0
$\begingroup$

When I throw a ball (system) up in the air, gravity (of the earth [external]) does negative work on the system. The system's total energy remains constant as it moves in the conservative field, potential and kinetic energies interplay. Here, work done by gravity is negative and we say that work is done by the system. What is the physical meaning of the system doing work (or in some cases work being done on the system)? Is there any such physical meaning?

$\endgroup$
3
  • $\begingroup$ What on Earth is "negative work"? When you throw a ball up, work is done (and energy conserved), and when it falls back down, work is done (and energy conserved). $\endgroup$
    – Steve
    Feb 2 '18 at 15:50
  • $\begingroup$ As far as I know, work done can be positive/negative. Friction, for example, does negative work. $\endgroup$
    – R004
    Feb 2 '18 at 16:01
  • $\begingroup$ But what about friction heating (such as in welding)? It seems odd to me that the useful effect might be called "negative work", and I've never come across a notion in mechanics that all work eventually adds up to zero (as it must if work has both a rigorous definition and negative elements). $\endgroup$
    – Steve
    Feb 2 '18 at 16:10
2
$\begingroup$

When you lift the ball up from the earth, you are doing work on the system by increasing the potential energy. This corresponds to negative work done by gravity (energy is added rather than used / removed).

By the way, the earth must be included in the system you are talking about. The potential energy is a property of both object and earth, not only of the object.

$\endgroup$
14
  • 1
    $\begingroup$ Simon, actually it is not so much about neglecting the Earth's mass, but rather about that mass already being included in $g$. I guess you, @R004, are thinking about the simple formula: $$U=mgy$$ You should be aware here that $g$ can be written as a collection of variables: $$g=-{GM \over r^2}$$ Earth's mass $M$ is already here! And therefore, the value is only the well-known constant $g=9.82\;\mathrm{m/s^2}$ on Earth's surface. If you did the calculation on the Moon, it would be different due to a new Moon mass (and a new distance $r$) $\endgroup$
    – Steeven
    Feb 1 '18 at 15:11
  • 1
    $\begingroup$ @Steeven I completely overlooked the $g$! This helps! $\endgroup$
    – R004
    Feb 1 '18 at 15:14
  • 1
    $\begingroup$ @R004 Good to hear. I would be surprised if you literally did see the phrase "Potential energy of the particle" in an established physics book. It is at best a sloppy wording. Without the Earth, the mass would not "have" any potential energy. That energy exists do to the presence of both object and planet, and is not "contained" inside one of 'em. $\endgroup$
    – Steeven
    Feb 1 '18 at 15:17
  • 1
    $\begingroup$ @R004 The universal, general potential energy formula always works for any two-body system: $$U=-G{Mm \over y}$$ The shorter version $U=mgy$ is this same formula, just with some of the parameters replaced with $g$. $\endgroup$
    – Steeven
    Feb 2 '18 at 6:28
  • 1
    $\begingroup$ @R004 Yes. Although there are other ways of adding energy to a system than as work. $\endgroup$
    – Steeven
    Feb 2 '18 at 18:19
1
$\begingroup$

If the system is the ball then the work done by the gravitational force on the system is $\vec F_{\rm gravity} \cdot \Delta \vec x$ where $\Delta x$ is the displacement of the force.

If the ball is moving downwards $\vec F_{\rm gravity} $ and $\Delta \vec x$ are in the same direction and the work done on the ball by the external force is positive which leads to an increase in the kinetic energy of the ball.

If the ball is moving upwards $\vec F_{\rm gravity} $ and $\Delta \vec x$ are in opposite directions and the work done on the ball by the external force is negative which leads to an decrease in the kinetic energy of the ball.
You can think of this as the ball doing the work.

When the system is the ball alone no mention must be made of gravitational potential energy as that is a property of the ball and the Earth not the ball alone.

So when

This question is related to your question where it is explained that it is usually only the work done on the ball which is considered since the Earth is so more massive than the ball the ball undergoes a much greater displacement than the Earth and so much more work is done on the ball than is done on the Earth.

$\endgroup$
4
  • $\begingroup$ In "An Introduction to Mechanics" by Kleppner and Kolenkow, there is not a single mention about the particle-earth system( not yet in the chapter that introduces these concepts ). "Potential energy of the particle" and "Total energy of the particle" are alone mentioned. $\endgroup$
    – R004
    Feb 1 '18 at 8:04
  • $\begingroup$ @R004 The (unwritten) assumption which is made is that one of the bodies has a mass which is very much larger (infinitely larger) than the other one and the subtlety of giving the Earth a finite mass is not discussed. It would have to be if one was considering conservation of momentum of the Earth & ball system when a ball is falling. $\endgroup$
    – Farcher
    Feb 1 '18 at 8:29
  • $\begingroup$ Why is GPE the property of ball-earth system and not just the ball? $\endgroup$
    – R004
    Feb 1 '18 at 9:23
  • $\begingroup$ @R004 Suppose the ball was another Earth which of the two “Earths” store the potential energy? $\endgroup$
    – Farcher
    Feb 1 '18 at 9:27
1
$\begingroup$

In a conservative system no work is really done, since the total energy is constant at every point. So then, the only work done is the imparted kinetic energy by the force applied in the upward movement of the hand for a given period of time. Once the ball leaves the hand, [the work done (added) to the system ceases.] A better example is a weight on a string. The work done is the initial displacement of the weight along the arc path, adding potential energy to the "system". Once released, the "conservative" system swings back and forth, converting P.E. to K.E. and back to P.E. No work is done within the system. If there is something called "negative" work, maybe it's the loss of energy as a "real" pendulum loses energy through the force applied by air resistance.

$\endgroup$
1
  • $\begingroup$ Don't you think it is better to say that in a conservative system no net work is done over a closed path instead of saying that no work is done? $\endgroup$
    – R004
    Feb 2 '18 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.