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I know the derivation that $W=Fd$, hence $F=mg$ and $d=h$ so energy gained by the body is $mgh$ considering the body on the ground to have $0$ gravitational potential energy.

But the definition of work is (as given in my book)

Work done is the product of force and displacement caused by it in the same direction.

That means work done on a body to lift it against gravity to a certain height should be equal to the potential energy gained by it, right? My book also states that:

$mg$ is the minimum force required to lift a body against earth's gravity(without acceleration).

But how does that make sense? Suppose a body is kept on the ground, and we apply a force $mg$ on it, won't the force of gravity and this external force cancel out and ultimately result in no movement of the body? How is the derivation of $U=mgh$ thus obtained?

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3 Answers 3

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Part of the problem is to distinguish between the work done by a particular force and the net work done by all the forces. The second is to notice that the work done on an object depends on the process undergone. The third is to understand that the relationship between work and potential energy is that the work done by a conservative force is proportional to the change in the potential energy.

Let's walk through the scenario. A block of mass $m$ sits on the ground at position $y=0$. There are two forces acting: the gravitational force downward and then normal force upward. Newton's 2nd Law tells us that $$ m\vec{a}=\vec{F}_{\textrm{net on object}} = \vec{F}_{\textrm{G, on object by Earth}} +\vec{N}_{\textrm{on object by ground}}\,. $$ We'll abbreviate these as $\vec{F}_{\textrm{net}}$, $\vec{F}_{\textrm{G}}$, and $\vec{N}$.

  1. In the case where the object is just sitting on the ground, the acceleration is clearly zero, and the normal and gravitational force cancel each other out. The block doesn't move, and so the net work done by either force must be zero: $$ W_{\textrm{by G}}=\int_i^f\vec{F}_{\textrm{G}}\cdot d\vec{r}=\vec{F}_{\textrm{G}}\cdot\Delta\vec{r} = 0\, $$ where the second equality holds beccause the gravitational force is constant near the surface of the Earth, and the third holds because the net displacement is zero.

  2. Now, someone grabs the block, accelerates it upwards, and then starts lifting the block upwards at constant speed. Ignoring the acceleration part, as the block moves up at constant speed, the net force on it must be zero, and so the gravitational force and normal force acting must cancel, as they did above, although now $\vec{N} = \vec{N}_{\textrm{by person}}$, which we'll just call $\vec{N}$. The work done by gravity and the work done by the person lifting the block can be computed as follows: $$ W_{\textrm{by G}}=-mg(y_f-y_i)\,, $$ where $y_f$ and $y_i$ are the initial and final heights of the object, and $$ W_{\textrm{by N}}=N_{\textrm{by hand}}(y_f-y_i)\,. $$ Note that these two works are equal and opposite, and so the net work done is zero, as it must because the kinetic energy isn't changing! However, the works done by the individual forces are non-zero.

  3. Looking at the $W_{\textrm{by G}}$, we can see that we can alternatively define it as $$ W_{\textrm{by G}} = -(U_f-U_i)\,, $$ where we define $U = mgy$ to be the potential energy when the object is at height $i$. Then, $U_f-U_i = mgy_f - mgy_i$ is just the change in potential energy as the object is lifted from height $y_i$ to height $y_f$. We could write this as $mgy_f - mgy_i = mgh$, where $h$ is the change in height, but this isn't a great way to do things, because $h$ could be negative (if the block moves downward), and it's easy to confuse a position with a change in position is if it's not notated correctly. I would write this as $mgy_f - mgy_i = mg\Delta y$.


To tie this in with the OP's specific questions, then, note that while the block is sitting on the ground, the potential energy is constant because its position doesn't change. The value of the potential energy itself is a meaningless quantity; it's only changes in potential energy that matter, via $W = -\Delta U$. We derive $U= mgy$ by considering the work done during a process in which the position of the object changes.


Last important note: the third bullet point requires a change in perspective, and without this change in perspective, things can go wrong (mixed up understandings and incorrect calculations). In our analysis above, we chose the system to be the ball, and we computed the change in kinetic energy of the ball by computing the works done by all forces acting on the ball. If these works cancel, then the net change in kinetic energy is zero.

If instead we move to a potential energy language, we have to reconsider what we call our system. Instead of thinking about the work done by the Earth via gravity on the ball, we consider a new system composed of both the Earth and the ball. In that case, we replace the work done by the Earth on the ball by the change in potential energy of the Earth-ball system, i.e., \begin{align} \Delta KE_{\textrm{ball}} &= W_{\textrm{N}}+W_{\textrm{G}} = \Delta KE_{\textrm{ball}} = W_{\textrm{N}}-\Delta PE_{\textrm{G}} \Longrightarrow \\ W_{\textrm{N}} &= \Delta KE_{\textrm{ball}} + \Delta PE_{\textrm{G}} \end{align} Since the kinetic energy of the Earth doesn't change, $$ \Delta KE_{\textrm{system}} = \Delta KE_{\textrm{ball}} + \Delta KE_{\textrm{Earth}} = \Delta KE_{\textrm{ball}}\,, $$ and so we can write $$ W_{\textrm{ext}} = \Delta KE_{\textrm{system}} + \Delta PE_{\textrm{system}}\,, $$ where $W_{\textrm{ext}}$ is the work done by objects outside the system on objects inside the system, or work done by external forces. In this case, that is the work done by the person in lifting the ball.

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    $\begingroup$ Everything is cool but the problem is i haven't learnt calculus yet as it's not in 10th grade, so i just gotta take your word for it now :P Thanks though, appreciate the effort $\endgroup$ Nov 18, 2021 at 18:00
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    $\begingroup$ You don't need calculus, though. The only part where that comes in is in the general definition of the work done. Since all these forces are constant, the work reduces to the product of the component of the force in the direction of the displacement times the displacement. So you can completely ignore the integral! Everything else is fine. $\endgroup$
    – march
    Nov 18, 2021 at 18:01
  • $\begingroup$ @JavaMonke Nope! The speed is in fact irrelevant, since when you compute $F_{\textrm{G},y}(y_f-y_i)$, the velocity (or even speed) doesn't even come into it! The "ignoring" the initial acceleration part was in service of understanding how the net work might be zero while the works done by individual forces aren't. $\endgroup$
    – march
    Nov 18, 2021 at 18:41
  • $\begingroup$ Alright, so $W_N= \Delta PE_G$ since $\Delta KE=0$? $\endgroup$ Nov 18, 2021 at 19:18
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    $\begingroup$ This is exactly why I recommended not writing down $W_N=\Delta PE_G$! That only holds for the very specific situation I was considering, in which the normal and gravitational forces are equal and opposite! What we have is that $W_G = -\Delta PE_G = -mg(y_f-y_i)$, no matter what the situation, and then $W_{F_H}=\Delta K +\Delta PE_G = \Delta K + \Delta PE_G$. In this case, you actually need to compute $W_{F_H}$, which means you need to know exactly what force $F_H$ was acting during the process. $\endgroup$
    – march
    Nov 18, 2021 at 19:36
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Suppose a body is kept on the ground, and we apply a force 𝑚𝑔 on it, won't the force of gravity and this external force cancel ...

Yes, they will cancel. Net force = 0, acceleration = 0.

...our and ultimately result in no movement of the body?

acceleration = 0 does not mean velocity = 0. If we could get the block moving for a moment, then the $mg$ force would be sufficient to maintain that motion against gravity. And since it is moving, there is now a non-zero displacement that can be applied.

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    $\begingroup$ I considered the same, but won't that minimum extra force required to give the body a certain velocity be considered too when measuring the energy gained by the ball? $\endgroup$ Nov 18, 2021 at 17:10
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    $\begingroup$ You can create scenarios where it does not. First of all there is no minimum speed, so you can make the KE arbitrarily small (much smaller than the potential energy gained). Also, you can apply it before consideration. Start the object moving before your reference point and only consider the work done after the reference point. $\endgroup$
    – BowlOfRed
    Nov 18, 2021 at 17:13
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    $\begingroup$ @JavaMonke Yes, and that extra force gives it kinetic energy. Then, in order for the object to stop at height $h$, you need to dampen the upward force a little, to convert that kinetic energy into potential energy. So from "unmoving at height $0$" to "unmoving at height $h$", the lifting force has gone from slightly above $mg$, to $mg$, to slightly below $mg$, averaging out to exactly $mg$. $\endgroup$
    – Arthur
    Nov 20, 2021 at 10:58
  • $\begingroup$ Thanks! @Arthur $\endgroup$ Nov 20, 2021 at 19:00
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But the definition of work is (as given in my book)

Work done is the product of force and displacement caused by it in the same direction.

That is called positive work. That's the work you do lifting the body because the force you apply is in the same direction as the displacement of the body. Positive work transfers energy to the body.

But at the same time you are doing positive work, gravity is doing negative work since its force is opposite to the direction of the displacement. Negative work takes energy away from the body. In this case, gravity takes part or all of the energy you supply the body and stores it as gravitational potential energy (GPE) of the Earth-body system.

That means work done on a body to lift it against gravity to a certain height should be equal to the potential energy gained by it, right?

Yes, if the body is to have no change in kinetic energy, i.e., $\Delta KE=0$, between $0$ and the height $h$. That will be the case if the body begins and ends at rest and the net work done is $mgh-mgh=0$. The underlying principle here is the work energy theorem which states: The net work done on an object equals its change in kinetic energyat the height.

My book also states that:

$mg$ is the minimum force required to lift a body against earth's gravity(without acceleration).

But how does that make sense?

Since the body starts at rest on the ground, in order to get it moving you need to apply a force $>mg$ to give it an initial acceleration. But in order for all your work to wind up as GPE, before reaching the height $h$ you need to apply a force $<mg$ to decelerate the body and bring it to rest at $h$ for a $\Delta KE$ of zero. What happens in between $0$ and $h$ does not matter, as long as the object begins and ends at rest so that $\Delta KE=0$.

Hope this helps.

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