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I have a question on the sign of the Work quantity. My understanding follows: $$ W = \int_C \vec F \cdot d \vec S $$

$$ W = -\Delta U, \qquad -W = \Delta U $$

For direction and sign: $$(-F, dx) = \Delta U = -W$$ $$ (F,dx) = -\Delta U = W.$$

It's my understanding that work is energy transferred to or from and object.

Question In the case of $-W$, does this mean that $W$ energy is transferred from the body applying force to the body receiving application of force? The body applying force loses $-W$ of $PE$?

I'm getting confused with the sign of Work. In the case of above question, I suspected that work would be positive for a transfer of energy, yet the definitions indicate it is negative.

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  • $\begingroup$ @BobD change in potential energy $\endgroup$
    – Nick
    Commented Nov 20, 2021 at 19:19
  • $\begingroup$ OK, and what does "C" mean in the integral? $\endgroup$
    – Bob D
    Commented Nov 20, 2021 at 19:29
  • $\begingroup$ @BobD line integral $\endgroup$
    – Nick
    Commented Nov 20, 2021 at 19:36

5 Answers 5

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Positive work indicates positive energy transfer from the e.g field to a mass. via the increase of kinetic energy

$\int F \cdot dr$

$\int F \cdot vdt$

$\int m (dv/dt) \cdot vdt$

$m \int d/dt(1/2 v^2)dt$

$1/2 mv^2$

Plugging in bounds from v0 to v1 where shows this is the difference of kinetic energy

so net positive work indicates a positive kinetic energy change, and net negative work equals negative kinetic energy change

The definition

W= -change in U

Means that if there is a positive change in U, then negative work is done

and if there's a negative change in U, positive work is done

There is no contradiction, as U is defined as the amount of work done against the field from A to B, if positive work is done against the field moving something from A to B then the field does negative work on the object moving from A to B meaning a decrease in Work done on the object

As change in U is defined as $\int_{a}^{b} -F \cdot dr $

This represents the work I would have to do against the field moving something from a to b. Taking the minus out,

$\int_{a}^{b} -F \cdot dr =- \int_{a}^{b} F \cdot dr $

The term on the right is - work done by the field so if Potential difference from A to B is positive then F.dr has to be negative, aka negative work done

EDIT:

To conceptualise this,

consider the equation.

$ke + \int_{a}^{x} F \cdot dr = 0$

Where ke is the amount of energy that the object has initially, e.g in the form of kinetic for this example.

This equation states that, given it has some kinetic energy, in moving from a distance (a to x) the field F does work on the object, such that the total kinetic energy when the object has reached a distance x is zero.

Rearranging for ke,

$Ke = - \int_{a}^{x} F \cdot dr $

This is the kinetic energy an object should have, if I want to reach a position x in the presence of a force field such that the object stops at x

This is what potential is defined as ,but instead of saying a particle has some initial ke, the energy could be given over a distance. same thing.

This is WHY this expression represents work done by an external force(me) to move something from a to b against a force field F , clearly if positive work is required to move something from a to b in the presence of forcefield F, then the forcefield F must do negative work on the object moving through that distance ( as I have to give it energy to move it there, because the field does negative work along the way)

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    $\begingroup$ "so positive work indicates a positive kinetic energy change, and negative work equals negative kinetic energy change". Shouldn't it read "net positive work" and "net negative work". Per the work energy theorem the NET work done on an object equals its change in KE. $\endgroup$
    – Bob D
    Commented Nov 20, 2021 at 19:42
  • $\begingroup$ Yes you are correct,I was thinking that but too lazy to reword . I'll change it now $\endgroup$ Commented Nov 20, 2021 at 19:46
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    $\begingroup$ You may be correct, but I don't understand what you are trying to say. Please define "field". I don't know what it means in this context. When you write $W$ I don't know if you mean internal work or external work. Maybe that would become clear if you define "field" $\endgroup$
    – garyp
    Commented Nov 20, 2021 at 20:08
  • $\begingroup$ @jensenpaul In your post: "Positive work indicates positive energy transfer from the e.g field to a mass. via the increase of kinetic energy" : yet in my OP note contradicts this stating $W = -\Delta U$. Are the definitions in my OP wrong? or perhaps I'm confusing the agent applying force over distance with the body receiving this force? $\endgroup$
    – Nick
    Commented Nov 20, 2021 at 20:14
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    $\begingroup$ Gary, OP talks about PE which is only made sense of if F is part of a force field, it is this to which I'm referring. and W being work done by the Force field on an object $\endgroup$ Commented Nov 20, 2021 at 20:24
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Asking whether negative work means energy added to or removed from a body, is the same as asking whether a negative speed means leftwards or rightwards.

You can't answer it without knowing the reference! Signs are mathematical inventions of no physical meaning - they purely relate to some reference, some definition, that often times can be arbitrarily chosen for different scenarios.

In the natural sciences there is not consensus over the reference in the case of work "direction". In e.g. chemical engineering, I've often seen positive work defined as work done on a system ($W$ is positive when the body gains energy). In e.g. thermodynamics, I've often seen the opposite definition of positive work defined as the work done by the system ($W$ is positive when the body loses energy).

The sign in any particular scenario thus has to be clearly defined by the author. If it isn't and the reader must guess the reference from the context, then he's done a poor job in his presentation.

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  • $\begingroup$ both on a system and by a system examples are helpful and are what I was looking for. In both cases, if work is done on a system what makes it different then being done by a system? I'm assuming Energy is conserved in the system so am curious what happens to PE. $\endgroup$
    – Nick
    Commented Dec 18, 2021 at 22:50
  • $\begingroup$ @Nick Work done on a system means energy added to that system. Work done by the system means energy removed from the system. $\endgroup$
    – Steeven
    Commented Dec 18, 2021 at 23:05
  • $\begingroup$ @Nick No, there won't be energy conservation within such system. The energy added or removed comes from somewhere external - the contained energy is not constant in such a situation. Think of stretching a rubber band with your hands. Its elastic potential energy increases (as well as its thermal energy). You are doing work on the rubber-band. If the rubber band is considered the system, then the energy of this system increases - its not conserved. For there to be energy conservation, you have to be included. A rubber-band-and-you system with no external energy entering or leaving. $\endgroup$
    – Steeven
    Commented Dec 18, 2021 at 23:07
  • $\begingroup$ Does this look correct? Rubber band: two systems, hand and rubber band. The hand does $+W$ on the rubber band system, resulting the hand's system $-PE$ and rubber band's system $+PE$. Second scenario: During the band's return to normal position, the hand system gains $+PE$ by the rubber band system doing $+W$. Finally, I often see this in equations where one displaces against/opposite a force field direction as, $-W$ done by the hand, $W=-PE$, therefore $+PE$ of the rubber band system. $\endgroup$
    – Nick
    Commented Dec 18, 2021 at 23:34
  • $\begingroup$ @Nick Keep in mind that there is no requirement of a potential-energy conservation. That something gains potential energy doesn't necessarily mean that something else loses potential energy. It just means that something else loses energy. So, to the first scenario, yes the rubber band gains $+PE$ and the hand loses an equal amount of energy. Whether that energy is potential, is not given. It might rather be chemical or biological. But other than the use of the label $PE$ for the hand, I agree with your energy math here. I did not fully understand your mast question about a force field, though? $\endgroup$
    – Steeven
    Commented Dec 19, 2021 at 0:40
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starting with Newton second law

$$m\,\frac{dv}{dt}=F$$

from here $$m\,dv=F\,dt=F\,\frac{dt}{ds}\,ds=\frac{F}{v}\,ds\quad \Rightarrow\\ m\int_{v_i}^{v_f} v\,dv=\int _{s_i}^{s_f}\,F\,ds=W$$

hence

$$\frac m2 (v_f^2-v_i^2)=\pm W\quad, s_f > s_i$$

is the final velocity $~v_f~$ greater then the initial velocity $~v_i~$ the work is positive , otherwise the work is negative .

potential and work

the kinetic energy is:

$$T=\frac m2 \,v^2$$ and the potential energy is

$$U=U(s)$$

with EL you obtain the Newton second law

$$\frac{d}{dt}\frac{dT}{dv}+\frac{\partial U}{\partial s}=0\quad\Rightarrow\\ m\frac{dv}{dt}=-\frac{\partial U}{\partial s}=F$$

hence the work is

$$W=-\int \frac{\partial U}{\partial s}\,ds=-U$$

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  • $\begingroup$ Appreciate the derivation. Where did $m/2$ come from? Also, when an external system performs $-W$ on a body, does that system experience a decrease of its potential energy? $\endgroup$
    – Nick
    Commented Nov 21, 2021 at 18:28
  • $\begingroup$ m/2 is tippo , $ W=-U$ thus if W is negative U is positive $\endgroup$
    – Eli
    Commented Nov 21, 2021 at 19:43
  • $\begingroup$ $v=\dfrac{ds}{dt}$ $\endgroup$
    – Eli
    Commented Dec 19, 2021 at 7:42
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Mathematically, it makes no difference whether you say $\Delta U=-W$ or $-\Delta U=W$, as one is obtained from the other by simply multiplying both sides of the equation by negative 1. If you are considering the work done by a conservative force (gravitational, electrostatic or elastic) you can simply state $\Delta U=-W$ as shown below.

If the direction of a force $\vec F$ is the same as the direction of the displacement $d\vec S$ then the work $W$ done by the force is positive. If the positive work is done by a conservative force it results in a decrease in potential energy (PE). An example is the positive work done by a gravitational force on a falling object. So we can say $\Delta U=-W$, since $W$ is positive making $\Delta U$ negative.

If the direction of the force $\vec F$ is opposite the direction of the displacement $d\vec S$ then the work $W$ done is negative. When a force does negative work on an object it takes energy away from the object. In the case of negative work done by conservative force, the energy taken from the object is stored as an increase in PE. An example is the negative work done by gravity on an object that has been lifted by an external (to the object-Earth system) force a height $h$ near the surface of the earth. Gravity takes the energy given the object by the external force and stores it as an increase in gravitational potential energy of $mgh$ of the object-Earth system. But we can still say $\Delta U=-W$ since in this case $W$ is negative making $\Delta U$ positive, as required.

Hope this helps.

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Some more information that may help. Your question deals with a system that is a solid body, not a fluid.

All of this discussion (question and earlier responses) deals with "work" as defined in classical mechanics as follows: the work done on a body by the net force on the body is the change in kinetic energy of the body. Work can be either + or - depending on whether the net force causes an increase or decrease in kinetic energy. In general, the change in kinetic energy includes rotational as well as translational motion. For a conservative force, the change in potential energy is defined as the negative of the work done by that force. The use of potential energy instead of evaluating the work for the conservative force simplifies many problems. For example, for the gravitational force near the earth the change in potential energy is simply $mgh$ where $m$ is mass $g$ is the acceleration of gravity and $h$ is the change in height, regardless of how complicated the actual physical path is in changing the height. $mgh$ is easy to evaluate and equals the negative of the work done by gravity $-\int_{r_1}^{r_2} \vec F_g \cdot d\vec r$ which is not always easy to evaluate. $\vec F_g$ is the force of gravity and $\vec r$ describes the path taken in changing the height by $h$.

The classical mechanics concept of work does not consider changes in the internal energy of a body. The assumption of a rigid body is frequently used in classical mechanics and a rigid body can have no change in its internal energy. In reality no body is perfectly rigid and for problems where changes in internal energy are not small, a broader definition of work is needed as provided in the first law of thermodynamics (conservation of energy). In thermodynamics work is defined as "energy that crosses the boundary of a system, without mass transfer, due to an intensive property difference other than temperature between the system and its surroundings". (In thermodynamics heat is defined as "energy that crosses the boundary of a system, without mass transfer, solely due to a difference in temperature between the system and its surroundings".)

Even for a system where the internal energy changes, the classical mechanics concept of work on the system is valid as applied to the center of mass of the system. That is, the change in kinetic energy of the center of mass of a system is equal to the work done on the system by the net force on the system. An example is a person jumping off the floor; see my response to If I jump off the ground is the net reaction force doing work on me? on this exchange. To avoid confusion, some have proposed using the term "pseudowork" for the classical mechanics concept applied to the center of mass, and reserve the term "work" to mean the broader thermodynamics definition of work. You can search for "pseudowork" on the web for more information.


The first law is applicable for fluids as well as solids and sometimes approximations are used in the mechanical energy balance on the fluid to account for dissipative effects. For example, for a flowing fluid a "friction loss" term is sometimes included in the Bernoulli mechanical energy balance to account for the dissipative effect of friction, but this assumes the change in internal energy of the fluid is small, so the "loss" causes a decrease in pressure as the fluid flows. If the fluid has a significant change in internal energy (e.g., change in temperature), a full first law energy evaluation is necessary (along with momentum and mass evaluations). Two good references on the first law and its applications are: Thermodynamics by Obert, and Transport Phenomena by Bird, Stewart, and Lightfoot.

Hope this helps.

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