated as heat.

Bottom line: The positive change in kinetic energy due to the positive work that gravity did on the object at the beginning equals the negative change in kinetic energy due to the negative work you did on the object to bring the object to rest. The fact that the object had kinetic energy between the start and stop is irrelevant because the overall net change in kinetic energy is zero. Kinetic energy is therefore accounted for.

Now let’s consider gravitational potential energy.

Since there is no net change in kinetic energy of the object, gravity does positive work on the object at the expense of the gravitational potential energy of the object/earth system, while at the same time you are doing an equal amount of negative work on the object transferring the energy gravity gave it to yourself. The net work done on the object is zero, but the gravitational potential energy of the object/earth system has been transferred to you.

Note that the reverse happens if you take an object at rest on the ground and place it at rest on the surface of the table. In this case, you do positive work on the object, transferring energy from you to the object, while at the same time gravity does an equal amount of negative work, transferring the energy you gave the object to the gravitational potential energy of the object/earth system. The net work done on the object is zero (no change in kinetic energy), but the external work you did on the object/earth system has increased the gravitational potential energy of the object/earth system.

I hope this different way of explaining what is going on will help you. The following link will perhaps further enhance this explanation.

https://courses.lumenlearning.com/physics/chapter/7-3-gravitational-potential-energy/

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  • $\begingroup$ The end result is the gravitational potential energy of the object/earth system is transferred to you (your muscles). Fine! Do you mean I become stronger? Or can I do without eating? $\endgroup$
    – Elio Fabri
    Apr 23, 2019 at 19:28
  • $\begingroup$ In your examples in the 2nd and 3rd paragraphs, the object starts and ends with 0 velocity, but at different heights. The object's total energy (just potential energy, since v=0) has changed, therefore, work has been done on it. When raising the object, you do more work than gravity, and when lowering it, gravity does more work than you. The part about transferring energy to your muscles is also a little misleading - you can't mechanically recharge your muscles like a spring, so that energy just gets dissipated as useless heat. $\endgroup$
    – Nuclear Hoagie
    Apr 23, 2019 at 19:37
  • $\begingroup$ @Elio Fabri of course not! It is likely dissipated as heat in your body $\endgroup$
    – Bob D
    Apr 23, 2019 at 19:44
  • $\begingroup$ @NuclearWang What do you mean different heights? For the second paragraph, I'm assuming it starts at rest at say, height $h$. Then you bring it down to rest gently at 0 height. Then I compare that with it starting at height 0 and you pick it up and bring it to rest at height $h$. When you raise the object you do not do more work than gravity if you bring it to rest at $h$. If you did more work than gravity, it would have a velocity and therefore kinetic energy at $h$. Do you understand the work energy theorem? In bringing it down, energy is transferred to your body. $\endgroup$
    – Bob D
    Apr 23, 2019 at 19:52
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    $\begingroup$ @NuclearWang Finally, the work energy theorem (or principle) states that the net work done on an OBJECT equals its change in kinetic energy. So there is no net work done on the object. The work you do in lifting the object from rest to rest is done on the object/Earth system where it is stored as gravitational potential energy of the object/Earth system. Hope this helps. $\endgroup$
    – Bob D
    Apr 23, 2019 at 20:17

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