Work done under constant velocity

[nDever]

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?

 

[Doc Al]

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.)

 

[nDever]

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]

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.

 

[nDever]

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 F_A 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]

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]

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]

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.

 

[Dr. Courtney]

Sounds like a few jobs I've had. Some bureaucrats applying the negative work, and employees applying the positive.

 

[enter]

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

 

[Dale]

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.

 

[vela]

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.

 

[nDever]

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]

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

 

[Doc Al]

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]

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]

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.

 

[OldYat47]

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.

 

[Dale]

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]

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.

 

[OldYat47]

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]

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]

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.

 

[Dale]

Pushing the box westward doesn't subtract from the work done pushing it eastward.

Obviously not, the work has the same sign in both directions. I think this may indicate that you have some confusion about the concept since this example does not involve any negative work at all.

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.

This is a better example. Here, the work you do lowering the weight is in fact negative. You can build a machine (a spring) that can lift and lower a weight without using energy precisely because the machine does positive work on the way up and negative work on the way down.

Take an example of box with some initial velocity sliding up and down a ramp with a small amount of friction. The work done by gravity is negative on the way up and positive on the way down, but the work done by friction is negative both ways. The total work done on the box is the sum.

 

[OldYat47]

Refer to the original post, confusing the direction the box is being pushed as positive in one direction, negative in the other. That's true for the vector directions as set up in the original post, but it's not negative work.
A spring is a source of energy. And in your example the spring is always pulling up with gravity always pulling down, right?

 

[Dale]

Refer to the original post, confusing the direction the box is being pushed as positive in one direction, negative in the other. That's true for the vector directions as set up in the original post, but it's not negative work.

The original post did have both positive work and negative work and the next post correctly identified each and came to the correct conclusion that the positive work and the negative work canceled each other for no net work. There was no confusion in the OP or the immediate responses

in your example the spring is always pulling up with gravity always pulling down, right?

Yes.

 

[OldYat47]

Remember that scalar quantities are directionless, so summing them up as if they were vectors is an incorrect concept.

In the original post it takes some force to get the block moving, some work is done. It takes some force to stop the block and get it moving in the other direction so some additional work is done, etc. through the various cycles. If we take one cycle the result is that the block is in the same place it started, but it took work on both ends to get it there. If you had to calculate how many calories the pushers on both ends burned that's not zero. The net displacement of the block may be zero but that's different from the amount of work it took to get it there. Otherwise you are describing a perpetual motion machine.

Similarly the spring is doing work as the weight oscillates up and down regardless of the instantaneous position of the weight.

 

[Dale]

Remember that scalar quantities are directionless, so summing them up as if they were vectors is an incorrect concept.

Nonsense. Addition is a perfectly valid scalar operation. In fact it was a valid scalar operation long before vectors were even invented. A scalar doesn’t need a direction to have a sign.

In the original post it takes some force to get the block moving, some work is done. It takes some force to stop the block and get it moving in the other direction so some additional work is done, etc. through the various cycles.

In the original post there is no cycle. The block begins at a velocity v and continues at v. It does not start nor stop, and there is no cycle. There is just a steady amount of positive work being done by one force and negative work being done by the other force.

Similarly the spring is doing work as the weight oscillates up and down regardless of the instantaneous position of the weight.

The spring is doing positive work on the mass while the velocity is up and negative while the velocity is down. Remember ##P=\mathbf f \cdot \mathbf v##. The force from the spring is always up, so when v is up the dot product is positive (positive work) and when v is down the dot product is negative (negative work)

@OldYat47 you clearly are confused on this material. Please do not continue posting this idiosyncratic approach. It is certainly not helpful for the OP.

 

[A.T.]

Remember that scalar quantities are directionless, so summing them up as if they were vectors is an incorrect concept.

Then try to convince your bank that they should stop counting the withdrawals as negative for your account's balance.

 

[OldYat47]

OK, I've spent some time stripping this to the simplest possible analysis, and this is my last post, promise.
"Work can be either positive or negative: if the force has a component in the same direction as the displacement of the object, the force is doing positive work. If the force has a component in the direction opposite to the displacement, the force does negative work."
Just like your spring.

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0. Suppose the observation stops when the block is at its original location. Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

So the sum of the work is (2 X work), not zero.

 

[jbriggs444]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

That is not the scenario under consideration. The situation under consideration (as described in the original post) contemplates two equal and opposite forces applied at the same time while the mass is moving. One is applied in the direction of motion. Since the other is opposite, it is clearly applied opposite to the direction of motion.

It is a mathematical fact that the work done by equal and opposite forces on the same object at the same time will be equal and opposite.

[For point-like objects anyway. There are caveats for non-rigid or rotating extended objects]

 

[Doc Al]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction.

OK.

The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

Not going to happen. Realize that even though the second push is in the -x direction, the block is still moving in the +x direction until it is brought to a stop.

Suppose the observation stops when the block is at its original location. Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

You need to rethink this.

 

[Dale]

Let's assume that one person pushes the block to some velocity "v" in the "+x" direction. The second person somehow pushes the block in the "-x" direction to some velocity "-v" without first bringing the box to v = 0.

First, this is impossible. If the v is in the +x direction and a force is applied in the -x direction to bring it to -v then the box must be brought to v=0 first.

It is possible to apply a force to bring it from v to -v without going to v=0, but in that case the force cannot be strictly in the -x direction. Instead, the force must turn the object, typically by always acting perpendicular to the velocity.

Positive work is done at both ends since the force at both ends is in the same direction as the displacement of the block.

This is not true in general.

Take the case of the turn. In that case the force is always perpendicular to the velocity so the dot product is 0 and no work is done at all.

Now, take the case of the straight line force. From when the velocity is v until it is 0 the force is in the opposite direction of the velocity, so the dot product is negative and negative work is done. Then from when the velocity is 0 until it is -v the force is in the same direction as the velocity, so the dot product is positive and positive work is done.

In all cases, the net work at the second end is 0. (Assuming no friction etc)

So the sum of the work is (2 X work), not zero.

No, in your notation the sum of the work is (1 x work).