Unraveling the Mystery of Energy Transfer: Lowering Objects Slowly

[dan1]

Hi everyone!

How do these two types of energy work for things that are being lowered down slowly? I've been told that whenever one drops something (let's say it's 1kg) from 5 meters, it will be traveling at 10 meters per second just before it stops at the ground. And I've also learned that this is true even when the object is sliding down a slide, it's just that the velocity may be going in another direction. And this is all expressed with the conservation of energy: PE at the top will be mgh, 1 * 5 * 10 (I'm using 10 for g just to make it easy for this question), and it will be the same as the KE at the bottom 1/2mv^2, 1/2 * 1 * v^2. So 50 = 1/2 * v^2 and v = 10, right? So this is how it relates to my question: How does the conservation of energy relate to this object at this height if the object is lowered down slowly?
If I exerted an upward force with my hand (let's say 5 Newtons) on the same object from the same height, the KE at the bottom wouldn't be 50 joules would it? So where did the original potential energy of the object go?
This question popped into my mind when I was thinking about a helicopter landing so if you want to use that as in example that would be great!
Thanks so much for taking the time t look over this, I really appreciate it!

 

[cepheid]

Welcome to PF!

Let's define upward to be the positive vertical direction, and downward to be the negative vertical direction.

1 mass of 1 kg weight 10 N (using your value of g). This force is directed downward, so it is -10 N. Meanwhile, if (as in the scenario you suggested), you are pushing upward with a force of +5 N, then there is still a net downward force on the object of -10 N + 5 N = -5 N.

So the object will still be accelerating downwards, just not quite at the same rate as it would be in free-fall. Its kinetic energy at the bottom would just be given by the work done, which is the force times the displacement, which is (-5 N)*(-5 m) = 25 J.

But I think the case you wanted to consider was lowering it in a perfectly controlled manner. In other words,suppose you applied enough upwards force to balance its weight entirely (i.e. +10 N). At the beginning, for a very short time, you'd apply slightly less than 10 N just to give the object a chance to accelerate up to a gentle speed. At the end, you apply just slightly more than 10 N to decelerate it back to rest. But we ignore these brief times at the beginning and the end where the force is slightly unbalanced. For the majority of the motion, the net force on the object is +10 N - 10 N = 0 N. There is no net force on the object. Therefore there is no work done on the object. Therefore, there is no change in its kinetic energy (it descends at a constant speed). So, with no work done on the object, you should really not be surprised that the object does not gain any kinetic energy during the motion.

So what happened to the potential energy that the object had initially? Well, don't forget that if you apply a 10 N upward contact force on it, then it applies a -10 N downward contact force on you. Not only that, but it applies this force over the displacement of - 5 m corresponding to the descent. So, it does 50 J of work on the part of you that is in contact with it. That's where that energy went.

 

[dan1]

Thanks for the reply!

I only have one question about this. Doesn't work done depend on only the net force and not the individual forces? If there was a 10 N force and a -10 N force, wouldn't the work just be 0 Joules?

 

[cepheid]

If you're talking about the work done on the object, then the answer is, "yes, and I already said that."

 

[cepheid]

Another way to answer your question is as follows. You can consider the work done on the object by the individual forces acting on it, as long as you add all these up at the end to get the total work done.

With that in mind, remember that by definition, the change in the gravitational potential energy (GPE) of an object is equal to the negative of the work done on it by gravity. That's all GPE really is. So, in our scenario, since the object has a downward gravitational force of 10 N applied to it over a downward displacement of 5 m, gravity does +50 J of work on the object. Therefore, the change in GPE is -50 J This is true both in the case of dropping it and in the case of lowering it at a constant speed. It's just that, in the second case, the +50 J of work done by gravity is countered by -50 J of work done by the person (since he/she applied an equal upward force over that downward displacement). Since the total work done is zero, the change in kinetic energy is zero, in accordance with the work-energy theorem.

 

[cepheid]

We can express this in equation form too. The simplest thing is to look at the work energy theorem, which says that the work done on an object is equal to its change in kinetic energy:

W = ΔKE

Since W = 0, ΔKE equals 0. Done. But, if you want to consider things in terms of kinetic and potential energy, then you have to split up the work done into its non-gravitational and gravitational contributions, because the definition of the change in GPE is the negative of the work done by gravity. (Notice how I emphasize this point, as before).

W = W_person + W_gravity = ΔKE

W_person - ΔPE = ΔKE

W_person = ΔPE + ΔKE

So, in this situation, where a non-conservative force (not just gravity) acts, we have a slight change in the conservation of energy equation. Instead of ΔPE + ΔKE = 0, we have ΔPE + ΔKE = W_person. The total change in energy is equal to the energy being added to the system in the form of work done by a non-conservative force. (Just like in situations with friction). So it's not that energy isn't being conserved here. It's just that you have to take into account the energy being added to (or in this case, taken away from) the system in the form of work done by the person doing the lowering. Since we've established that ΔKE = 0, we're left with:

W_person = ΔPE

So, in the case where you dropped the object, the potential energy lost would be converted into kinetic energy gained by the object. In this case, however, that energy is instead removed from the system (object) entirely, because there is a person doing negative work on the object (and having positive work done on him).

 

[Studiot]

Try this:

Suppose you lower your block slowly so that when it lands it looses 50 Joules of potential energy.

How would you do this?

A simple method might be to tie a rope and pay the rope out slowly.

How do you pay the rope out slowly?

Well you wind it round something.

In other words you use friction to control it (slow it down).

The work done against friction will be equal to the loss of potential energy.

There is always something in the lowering mechanism that absorbs the potential energy.