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So I have some confusion here, I am sure I knew this at some point. Let's say an object of 10 kg is dropped from a height of 10 m. When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m. That gives 980 joules of work done on the object by gravity.

But the object did not gain 980 joules of mechanical energy. It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level).

Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.

So

a) am I right that no net work was done on the car by sum of all forces?

b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?

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3 Answers 3

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The confusion here comes from the fact that your choice of system is not clearly defined.

If the system is the earth plus the object, then there is no external force, and therefore no change in total energy. The potential energy of the system is transfered into kinetic energy. No external work done, and external work is what adds or removes energy of the system.

If the system is the object, then gravity does external work on the system, adding energy, increasing its kinetic energy. Potential energy is not defined for a single object. There is no potential energy with this choice of system. Potential energy is always defined for pairs of interacting objects. With this system, there is work done.

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  • $\begingroup$ So since I am using gravity to do work, and the gravity is "coming" from outside the system that is having work done on it, it is not correct to say that the 10 kg object has gained potential energy because if the object is the system (exclusive of the larger gravitating mass) then there is no change in potential energy of the system that I have chosen? (Is that what you are saying?). Anyway, wouldn't the atoms of the object be vibrating differently at a different point in the gravitational field? Or wouldn't the object have less gravitational flux passing through it? $\endgroup$
    – Joseph Hirsch
    Oct 25, 2016 at 1:27
  • $\begingroup$ No, that's not what I'm saying. I'm saying that potential energy is not even defined if you system is the object only. It's incorrect to say that it does not change. It doesn't exist. I have two things to say about your comment about atoms. For simplicity we often take a point particle model of an object, and that's what I had in mind implicitly. Second, in principle, tidal forces could deform an extended object and raise its temperature. But tidal forces on "normal sized" (~ 1 m) object above the surface of the earth would be very small, and the effect would be very hard to detect. $\endgroup$
    – garyp
    Oct 25, 2016 at 2:48
  • $\begingroup$ This answer is wrong. Gravity does not add energy to a falling body. $\endgroup$
    – John Duffield
    Nov 5, 2016 at 12:24
  • $\begingroup$ @JohnDuffield Can you elaborate? If your choice of the system is the body, and gravity is acting on it, and the body displaces in response to gravity, then the force of gravity does work on the object. By the work-kinetic energy theorem, the kinetic energy of the system increases. $\endgroup$
    – garyp
    Nov 5, 2016 at 19:14
  • $\begingroup$ @garyp : the external kinetic energy of the body increases, but only because its mass-energy, which is internal kinetic energy, is decreasing. Once the body hits the ground the external kinetic is dissipated and you're left with a mass deficit. Conservation of energy applies. $\endgroup$
    – John Duffield
    Nov 6, 2016 at 10:19
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This is a common point of confusion that boils down to the fact that there are two physically equivalent but conceptually different ways of viewing this situation.

You can either look at gravitational potential energy (GPE) as an "internal" form of energy that your 10 kg object can have or you can look at the gravitational force (Fg) as an external force acting on the object. What you cannot do is look at the situation in both ways at once.

If you choose to see GPE as a form of energy, what is happening in this situation is that GPE is turning into kinetic energy (KE) as the object falls. If you choose to see Fg as an external force, what is happening is that Fg is doing work on the object, which increases its kinetic energy.

Either way, the amount by which the kinetic energy increases when the object reaches the ground is mgh, or 980 Joules in this case.

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  • $\begingroup$ So let's say that instead, I lift the 10 kg object 10 meters. I have done work to increase its potential energy by 980 J. (Also I exerted a force in a direction of motion). Did gravity also do work or negative work while I raised it, since gravity was a force opposite the direction of motion? If gravity did not do negative work then the object would reach 10 m with a velocity gained by the force that I exerted on it, so it seems that I did work, but gravity also did work such that the object had not velocity when it reached 10 m. $\endgroup$
    – Joseph Hirsch
    Oct 23, 2016 at 2:21
  • $\begingroup$ I think I've got it now based on Garyp's answer. The falling object has work done on it by gravity because there is a force acting over a distance and its KE increases, but the object itself does not possess potential energy, potential energy is possessed by the object and the source of gravity as a system. $\endgroup$
    – Joseph Hirsch
    Oct 25, 2016 at 1:43
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So a) am I right that no net work was done on the car by sum of all forces? b) is the work done by gravity equal to force time distance, or is it equal to the change in energy of the object which is zero?

You defined work correctly for your example:

$$W=F\Delta h=mgH$$ The Work-Energy theorem tells us that the change in potential energy $\Delta U$ is equivalent to the work done, so:

$$W=\Delta U$$

In your case:

$$\Delta U=mgH$$

So it tallies perfectly!

It's clear you're confusing conservation of energy with work-energy equivalence. Work was done but overall the total energy of the system hasn't changed: only potential energy, $U$, has been converted to work, $W$.

Edit: in answer to OP's last comment.

1) a 10 kg object is slowly raised to a height of 10 m. Its potential energy has increased by 980 J, it is motionless, and so I assume that you did 980 J of work on the object to raise it. 2) You toss the object up, it gains enough kinetic energy to rise to a height of 10 m before it stops rising. It is in the same final state as situation 1, but didn't gravity do -work to decelerate it? You do the same work to accelerate it so that it rises to 10 m and motionless as if you raised it slowly.

Case 1):

In order to get to get it up there, you need to provide work against gravity, so:

$$W=mgH=\Delta U$$

Case 2):

You toss the object up and it just reaches $H$.

To do so, you will have to impart kinetic energy $\Delta K$ to the object, equivalent to $\Delta U=mgH$, so $\Delta K=\Delta U$. During 'flight' this kinetic energy is then converted to potential energy and the object ends up with $K=0$ because $v=0$.

To impart that kinetic energy $\Delta K=\frac12 mv_0^2$ ($v_0$ is the launch velocity), you need to perform work:

$$W=\Delta K=\frac12 mv_0^2$$

And since as $\Delta K=\Delta U$, then:

$$W=\Delta U=mgH=\frac12 mv_0^2$$

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  • $\begingroup$ You say that the work-energy theorem states that change in potential energy is work done. What about when a horizontal force accelerates an object horizontally? The potential energy hasn't changed, but work has been done to change the kinetic energy, right? $\endgroup$
    – Joseph Hirsch
    Oct 23, 2016 at 2:14
  • $\begingroup$ It is confusing to me that for the falling object it does not seem to matter whether the object reaches the ground with kinetic energy or not, so long as the potential energy change is the same. $\endgroup$
    – Joseph Hirsch
    Oct 23, 2016 at 2:16
  • $\begingroup$ The work done is also equivalent to the change in kinetic energy: $W=\Delta U=\Delta K$, in the case of your post. In the horizontal case: $W=\Delta K$. $\endgroup$
    – Gert
    Oct 23, 2016 at 2:25
  • $\begingroup$ Thanks. So what about this. If I lift a 10 kg object 10 meters (and motionless), did GRAVITY do work on it while I lifted it (since gravity was a force that acted over the distance and parallel to the motion?) $\endgroup$
    – Joseph Hirsch
    Oct 23, 2016 at 2:38
  • $\begingroup$ If you start motionless and end motionless, then $\Delta K=0$. In that case too: $W=mgH=\Delta U$. But it's not gravity that does the work: it's the lifting force against gravity that does the work. That's important to get the right sign. (Logging off now: very late where I am). $\endgroup$
    – Gert
    Oct 23, 2016 at 2:46

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