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If an object is stationary on ground,then it's potential energy is $0$ and kinetic energy is $0$ as well. So it's mechanical energy is $0$. But we know energy is always conserved. Then how does that same body when thrown upward with a velocity gain potential+kinetic energy? That does not equate to $0$ which was initially the energy of the object.

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  • $\begingroup$ Whatever mechanism throws the object upwards loses energy so it balances out. $\endgroup$
    – AfterShave
    Commented Jul 16, 2022 at 4:48
  • $\begingroup$ Energy is not always conserved for any given system, because it can exchange energy with its environment. For example, an external force (force due to the environment) can change the energy of the body (system) it acts on. $\endgroup$
    – hft
    Commented Jul 16, 2022 at 5:22
  • $\begingroup$ @AfterShave sorry i didn't understand what you said,aren't we talking about the energy of the object here? The object's initial energy was $0$ and now it has some positive mechanical energy. $\endgroup$
    – madness
    Commented Jul 16, 2022 at 5:57
  • $\begingroup$ Well things don't move on their own, there has to be an external mechanism involved. If you don't include that mechanism in your model then yes energy won't be conserved, but that's only because you'ved ignored some parts of the problem. $\endgroup$
    – AfterShave
    Commented Jul 16, 2022 at 6:07

3 Answers 3

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So, right now, you are considering the ball as a part of the system, which obviously will lead to contradictions in the conservation of energy in a similar, but not the same way as you have told. See, firstly, the GPE (gravitational potential energy) and KE summed together equals a constant number, that denotes the energy supplied to a ball. I want you to abandon the idea of systems for now, and just think of someone throwing the ball in the air, upward, with energy supplied = 100J.

TO keep it extremely simple, I am ignoring the formulas for KE and GPE and the exact speed of the ball (varying for different masses).

At the instant the ball leaves your hand, it has 100% KE, which means 100J of kinetic energy. As it goes upward, it is gaining height and slows down, therefore, the losses in KE are made up for by increases in GPE. At a specific height h where it has a speed of 0 (h=v^2/2g, g=gravitational acceleration), it has all 100J in GPE, and 0J in KE (because v=0). Ignore all air friction here.

Now coming to the main part of your question of how the energy can be conserved if the ball moves upward, then see this reasoning.

You are absolutely right in saying that the ball has energy while moving upward in the forms of GPE and KE.

But, where did that energy come from? Obviously, a source like a person who threw it upward, and imparted some energy to the ball. So, we need to include the source of the energy in the system to ensure the conservation of energy. Then, if I say that the GPE+KE=100J, then I must have lost 100J to the ball, or I have worked on the ball transferring 100J of energy.

So, the net result is: My energy goes down by 100J, and the ball's energy goes up by 100J. -100J+100J=0 Therefore, the energy conservation principle still holds.

Any suggestions or corrections are welcome.

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  • $\begingroup$ Thanks a lot for replying,actually i wanted to know that energy is conserved BUT energy of WHAT is conserved? As per your answer,it is the energy of me and the stone,but typically we deal with the energy of object only. $\endgroup$
    – madness
    Commented Jul 16, 2022 at 7:47
  • $\begingroup$ Yes, that is because we simplify the situation. In reality, the Potential energy at the height h of the ball is not of the ball alone, but the ball-earth system. Think of a system as a combination of two physical objects, chosen for analysis, and the rest outside of it as it's environment. To show you why taking me and the stone together is correct than only the stone, consider the universe as a system. Now, since the universe by definition includes every physical object, there is nothing beyond it. Thus, energy must be conserved inside the system, everywhere, so energy conservation applies. $\endgroup$
    – Aveer
    Commented Jul 16, 2022 at 17:29
  • $\begingroup$ Now, if I isolate the ball as an example, of course, energy conservation will be violated. If I am not wrong, all the environments of a system are to be ignored except their impacts (here-to give energy to the ball by throwing it in the air). Since energy is not conserved upon the application of external forces, which would mean that I consider dealing with the energy of the object only, their is no problem in the argument $\endgroup$
    – Aveer
    Commented Jul 16, 2022 at 17:32
  • $\begingroup$ Check this link- physics.stackexchange.com/questions/491525/… (1st para of 1answer) $\endgroup$
    – Aveer
    Commented Jul 16, 2022 at 17:33
  • $\begingroup$ and check these answers too (dont get afraid by the long qs) physics.stackexchange.com/questions/255928/… $\endgroup$
    – Aveer
    Commented Jul 16, 2022 at 17:34
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Energy is NOT always conserved.

Energy is conserved ONLY in closed systems.

If you take up a stone in your hand, the stone is not a closed system. If you put energy into a stone by accelerating it, it is obviously not a closed system.

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  • $\begingroup$ Thanks for replying,could you please tell me what a closed system is? And when can't we treat an object as a closed system like it didn't in my problem? $\endgroup$
    – madness
    Commented Jul 16, 2022 at 7:45
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But we know energy is always conserved

For mechanical systems, a more general statement is that the work of the net force is equal to the change of the kinetic energy. For example, if the object is stationary on the ground, the gravitational force is balanced by the Normal force. The net force is zero and also the work. So the kinetic energy doesn't change.

If the upward force is greater than the weight, the object accelerates upward, and the work of the net force result in an increase of the kinetic energy.

If the upward force is turned off while the object is going up, the net force is gravity, that is conservative, and so can be expressed as minus the derivative of the potential $\phi = mgh$. In this case, the statement that work is the change of the kinetic energy can be written as: $$ w = F\Delta h = -\frac{d\phi}{dh}\Delta h = -mg\Delta h = \Delta\left(\frac{1}{2}mv^2\right)$$

which implies: $$mgh + \frac{1}{2}mv^2 = constant$$ If we call the left side total energy, we can say that it is a conserved quantity.

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