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Imagine to lift an object with mass $m$ from height $h_1$ to height $h_2$ and neglect the friction with air. How much work have you done on the object?

My answers (big doubt in the second one!):

Answer 1. Let's assign $0$ to the gravital potential energy of the object at height $h_1$. Since it is still by hypothesis, its mechanical energy $E_1$ is purely potential hence in this situation is $0$:

$$E_1 = 0$$

At the end the object is still hence its mechanical energy $E_2$ is purely potential, that means it is equal to:

$$E_2 = mg(h_2-h_1)$$

Hence, the work done on the body is:

$$W = E_2 - E_1 = mg(h_2-h_1)$$

Note: I'd say this quantity is not equal to the work performed by the lifter, as he should replace the quantity $m$ with the total mass of the object plus the part of the body that the lifter is moving together with the object to perform the lift.

Answer 2. Let's evaluate the work through its definition and not through the energy-work theorem:

$$W = \int_{h_1}^{h_2} \vec F \cdot d \vec s$$

First doubt: At this point, to get the same result of the Answer 1, I have to assume that $\vec F$ and $d \vec s$ are parallel. That means that, since by hypothesis you are lifting the object with a vertical path, the force must be vertical. Why does Answer 1 ignore such an assumption?

Let's continue the evaluation with such an assumption:

$$W = \int_{h_1}^{h_2} F ds$$

Now, I should stop. I get the same amount of work of answer 1 if I make the substitution $F = mg$. But that's not correct. In fact, if the object is moving from $h_1$ to $h_2$, that's because the force $F$ is higher than the object weigth force. And then, such a force reduces to the object weight as the object is kept still at height $h_2$. In other words, I'd say that the force $F$ is not constant along the movement ($F = F(s)$):

$$W = \int_{h_1}^{h_2} F(s) ds$$

At the beginning it is higher than $mg$, at the end it is equal to $mg$.

Second doubt: how can the area of F (integral) in the path beingh equal to the work done by $mg$ if F is higher at the beginning and equal to $mg$ at the end? There should be a moment when $F(s)$ becomes lower than $mg$. I imagine that this could happen before the end, like a sort of attempt to decelerate the object. If we were to accelerate the object with $F > mg$ and then applying a force $F = mg$, it would continue moving because of the Newton first law. So, I'd say that $F$ must grow, decrease, and stop at $mg$, with the property of having the same area of $mg$.

I think it may be a correct analysis. But I'm not sure, as I cannot perceive this "ondulating" behaviour of my force $F$ when I lift an object :(

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  • $\begingroup$ You are absolutely correct but assume if the force you apply $F=mg+\epsilon$ but in this way you are giving the body a kinetic energy $\epsilon(h_2-h_1)$ but we assume thst the process occurs so slowly that the $\epsilon$ is very small compared to the force. $\endgroup$ Sep 18, 2021 at 8:29
  • $\begingroup$ If you want to let ds stay an arbitrary differential element of path length, you have to use a path integral (and a computer to solve it, unless your path has a convenient symmetry). If you're just integrating over height, $d\vec s$ can only be the differential element of height, so $\vec F \cdot d\vec s$ is $Fsin(\theta) dh$ $\endgroup$
    – g s
    Sep 18, 2021 at 8:35
  • $\begingroup$ @g s I've implicitely assumed F being conservative... Is it wrong? $\endgroup$
    – Kinka-Byo
    Sep 18, 2021 at 8:38

2 Answers 2

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The object could be lifted from $h_1$ to $h_2$ slowly, without creating much kinetic energy (blue line), here the force matches the weight. The answers 1) and 2) would be the same.

If a higher force than necessary was used at the start (red line), then the object would gain lots of kinetic energy at first, so that the force could then be reduced, if the object is to finish at $h_2$ with no kinetic energy.

enter image description here

Or the yellow line might be a realistic case, some kinetic energy is created, but not much.

If the area under the lines is the same, then the object will finish at $h_2$ with no kinetic energy in each case.

The area under the lines represents the work done on the object.

So the work done in the 'red lift', for the first half of the lift, is greater than in the blue lift. As the object reaches the same height at the halfway point in both cases, kinetic energy was created in the red case during the first half of the lift.

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That means that, since by hypothesis you are lifting the object with a vertical path, the force must be vertical. Why does Answer 1 ignore such an assumption?

It ignores this because you are actually assuming the force you use is exactly equal and opposite to the weight force. But technically, approach 1 is only looking at the work done by gravity, not by you as you lift the object.

So, I'd say that $F$ must grow, decrease, and stop at $mg$, with the property of having the same area of $mg$.

If you first apply a force larger than $mg$, then at some point you would need to apply a force less than $mg$ to cause the object to come to rest.

You have uncovered the problem with questions like these that is really confusing to students (and teachers) who are actually paying attention and not just "plugging and chugging": We are not told the final state of the object being lifted. We could just fling the object upwards, and so at $h_2$ we have done a huge amount of work. The problem should either specify that the object starts and ends at rest, or it should ask what is the minimum amount of work you have to do to lift the object from $h_1$ to $h_2$. This is most likely what the author of the question had in mind to where the actual correct answer is $mg\Delta h$


Note: I'd say this quantity is not equal to the work performed by the lifter, as he should replace the quantity m with the total mass of the object plus the part of the body that the lifter is moving together with the object to perform the lift.

We can ignore the masses of the body parts; no need to make things more complicated than needed for exercises like these focusing on a specific concept.

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  • $\begingroup$ Thank you for the explanation. About the fact that "Approach 1 is only looking at the work done by gravity, not by you as you lift the object", can you tell me which may be an "energy conservation way" to evaluate, in theory, the work done by the lifter? $\endgroup$
    – Kinka-Byo
    Sep 18, 2021 at 17:42
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    $\begingroup$ @Kinka-Bye $$\Delta K+\Delta U=W_\text{ext}$$ $$\frac12m\Delta(v^2)+mg\Delta h=W_\text{lift}$$ $\endgroup$ Sep 18, 2021 at 18:32
  • $\begingroup$ Won't this equation lead to the same work done by gravity in case the object is lifted and then kept still at the final height? $\endgroup$
    – Kinka-Byo
    Sep 18, 2021 at 18:36
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    $\begingroup$ @Kinka-Byo Yes. Sorry I had missed your assumption about starting and stopping at rest earlier $\endgroup$ Sep 18, 2021 at 19:05

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