Work done in raising an object to a height

When we raise an object to a height $$h$$, it is said that the potential energy of the object is increased by $$mgh$$. But isn't the work done by gravitational force $$-mgh$$? Then that will essentially mean $$Work= mgh+(-mgh)=0$$ How is the increased in gravitational potential energy justified?

• The total work done on the object is the change in its kinetic energy. What you obtained (net work done = 0) makes sense as the object was initially at rest, and is also at rest after raising it, and its kinetic energy does not change. Commented Dec 5, 2023 at 19:06
• @ArchismanPanigrahi isn't the work done equal to change in total mechanical energy, which includes the increase in potential energy by $mgh$? Commented Dec 5, 2023 at 19:09
• That's when you consider work done by external agents, and don't take the (conservative) gravitational force into account. Let's say, you don't know about the existence of gravity, and have done work $mgh$ on the system to raise it to a height $h$. But hey, you have done some work, then why didn't the kinetic energy increase? That's because you did not take the work done by the gravitational force into account, and you can deal with this by assuming the work you did went into some (fictitious) potential energy. Commented Dec 5, 2023 at 19:12
• @ArchismanPanigrahi thanks got it. Commented Dec 5, 2023 at 19:14
• Here's a PSE answer to a similar question. Maybe it would help clarify things. Commented Dec 5, 2023 at 19:47

(N.B- Boldface characters represent vectors, italicized characters represent scalars or vector magnitudes)

I would like to address this query from a mathematical point of view, just to deeply clarify the ideas of potential energies for conservative force fields.

As we know from the Work Energy Theorem, a small change dK in the kinetic energy of a system due to a small displacement dr is related to the work performed on the system by a number of force fields as:

dK = ∑dW = ∑F⋅dr

Suppose all the force fields performing nonzero work on the system are conservative. Thus, the change in kinetic energy over an extended displacement can be deduced via integration over the entire path of the system as follows:

dK = ∆K = ∫dW = ∫F⋅dr

and thus,

∆K + ∫-F⋅dr = 0

Interestingly, since the forces are conservative, ∫F⋅dr is dependent only on the initial and final states of the system and thus the change in kinetic energy of the system is dependent solely on the initial and final states of the system and is completely independent of the path taken. This key property differentiates conservative forces from nonconservative ones and motivates the definition of the quantity F⋅dr for conservative force fields. We define a change in potential energy of a conservative force field as dU = -F⋅dr = -dW. The equation just derived can then be rewritten in terms of dU as:

∆K + ∫dU = 0 ⇒ ∆K + ∆U = 0

Which is popularly known as the law of conservation of Mechanical Energy.

Now let's look at the problem you presented.

When we raise an object to a height h , it is said that the potential energy of the object is increased by mgh .

Let's see. Suppose h is the position vector of the object from the centre of the Earth. At small heights above the surface the geo-gravitational force on a given mass is nearly of constant magnitude and is given by:

F=m g (g towards the centre of the Earth)

And so the change in geo-gravitational potential energy is given by dU = -F⋅dh = m g dh.

Integrating, we get ∆U = m gdh = m g ∆h. So if you raise an object by a height h,the geo-gravitational potential of the body is raised by m g h.

True indeed. So far so good.

But isn't the work done by gravitational force −mgh ?

Since dW = -dU, ∫dW = W = -∆U. We just established that ∆U = m g ∆h, so W = -m g ∆h

Checks out. Let's proceed

Then that will essentially mean Work=mgh+(−mgh)=0

Or does it?

This is the flaw in your argument. The change in potential energy is DEFINED equal to the negative of the work done by the conservative force field. They describe the exact same thing, just with different signs. They're two faces of the same coin. So, if the conservative force field did some work W to cause a change ∆U in the potential energy, it does NOT mean it did a net work of (W+∆U).

Put simply, you can use either of the two equations: ∆K + ∆U = 0 OR ∆K = ∫dW, both of them are perfectly valid.

What you've written here though is essentially:

∆K = ∆U + ∫dW

Which is wrong.

The solution is all about which system you are considering.

I prefer this version of the work-energy theorem for mechanical systems:

$$W_\text{net,ext} = \Delta K + \Delta U$$

The subscripts on the left-hand side mean only add up the works by external forces, i.e. by forces from objects outside of our chosen system.

System choice: Earth and ball

• There are no external forces, so the left-hand side is zero.
• Gravitational potential energy is an internal form of energy, so since Earth and the object are both in the system, we may safely use $$\Delta U = mgh$$.

System choice: Only the ball

• Now there is an external work done $$W = -mgh$$ by Earth.
• Since Earth is external, there is no longer any internal gravitational potential energy. Consider this similar to having a spring 'outside' of our system; there would not be any elastic potential energy.-

Let's say a hand is raising the object of interest. The two equations would then yield $$W_\text{hand}=\Delta K + mgh$$ and $$W_\text{hand}-mgh=\Delta K + 0$$, both of which are algebraically equivalent.

When we raise an object to a height $$h$$, it is said that the potential energy of the object is increased by $$mgh$$.

Although we commonly say that, it is not technically correct. We have increased the gravitational potential energy of the Earth-Object system, not the object alone. Gravitational potential energy, like other forms of potential energy, is a system property.

But isn't the work done by gravitational force $$-mgh$$? Then that will essentially mean $$Work= mgh+(-mgh)=0$$

The net work done on the object being zero only means that there is no change in kinetic energy of the object (such as if the object begins and ends at rest). It doesn't mean there is no change in gravitational potential energy.

Positive work transfers energy to an object. Negative work takes energy away from an object. In this case, the negative work done by gravity takes aways the energy transferred to the object by the positive work raising it and stores it as gravitational potential energy in the Earth-Object system.

Hope this helps.

Work is the transfer of energy from one energy-holding abstract object to another. If A gives B 5 joules, B gives A -5 joules. We don't sum them to find out how much was transferred; they're two ways of saying the same thing. An energy holding object is a kinetic energy associated with some bodies' relative motion or a potential energy associated with some configuration.

The work done on an object by a net force equals the sum of change in kinetic energy of the object and the potential energy acquired by the object: W net = ΔKE + ΔPE . (1) Hence it can be said that since since both KEi and KEf are zero, hence the total work done done on the object by external forces is equal to the change in potential energy of the object. Even more interestingly, you can see this post (https://physics.stackexchange.com/questions/262125/work-done-relation-to-potential-energy#:~:text=I%20know%20work%20done%20is,system)%20decreases%20its%20potential%20energy.) that actually the rise in potential energy is nothing but the negative of work done by gravity. And hence you can write +mgh in the right hand side of the equation (1) [as the rise in potential energy] or on the left side of equation (1) as the work done by gravity. They are basically the same thing.

• The way I understand it, we only have to include the work done by the non conservative forces while using said definition of work done, right? Also, I think you post half-a-URL after the first one. Commented Dec 5, 2023 at 19:25
• No, not right...
– hft
Commented Dec 5, 2023 at 19:28
• @hft how would you explain it then? Commented Dec 5, 2023 at 19:29
• "The work done on an object by a net force equals the sum of change in kinetic energy of the object and the potential energy acquired by the object..." No. This is wrong. It literally contradicts the well known work kinetic energy theorem. The blind are leading the blind over here...
– hft
Commented Dec 5, 2023 at 19:33
• So, like: $\int\vec F_{net, non-conservative}\cdot \vec {dx} = \Delta (T+U)$, where $T$ is the kinetic energy and $U$ is potential energy.
– hft
Commented Dec 5, 2023 at 19:42