For what reason is the difference of potential energy $\Delta U=-W$ equal to the *opposite* of the work done?

Question

In my classical mechanics physics textbook (a translation of the Walker-Halliday-Resnick Fundamentals of Physics) the difference of potential energy is defined as

$$ \Delta U = -W \qquad (1) $$

I have done extensive research (taking me 5+ hours) and I claim to have a reasonable understanding of this model. In particular, I understand that if we throw a solid object in a straight upward direction then the work (i.e., the quantity of kinetic energy conveyed or subtracted from a body) exerted by the Earth's gravitational force is negative because they act on opposite directions: $W = \vec{F} \cdot \vec{d} = F \cdot cos(\phi) \cdot d$, where $cos(\phi) = -1$ due to $\phi$, the angle between the movement and the gravitational force, being $180°$.

However, I couldn't find anywhere an explanation for this. I was demonstrated that for a conservative force $\vec{F}$ doing work along a path $ab$, $W_{ab} = -W_{ba}$, and I also know that we can always associate a potential energy to a conservative force. But I'm still missing a link, and not knowing how the negative work of a force relates to its potential energy gives me brain fog.

Can you please provide an explanation, or an appropriate proof, for $(1)$? Please note that my physics knowledge only extends up to what is taught in university-level Physics I and Physics II courses.

Answer 1

The potential is defined as a function $U$ such that the conservative force $\vec F$ that we are studying is given by the gradient $\vec F = -\nabla U.$ Since you probably have not seen vector calculus yet, let me be very careful to write this out as the components, $$F_x = -\frac{\partial U}{\partial x},\\ F_y = -\frac{\partial U}{\partial y},\\ F_z = -\frac{\partial U}{\partial z}.$$These "partial derivatives" are evaluated as normal derivatives treating the other variables as constant, so for example the potential $U=k~x^2~y + p~z^2$ would generate $F_x = -2k~x~y, F_y = -k~x^2, F_z = -2p~z.$

Partial derivatives are the natural way to understand calculus on a function of many variables. In single-variable calculus, you were trying to approximate a curve with a tangent line; in this multi-variable calculus we are trying to approximate surfaces with planes. In particular, if you can make this approximation then it means that a function can be expanded around a point, $$f(x + \delta x, y+\delta y, z+\delta z)\approx f(x, y, z) + \frac{\partial f}{\partial x}~\delta x+ \frac{\partial f}{\partial y}~\delta y+ \frac{\partial f}{\partial z}~\delta z.$$Here by $\delta q$ I just mean "a little change in $q$", whatever $q$ is. One could also move this term $f(x,y,z)$ to the left-hand side and refer to that difference as $\delta f$, if you'd like. This understanding of expanding out a multivariable function will be important.

The power exerted by a force on a particle is the dot product of that particle's velocity with the force, and the work done over the path is the time integral of power exerted by that force. It is common to denote the position of the particle as a vector $\vec r(t)$ with components $r_{x,y,z}(t)$ and then this is: $$ P(t) = F_x~\frac{dr_x}{dt} + F_y~\frac{dr_y}{dt} + F_z~\frac{dr_z}{dt}\\ W = \int_{t_0}^{t_1}dt~P(t).$$ The combination of these two definitions is what you seem to be asking about: but it is not very complicated at all. Combine the two and then stare for a second at the following: $$P(t)~\delta t = -\frac{\partial U}{\partial x} ~\frac{dr_x}{dt}~\delta t -\frac{\partial U}{\partial y} ~\frac{dr_y}{dt}~\delta t -\frac{\partial U}{\partial z} ~\frac{dr_z}{dt}~\delta t.$$What should now stand out to you is that this is very much like the above expression for $\delta f$ above, if we defined $\delta x = \frac{dr_x}{dt}~\delta t$ and so on for $\delta y, \delta z.$ And those are very natural definitions, as $r_x$ represents an $x$-component of position and if we take this time-derivative we get a component of velocity, and multiplying against a short time $\delta t$ we get a small change in this $x$-component due to the particle's current motion.

There is a more formal way to do this and it is to invoke the chain rule, which says that when we apply some function $U(x, y, z)$ to these time-varying components $x = r_x(t)$ and so forth, we find that:$$\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) = \frac{\partial U}{\partial x}~\frac{dr_x}{dt} + \frac{\partial U}{\partial y}~\frac{dr_y}{dt} + \frac{\partial U}{\partial z}~\frac{dr_z}{dt}.$$ Therefore what we have found above is simply, $$P(t) = - \frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big).$$The work is the time integral of power, but integrals perfectly undo derivatives, and therefore when we do this definite integral we get from the fundamental theorem of calculus, $$\begin{align} W &= -\int_{t_0}^{t_1} dt~\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) \\ &= -\Big(U\big(r_x(t_1),~r_y(t_1),~r_z(t_1)\big) - U\big(r_x(t_0),~r_y(t_0),~r_z(t_0)\big)\Big)\\ &=-\Delta U.\end{align}$$ That's really all there is to it: for conservative forces $\vec F = -\nabla U$ the power $\vec F \cdot \frac{d\vec r}{dt}$ is immediately seen to be a chain rule expression $\frac{d}{dt} U(\vec r) = \nabla U\cdot \frac{d\vec r}{dt}$, which identifies it as a total time derivative, and therefore the work, which is just the time integral of the power, must be the overall change in the quantity: in this case the quantity is $-U$ and so $W = \Delta(-U) = -\Delta U.$

Answer 2

I think you are struggling with the boundaries of the system.

When it is thrown in the air, straight up, at from the moment of release gravity is doing work in the direction opposite (or, I should say normal to) the planes of equal potential energy, which increase in value with height. Thus, if the KE is 100 at the bottom, and 0 at the peak of the arc, the change in PE is the negative of the work = $-(0-100) =100$. This system has no energy entering or leaving.

But suppose we had accomplished this by firing from a cannon with compressed air. During the launch, it seems the work goes the other way: since work = force times distance, it seems positive work was done, and the PE also increased. So the total energy of the 'system' seems to have grown.

What is getting lost is that somehow, somewhere, that air got compressed. Maybe, for example, a large rock was released onto a piston over a large column of air. Now the rock started with a higher PE and ended at a lower one; which means positive work got done and energy was stored (which gave us our blast later).

This abstraction with signs can drive you nuts, but it is helpful in the end.

Answer 3

OP is asking

What is the reason for the minus sign in the formula $$\Delta E_{\rm pot}~=~\color{red}{-}W_{\rm c} \tag{1}$$ for the work $W_{\rm c}$ done by conservative forces ${\bf F}_{\rm c}$?

Answer: Imagine that we have grouped the world into 2 "accounts": $$\text{system}\quad +\quad \text{environment} $$ and we want to keep track of changes of energy between the 2 accounts. The minus sign in eq. (1) can be viewed as re-assigning the conservative forces ${\bf F}_{\rm c}$ to the opposite account.

In more details: Recall the work-energy theorem $$ \Delta E_{\rm kin}~=~W_{\rm tot}~=~W_{\rm c}+W_{\rm nc}. \tag{2}$$ Next define the mechanical energy $$ E_{\rm mech}~:=~E_{\rm kin}+E_{\rm pot}\tag{3}$$ as the sum of kinetic and potential energy. Eqs. (1)-(3) imply that the change in mechanical energy $$ \Delta E_{\rm mech}~=~W_{\rm nc} \tag{4} $$ is given by the work $W_{\rm nc}$ done by non-conservative forces ${\bf F}_{\rm nc}$. That's pretty nifty: If there are no non-conservative forces, then the mechanical energy is conserved!

Answer 4

According to Work-Energy theorem "Work done by all forces on a particle is equal to change in its kinetic energy". Suppose there is only one conservative force, such as Gravitational, is acting on a particle. After elapse of some time, the particle gains some speed from rest and moves to some distance. If we need to define a potential energy for the same conservative force that ultimately converts to the kinetic energy, this is how we define it. From work-energy theorem, W_f = KE₂ - 0. Also from conservation of energy, PE₁ + KE₁ = PE₂ + KE₂ OR, PE₂ - PE₁ = - KE₂ = - W_f. Always remember, you can only calculate change in potential energy and not absolute potential energy. In earth-particle system, potential energy at very far (infinity) from earth is assumed to be zero and so potential energy at earth is defined.

Answer 5

In models of physics equivalent concepts can be described differently. Take it as an axiom of terminology of physics that 'work' is same as 'energy' whether energy is called potential energy (PE) or kinetic energy (KE). Energy is a scalar concept. It has no direction. KE resides in moving objects and PE resides at locations along a force field. If KE decreases in mass moving along direction of force vector then PE increases and vice versa. These are the logic of concepts appearing in the models of physics. If concept of work (W) refers to movement against a force then KE will decrease and in that case body or mass has moved to a higher PE location. If W refers to movement by the force over a distance then KE is gained. The location of the body or mass has then moved to a lower PE location. Algebraically where change in PE designated as 'dW' is equated to W it is the first case of moving to higher PE location.; if designation is '-dW' equation is with second case of movement to lower PE location. Mathematical equations are descriptions by humans in models of physics and they are not physical events happening mathematically. One cannot allow mathematics in service of physicist to become his or her master.

Take case of water mill. Water moves from higher PE to lower PE location. It has gained equivalent KE. The KE works the grain mill to convert KE to work hiding in ground wheat. Economically the work is in goods or service paid for to the miller. Falling water is an economic resource because KE becomes work. God willing what can water falling down Niagara falls not achieve? But then transfer of resources from tourists to ??? will not be there. Where gain in KE becomes generation of voltage driving current, kilo VA or kilo Wattage working for an hour is work called electrical energy saleable say for 10 cents.

Answer 6

I think your confusion pertains to what we're measuring and how we measure it.

So Work is not a property of a given system (like kinetic energy, potential energy, temperature, pressure, mass, volume, etc.). Instead, Work is a process, meaning that a single number doesn't actually tell the whole story.

Let there be a ball of mass m being dropped from a height h from the ground with a gravitational acceleration constant g. We know that the potential energy of the ball at the height h is equal to U1 = mgh. If we let the height of the ball at the ground equal zero, then we know the potential energy at the ground to be U2 = mg(0) = 0. Therefore, the change in potential energy of the ball is equal to U2 - U1 = -mgh. The resultant change in potential energy is negative, because potential energy was lost from the ball.

Now let's examine if there was any work being done. The ball exerted no forces on anything; therefore, it's work is equal to zero. The force of gravity, however, exerted a constant force on the ball in one direction. This force was equal to F = mg, and the distance the force moved the ball was d = h. The work then done by gravity is equal to W12 = F * d = mgh. Note that in this case, the work done by gravity is equal to the negative of the change in potential energy of the ball. Does this mean that W12 = -(U2 - U1), or could this just be a coincidence?

Consider the converse example:

A machine carries the same ball up from height zero to height h. The machine must now exert a force mg to do this. And since it is doing so for a distance h, the work here is again equal to mgh. The ball still exerts no practical forces and therefore does not do any work. It's potential energy has increased, though, from zero to mgh once again. What then about gravity? Well, we see that gravity is still exerting a force mg, and the ball is still traveling a distance h. Does this mean that the work being done by gravity is equal to mgh once again? No. Recall that Work is a process and not a property. Therefore, you can not have a change in Work - that doesn't mean anything or make any sense. So, a negative value for a process like Work, does not mean that we have lost Work, because Work isn't a thing we have, it's a something we have done, or have the potential to do. What, then, does a negative sign imply when talking about Work? It simply implies the same thing it would imply when talking about number lines or other coordinate systems of space/time: direction.

In the first example, the force of gravity acted downward, and it accomplished its goal of moving the ball downward. Therefore, since both the force and the motion of the object acted in the same direction, we give the resultant measurement of Work done a positive sign (i.e. +mgh, or just mgh). If the direction of force is opposite to the direction of motion, we give the value a negative sign (-mgh). So for our second example, the work done by gravity equals -mgh. Wait, but does this mean that the law that the negative change in potential energy equals the work done??? Not quite...

That the negative change in potential energy of the ball was equal to the work done by gravity was in fact a coincidence. The reason for this coincidence is that all of the potential energy that the ball has is coming from gravity. So, it makes sense that the work done on the ball by gravity would be equal in magnitude. The reason it's equal to the negative of the potential is because the potential is a measure of how much energy the ball has ready to use, and Work is a measure of how much is used. Since the ball lost energy for gravity to use it, it makes sense that the work done equals mgh and the potential energy gained equals -mgh.

Consider the following:

The same ball is held up at a height h before being dropped, but now this time has a spring fixed to the ground pulling on it. Let the spring have a constant k and an original, unstretched length L. Now, the potential energy of the ball at height h equals mgh + k(h-L), but the work done by gravity after it falls is still equal to mgh, as now the spring is also doing work equal to k(h-L).

This will become more clear if/when you take a Thermodynamics class.

Answer 7

This answer will be a little long but I think you’ll understand by the end of it

If you throw a ball up in the air. You have imparted initial Kinetic Energy (KE) right at the beginning. Observe the ball move up.

As the ball moves up, you see the velocity of the ball is reducing since the force of gravity is acting against it. In Physics, we say that this force of gravity is doing negative work on the ball.

The ball has now reached the top, its velocity is zero. Basically force of gravity has done enough negative work to reduce the velocity to zero and therefore its KE has also becomes zero. Let’s say this total work done by gravity in upward journey is W1 (it would be a negative sign e.g. -4J or -10J)

But what has happened to the initial KE. The KE of the ball keeps reducing as it moves up but another form of energy keeps increasing. This other form of energy is Potential Energy (PE). Thus the PE at start is zero and keeps increasing till all KE has converted to PE by the time it reaches the top of the flight. The gravitational force that did negative work on the ball and decreased its KE has in the process increased the PE of the ball. Thus negative work (W1) has resulted in positive change in PE. According to work KE theorem,

Delta KE = W1 ———- Eq. 1

but since Mechanical energy has to be conserved

Delta PE + Delta KE = 0 ———- Eq. 2

Use Eq. 1 to substitute Delta KE as W1 in equation 2, we get

Delta PE + W1 = 0

or

Delta PE = - W1

As an example, if work done is say -10 J and the change in PE is from say PE (initial) = 0 J to PE (final) = 10 J, then-

Delta PE = -W

or PE (final) - PE (initial) = -W

or 10 J - 0 J = - (-10 J)

10 J = 10J

Watch this video that I have created to complete your understanding of conservative (and non conservative) forces-

Why is Delta U = -W