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Basic equation of work is given by $F\cdot s$. When work is done, the energy is stored either in form of potential or kinetic. My question arises when we look at a case of applying $m g$ of force upwards on a box of mass $m$. If the box was initially stationary, with application of that force, the net force will be 0 thus the box will remain stationary with no change in potential energy. But if the box initially had some velocity, with application of $m g$ upwards, the net force will be 0 again but this time the box will be constantly moving up. If the box constantly moves up, it will be gaining potential energy as well. It requires same amount of energy to generate $mg$ upwards but in the first case, no energy is stored while in the second, some energy is being stored. So whats going on in this case? Thanks

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2 Answers 2

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When work is done, the energy is stored either in form of potential or kinetic.

Just to be clear, it is an inclusive "or". When work is done all the energy may be stored as potential energy, or it may be stored partly as potential and partly as kinetic energy, or it may be stored completely as kinetic energy, as illustrated in the following 3 scenarios that may help you in answering your questions.

Scenario 1: A net constant horizontal force $F$ is applied to an object thorough a horizontal distance $d$ in the absence of friction in the Earth's gravitational field. The net work $Fd$ will equal the change in kinetic energy $\Delta KE$. All of the work done is stored as kinetic energy and none as gravitational potential energy. I realize this is not the situation you are talking about, but I mention it only because of the generalization implied by your introductory statement.

Scenario 2: An object initially at rest on the ground is raised and brought to rest at a height $h$ above the ground. The net change in kinetic energy is zero. All the work is stored as potential energy. To make this happen, initially, an upward force greater than $mg$ is needed to give it an initial acceleration to start it moving upward. So it initially acquires kinetic energy. Then let the force be quickly reduced to exactly equal $mg$. Now the net force is zero, but the object continues to move up at constant velocity.

Before reaching height $h$ the upward force is reduced to less than $mg$ so that the object decelerates and comes to rest exactly at height $h$. It has lost kinetic energy exactly equal to what it initially acquired by the time it reaches the height $h$. The overall change in kinetic energy is therefore zero. But the change in potential energy is $mgh$. All of the work done is stored as gravitational potential energy.

Scenario 3: Same as scenario 2 except that before reaching $h$ the force is not reduced so that the object continues to have a constant velocity $v$ when it reaches $h$. In this case, when the object reaches $h$, it has both stored potential energy $mgh$ and stored kinetic energy $\frac{mv^2}{2}$.

Given the above, in answer to your specific questions:

If the box was initially stationary, with application of that force, the net force will be 0 thus the box will remain stationary with no change in potential energy.

Correct. If the upward applied force equals, but never exceeds $mg$, it will never acquire an upward acceleration to start it moving. In order to start upward movement, the upward force must at least momentarily exceed $mg$ to give it an initial acceleration. Once in motion, the upward force can be reduced to equal $mg$ and the box will continue to move up at constant velocity. This is scenario 2 above.

But if the box initially had some velocity, with application of 𝑚𝑔 upwards, the net force will be 0 again but this time the box will be constantly moving up. If the box constantly moves up, it will be gaining potential energy as well.

Correct, but only if initially the upward force was greater than $mg$, even if briefly, in order to get the box moving. This is analogous to scenario 3 above. The box has acquired kinetic energy based on at least a momentary upward force of greater than $mg$ to get it started. With the velocity constant, its kinetic energy is constant. But it continues to acquire potential energy as it moves upward as long as the upward force equals the downward force of gravity so that it maintains its constant velocity when it reaches height $h$.

It requires same amount of energy to generate 𝑚𝑔 upwards but in the first case, no energy is stored while in the second, some energy is being stored. So whats going on in this case? Thanks

Based on the above you should now realize that an initial upward force greater than $mg$ is necessary to get the box moving and do work. Energy does not create $mg$ upward because $mg$ upward is a force, not energy. Energy is involved only if $mg$ upwards causes movement of the box. And that only happens if there is initially a net upward force to get the box moving, that is, only if the initial upward force momentarily exceeds $mg$.

Hope this was not too long and helped.

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  • $\begingroup$ Thanks for the reply! But consider the cases of rocket for scenario 2 and 3 starting at the instant net force is 0. For scenario 2, energy is stored in gav-PE and rocket will be held mid air. For scenario 3, the rocket has both PE and KE at h. Lets consider fuel for propulsion as chemical potential energy. For case 2, fuel will burn (Chem PE decreases) while grav-PE and KE doesn't change. For case 3, chem-PE decreases while grav-PE increases and KE remains. So how is this discrepancy explained? Where does energy go for scenario 2? Thanks again! $\endgroup$
    – VVC
    Oct 2, 2019 at 2:59
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    $\begingroup$ Too many rocket scenarios to respond to in comments format. But just because fuel is burned doesn’t mean work it has to be accounted for as work, PE or KE $\endgroup$
    – Bob D
    Oct 2, 2019 at 3:31
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    $\begingroup$ @Patrick the total energy burned by the fuel goes into increasing the KE of both the rocket and the fuel. physics.stackexchange.com/questions/428952/… $\endgroup$
    – BowlOfRed
    Oct 2, 2019 at 5:33
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It requires same amount of energy to generate mg upwards but in the first case, no energy is stored while in the second, some energy is being stored.

Here is the key mistake. The energy required is not the same in the two cases. In the first case no energy is required, this matches with no energy being stored. In the second case energy is required, this also matches with energy being stored.

There are many examples of forces that do not require energy. For example, a table can hold up a book indefinitely, without ever using or requiring energy. In contrast, an elevator does require energy to lift the book.

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  • $\begingroup$ But what about the case of a rocket in mid air? If +mg is applied when v=0, it will be held in mid air with propulsion. But when +mg is applied when v=c, it will be constantly moving upwards with propulsion, thus gaining potential energy. But same amount of energy is used to generate mg propulsion. $\endgroup$
    – VVC
    Oct 2, 2019 at 1:49
  • $\begingroup$ In that case the energy goes into accelerating the exhaust particles. As the rocket moves upwards there is less energy that goes into the exhaust and more that goes into the rocket. $\endgroup$
    – Dale
    Oct 2, 2019 at 2:15
  • $\begingroup$ What did you mean by the second sentence? Isn't what accelerates the rocket the reaction force from accelerating the air particles (Newton's 3rd Law)? So to generate the same amount of a, the same amount of energy has to be put into accelerating the air particles no? $\endgroup$
    – VVC
    Oct 2, 2019 at 2:24
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    $\begingroup$ If you were interested specifically in rockets you really should have put that into your question. Comments are not designed to answer new questions, they are meant only to clarify or improve the questions and answers. The short answer to your new question is no, the energy that goes into the exhaust is not the same but depends on the speed of the rocket. See physics.stackexchange.com/questions/428952/… $\endgroup$
    – Dale
    Oct 2, 2019 at 10:39

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