# Work done under constant velocity

• B

## Main Question or Discussion Point

Depending on the answer to this question, I may have others.

Consider a one-dimensional ice rink. Rightwards is taken as the +x direction. A box of mass m slides leftward on the surface with a constant velocity v.

Two people go to opposite ends of the box and apply equal and opposing forces, and so, v remains constant.

Clearly, the net force on the box is zero and there is no net work done on it. However, the box is still being displaced and there are forces applied. Is work still being done on the box by the people individually in opposite directions?

Related Classical Physics News on Phys.org
Doc Al
Mentor
Is work still being done on the box by the people individually in opposite directions?
The person pushing left does positive work on the box; the one pushing right does negative work. (Essentially they are just working against each other -- the box is just a conduit.)

• EVERYTHINGISNOTHING
Perfect. So then it follows that if the direction of v is reversed, then the person pushing left would be doing negative work and the person pushing right would be doing positive work?

Doc Al
Mentor
Perfect. So then it follows that if the direction of v is reversed, then the person pushing left would be doing negative work and the person pushing right would be doing positive work?
Sure. Whoever pushes in the same direction as the velocity does positive work.

Ok then, I have this question.

To compute the gravitational potential energy of a mass m2 due to the field of some other mass m1, we calculate the negative work done in bringing m2 from infinitely far away (where the field has no affect) to some other point in the field without accelerating the mass.

For this example, m1 will be at x = 0, and m2 will be located significantly far away to the right. At the initial moment, m2 has an infinitesimally small velocity to the left. Now, with gravity pulling leftward on m2, I have to apply a rightward force FA to prevent it from accelerating. So, I'm doing work against the force of gravity to sustain a constant velocity on m2.

The work I'm doing would have to be negative because the displacement is to the left, but the force I'm applying is to the right. But then, that would mean that the change in gravitational potential energy is positive. This contradicts the equation Ug = -GMm/r.

What am I thinking wrong?

ZapperZ
Staff Emeritus
Ok then, I have this question.

To compute the gravitational potential energy of a mass m2 due to the field of some other mass m1, we calculate the negative work done in bringing m2 from infinitely far away (where the field has no affect) to some other point in the field without accelerating the mass.
Why is it necessary to move the mass without accelerating?

It appears that you are calculating the work done. Why can't you just let go of m1, and let the gravitational force do the work? In calculating the work done, all you have is the integral of F dot dx. There is no requirement that it moves with constant velocity. The amount of work done at the end will be the work done BY the gravitational force, and this is equal to the change in gravitational potential energy.

Zz.

Doc Al
Mentor
The work I'm doing would have to be negative because the displacement is to the left, but the force I'm applying is to the right.
Correct.

But then, that would mean that the change in gravitational potential energy is positive.
Why do you think that?

Take a simpler example. Lift a book. You do positive work to lift the book--the change in gravitational PE of the book is positive.

Dale
Mentor
The work I'm doing would have to be negative because the displacement is to the left, but the force I'm applying is to the right. But then, that would mean that the change in gravitational potential energy is positive.
You have this backwards. The work you do is negative, therefore your change in energy is positive. But the work done by gravity is positive, therefore the change in gravitational PE is negative.

• nDever
Dr. Courtney
Gold Member
Sounds like a few jobs I've had. Some bureaucrats applying the negative work, and employees applying the positive.

• Dale
But from a frame of reference where the box is still, would there be any work applied?

Dale
Mentor
But from a frame of reference where the box is still, would there be any work applied?
No, there would not be any work either on or by the box in that frame.

• enter
vela
Staff Emeritus
Homework Helper
The work I'm doing would have to be negative because the displacement is to the left, but the force I'm applying is to the right. But then, that would mean that the change in gravitational potential energy is positive. This contradicts the equation Ug = -GMm/r.

What am I thinking wrong?
I'm guessing the negative sign you're thinking of is the one that appears in equations like $\Delta U = -W$ and $F = -dU/dx$. If so, you're looking at the wrong force in your example. $W$ is the work done by the conservative force $F$, gravity in this case, not the work done by you to keep the mass from speeding up. In your example, gravity does positive work on mass 2, so the change in potential energy is negative.

• nasu
After some time of thinking, I've zeroed in a bit more on the source of my confusion, and I believe it is the misuse of the work integral.

So, in the general form, the work done BY a force on a mass along a path is:

$$W=\int_{C} {\mathbf F \cdot d\mathbf s}$$

If the magnitude and direction of F is constant but not directed along the same line as the path, then:

$$W=\int_{C} {F \cos \theta}\, ds=F \cos \theta \int_{C}\, ds$$

In the case of motion in a straight line, the curve C is the displacement:

$$W=F \cos \theta \int_{x_i}^{x_f}\, ds$$

If we use this expression for a scenario when a force causing a moving mass to decelerate, we expect negative work to be done, but it comes out positive because cos theta is -1 and the integral is negative. Where is the flaw?

vela
Staff Emeritus
Homework Helper
You have to be more careful evaluating the dot product: $\vec{F}\cdot d\vec{s} = \| \vec{F} \|\,\| d\vec{s} \|\cos \theta$.

• nDever
Doc Al
Mentor
cos theta is -1
OK.

and the integral is negative
Why? $\int_{x_i}^{x_f}\, ds$ is positive.

• nDever
What if the displacement is from right to left? xi > xf.

vela
Staff Emeritus
Homework Helper
Does the sign of $\int \| d\vec{s} \|$ depend on the path? The integrand is always positive.

• nDever
You have to be more careful evaluating the dot product: $\vec{F}\cdot d\vec{s} = \| \vec{F} \|\,\| d\vec{s} \|\cos \theta$.
Yes.

Does the sign of $\int \| d\vec{s} \|$ depend on the path? The integrand is always positive.
Eureka. Details, details.

So then, how do I interpret the notation? Do I evaluate the integral and remove the sign, flip the limits without negating the integral, .....?

vela
Staff Emeritus
Homework Helper
For the one-dimensional case, you can parameterize the path as $\vec{s} = x\hat i$ so that $d\vec{s} = dx\,\hat i$. If you're going from left to right, $dx > 0$. If you're going from right to left, $dx<0$.

Channel your inner mathematician and be careful with the notation. You will find it all works out consistently.

• nDever
For the more engineering oriented, the total work done to the box is the sum of work done on both ends of the rink. That work is the (force applied by the pushers decelerating the moving block X the distance it takes to bring the block to zero velocity) plus the (force applied by the pushers X the distance they are in contact with the box). The direction the box moves may or may not be relevant depending on what your goal is with the pushing and sliding. It may be relevant if you need to know the position or direction of the box at any time. It may be irrelevant if you need to figure out (for example) how much fuel you need to accomplish a certain number of slides.

I detest the term "negative work", preferring to keep it in terms of vectors.

• EVERYTHINGISNOTHING
Dale
Mentor
I detest the term "negative work", preferring to keep it in terms of vectors.
But work isn't a vector. It is a scalar and mistakenly thinking of it as a vector can cause problems.

ZapperZ
Staff Emeritus
I detest the term "negative work", preferring to keep it in terms of vectors.
Why would you "detest" such a thing? After all, based on its definition in physics (and engineering), there is a clear conceptual description of what negative work is. The concept of negative energy, even in introductory physics, is well-known.

And as @Dale has stated, you're making a puzzling mistake here, considering that work is a scalar, not a vector. I would think that you would "detest" such error more than the use of negative work, which isn't an error.

Zz.

• weirdoguy
Work is a scalar and so it has no direction. When I mentioned vectors I was referring to the forces involved. I don't like using the term "negative work" because it is so often misapplied. The original post a good example.

Suppose we push a box across a floor, say, eastward. That takes some force over some distance, so some work is done. Suppose there is enough friction between the box and the floor that the box stops moving when you stop pushing. Now we push the same box the same distance westward. More work is done. Pushing the box westward doesn't subtract from the work done pushing it eastward. The total work done is the sum of the two slidings of the box across floor. The result is that the box winds up in its original location, which results from the force vectors. But the sum of the work isn't zero as far as energy expended, calories burned, etc. Neither direction of sliding is "negative work", it's work done in a different direction.
Suppose you lift a barbell. You do some "positive" work. If you lower the barbell at a controlled rate (rather than dropping it) you are doing "negative" work. You are still pushing up but with insufficient force to keep the barbell from descending.

Doc Al
Mentor
I don't like using the term "negative work" because it is so often misapplied.
Just apply it correctly!

Suppose we push a box across a floor, say, eastward. That takes some force over some distance, so some work is done.
You do positive work on the box.

Suppose there is enough friction between the box and the floor that the box stops moving when you stop pushing.
The friction (floor) does negative work on the box. Total work done = 0.

Now we push the same box the same distance westward. More work is done.
Again, positive work.

Pushing the box westward doesn't subtract from the work done pushing it eastward.
Not sure what you mean.

The total work done is the sum of the two slidings of the box across floor.
Again, the floor does negative work.

The result is that the box winds up in its original location, which results from the force vectors. But the sum of the work isn't zero as far as energy expended, calories burned, etc. Neither direction of sliding is "negative work", it's work done in a different direction.
In both directions, you perform positive work while the floor does negative work. It all cancels out, leaving the box with no kinetic energy. (Internal energy has surely changed.)

ZapperZ
Staff Emeritus
Work is a scalar and so it has no direction. When I mentioned vectors I was referring to the forces involved. I don't like using the term "negative work" because it is so often misapplied. The original post a good example.

Suppose we push a box across a floor, say, eastward. That takes some force over some distance, so some work is done. Suppose there is enough friction between the box and the floor that the box stops moving when you stop pushing. Now we push the same box the same distance westward. More work is done. Pushing the box westward doesn't subtract from the work done pushing it eastward. The total work done is the sum of the two slidings of the box across floor. The result is that the box winds up in its original location, which results from the force vectors. But the sum of the work isn't zero as far as energy expended, calories burned, etc. Neither direction of sliding is "negative work", it's work done in a different direction.
Suppose you lift a barbell. You do some "positive" work. If you lower the barbell at a controlled rate (rather than dropping it) you are doing "negative" work. You are still pushing up but with insufficient force to keep the barbell from descending.
But what if I lift the box to a certain height h, and then I let go of the box and let it fall back to its original position? What is the total work done then?

The example you gave is NOT a good example to illustrate the concept of "negative" work done, because there isn't any when you only look at work done by you.

When I lift the box, work is done by me, but work is done ONTO the box, or in this case, against the gravitational field. So work done by you is positive, but work done onto the field, or by the box, is negative. When I let go of the box, the field or box does positive work in moving the box back to its original location. The total work done by the box is now zero because both the negative work and the positive work are the same in magnitude.

It is in the concept of what is doing the work, and what is "receiving" the energy or work, that allows the conceptual idea of positive and negative work. I do not see this as being problematic or anything to detest about.

Zz.

• weirdoguy