# Impulse without doing work?

physicsforum7
Hi.

I have one general question and one specific question (which is related by example.)

Specific: If I am sitting in a chair with wheels and I push off against a wall, I accelerate. But the force was not applied through any distance. Did I do any work? (I know that an impulse was applied, for I pushed off against the wall.) If so, how?

General: Is it possible for work to be done without applying an impulse, or vice-versa?

(Note: I'm pretty confident that work cannot be done without an impulse; for I do not know of any way in which an object can travel through a distance in no time. But the "specific" question makes me ask whether the converse is true.)

Thank you.

Mentor
Specific: If I am sitting in a chair with wheels and I push off against a wall, I accelerate. But the force was not applied through any distance.
That is not correct. You apply the force by moving your limbs. If you push off with your legs, you may apply a force through a distance of up to a meter. So:
Did I do any work? (I know that an impulse was applied, for I pushed off against the wall.) If so, how?
Yes. You applied the force through a distance, hence: work.
General: Is it possible for work to be done without applying an impulse....
No.
...or vice-versa?
Yes.

Naty1
An impulse is is force applied over a time period.

FT = MV and W = Fd.

In classical mechanics, impulse (abbreviated I or J) is defined as the integral of a force with respect to time.

http://en.wikipedia.org/wiki/Impulse_(physics [Broken])

edit: I see Russ posted while I was composing...and we are consistent

Last edited by a moderator:
Mentor
Just to be clear, even though the wall exerts a force on you, it does no work on you. (There is no displacement at the point of application of the force.)

Mentor
Just to be clear, even though the wall exerts a force on you, it does no work on you. (There is no displacement at the point of application of the force.)
I didn't think of that: you may be right that that's the point of confusion.

salzrah
While you are in contact with the wall, it is doing work on you because if the wall applies an impulse on you for a given time interval T, then during T you were also being pushed back continuously through a distance, however small. But after you let go, no work is done on you.
I think an impulse cannot exist without work being done because if a force is applied in a time interval (and the object is movable) then there has to be a distance associated with that time if the object's momentum changes.

Mentor
While you are in contact with the wall, it is doing work on you because if the wall applies an impulse on you for a given time interval T, then during T you were also being pushed back continuously through a distance, however small. But after you let go, no work is done on you.
No, the wall doesn't move and thus does no mechanical work on you. It is not a source of energy. All the energy that ends up as your kinetic energy comes from within your body.

The wall certainly exerts a force and an impulse on you, but it does no work.

Homework Helper
There are several ways of looking at the situation.

You can close your eyes and pretend that both you and the wall are rigid bodies and that there is an abstract force that pushes you off.

Under that interpretation, your body moves and the abstract force does work given by the magnitude of the force multiplied by the displacement of your center of gravity.

Or you can open your eyes and look at a particular force acting across a particular interface.

Under this interpretation, the point of contact does not move and the force does no work.

Or you can compromise and model the situation as a rigid wall, your rigid body and the non-rigid arms that you use to push off the wall. Then you could discuss the work done at the hand-wall interface, the work done at the body-arm interface and the energy supplied within the non-rigid arm.

Mentor
There are several ways of looking at the situation.

You can close your eyes and pretend that both you and the wall are rigid bodies and that there is an abstract force that pushes you off.

Under that interpretation, your body moves and the abstract force does work given by the magnitude of the force multiplied by the displacement of your center of gravity.
It's perfectly legitimate and very useful to multiply the net force times the displacement of the center of mass. (No need to close your eyes to apply Newton's laws!) That quantity, though it looks like work, is not really the work done by the wall force since Δxcm is not the displacement through which that force moves. (Some call that quantity pseudo-work; some do call it 'center of mass work'.)

Or you can open your eyes and look at a particular force acting across a particular interface.

Under this interpretation, the point of contact does not move and the force does no work.
Sure.

Or you can compromise and model the situation as a rigid wall, your rigid body and the non-rigid arms that you use to push off the wall. Then you could discuss the work done at the hand-wall interface, the work done at the body-arm interface and the energy supplied within the non-rigid arm.
It's all good.

All three of these views are completely legit.

salzrah
My interpretation of what Doc Al said is that if the force doesn't move through some displacement itself, then no work is done. This seems like a reasonable conclusion. I was under the impression that the displacement quantity in W=Fd, was for the object, but it is actually for the force. Great explanation Al!

Gold Member
To be really clear, forces don't move, objects do, so the "d" in Fd is always the distance an object moves, never the distance a "force moves". However, what you choose to regard as "the object" is where the confusion comes in-- the wall doesn't push on your center of mass (not being a gravity force), it pushes on the base of your feet. The base of your feet is then the "object" the wall is pushing on, and that object does not move while there is contact with the wall, and that is why the wall does no work. Your feet push on the rest of you, and the rest of you does move, so your feet do work on the rest of you (which comes ultimately from energy in your muscles as has been pointed out).

Note we could also analyze the work done by the gravity of the wall, even though it is of course extremely tiny. The wall does do gravitational work on you (which is negative), because the wall exerts that force (in effect) on your center of mass, which does move. So even though the wall does not move, it can still do work on you, if it is exerting a force on some part of you that moves.

Mentor
My interpretation of what Doc Al said is that if the force doesn't move through some displacement itself, then no work is done. This seems like a reasonable conclusion. I was under the impression that the displacement quantity in W=Fd, was for the object, but it is actually for the force. Great explanation Al!
Yes, but keep in mind that the force can be analyzed at yout feet or at your hip. Your feet don't move when you push off a wall, but your hips do and the force is the same. I'd model this as a mass with a spring pushing it away from the wall. At the point of contact with the wall, no work is being done. At the connection to the mass, work is being done.

Mentor
To be really clear, forces don't move, objects do, so the "d" in Fd is always the distance an object moves, never the distance a "force moves".
Right. That's why I usually specify the displacement of the point of contact. (But sometimes I get lazy.)

Mentor
Yes, but keep in mind that the force can be analyzed at yout feet or at your hip. Your feet don't move when you push off a wall, but your hips do and the force is the same. I'd model this as a mass with a spring pushing it away from the wall. At the point of contact with the wall, no work is being done. At the connection to the mass, work is being done.
Right. My point, which I thought was part of the OP's question, was that the wall does no work on you and is not a source of energy. All work done is internal to your body. (I think we're saying the same thing.)

Gold Member
Yes, if the question has not yet been completely answered, then any remaining difficulty must lie in a confusion between the work done by the wall, and the work that is associated with the impulse on the person. In the simplest terms, impulse is F*t, work is F*d, so the ratio of the work to the impulse is characterized by d/t, the velocity the object acquires. So in that sense, there cannot be impulse without work, unless the object acquires no motion (say, if the object were infinitely massive). However, it is a separate issue to determine by what cause the work is done.

Homework Helper
Yes, if the question has not yet been completely answered, then any remaining difficulty must lie in a confusion between the work done by the wall, and the work that is associated with the impulse on the person. In the simplest terms, impulse is F*t, work is F*d, so the ratio of the work to the impulse is characterized by d/t, the velocity the object acquires. So in that sense, there cannot be impulse without work, unless the object acquires no motion (say, if the object were infinitely massive). However, it is a separate issue to determine by what cause the work is done.

Just to quibble, d/t is not the velocity the object acquires. It is (for an appropriately symmetric force profile) the average of the velocity the object had previously and the velocity the object acquired after.

Darwin123
Just to be clear, even though the wall exerts a force on you, it does no work on you. (There is no displacement at the point of application of the force.)
The corporate system of "you" is an extended body. The boundary of "you" does not enclose only one point. It includes all your atoms. The spatial position of "you" is your center of mass, not the position of your feet. The wall exerts an external force on "you".
By the work-energy theorem, the kinetic energy of displacement of a corporate body is determined by the velocity of the center of mass. The internal forces within the body cancel out due to Newton's third law. Therefore, the state of any atom of your body doesn't count except that it affects the center of mass.
When you push off against the wall, the position of your center of mass is changing continuously. The force of the wall is indirectly acting on the center of mass, even though it isn't in contact with the center of mass. So the displacement of your center of mass times the force on on your feet is the work done on "you", the corporate body.
It would be fair to say no work is done on the atoms of the soles of your feet while pushing off from the wall. However, "you" are not the soles of your feet.
Boundaries, boundaries, boundaries. Physics theorems always have boundaries. The tricky part of your problem is recognizing that the system of "you" is the surface of your body, not the atoms upon which the external force that is directly acting. A problem in English rather than physics, really.

Mentor
The corporate system of "you" is an extended body. The boundary of "you" does not enclose only one point. It includes all your atoms. The spatial position of "you" is your center of mass, not the position of your feet. The wall exerts an external force on "you".
By the work-energy theorem, the kinetic energy of displacement of a corporate body is determined by the velocity of the center of mass. The internal forces within the body cancel out due to Newton's third law. Therefore, the state of any atom of your body doesn't count except that it affects the center of mass.
When you push off against the wall, the position of your center of mass is changing continuously. The force of the wall is indirectly acting on the center of mass, even though it isn't in contact with the center of mass. So the displacement of your center of mass times the force on on your feet is the work done on "you", the corporate body.
It would be fair to say no work is done on the atoms of the soles of your feet while pushing off from the wall. However, "you" are not the soles of your feet.
Boundaries, boundaries, boundaries. Physics theorems always have boundaries. The tricky part of your problem is recognizing that the system of "you" is the surface of your body, not the atoms upon which the external force that is directly acting. A problem in English rather than physics, really.
It is somewhat a matter of semantics, but there is some important physics here.

If you want to know the work done by the force of the wall, you must consider that that force is pushing on your feet, which do not move. No work is done by that force. But you are correct that that external force does affect the motion of the center of mass.

As I said before, you can certainly apply Newton's laws and compute Fnet*Δxcm, which does look like a 'work' term. And it will give you the translational KE of the center of mass. But the the wall does not put energy into the system. All that energy is provided by your body, via your muscles.

Gold Member
Yes, it's all the difference between work done "on" vs. work done "by." The OP is answered by saying that the impulse the person receives is indeed associated with work done "on" them, and even though the impulse does come from the wall, the work does not. Thus we can say there can be no change in velocity without impulse, and there can be no change in velocity without work either, but there can be a source of the impulse that is different from the source of the work that the impulse is paired with. In the OP, the source of the impulse is the wall (no wall, no motion), but the wall is not where the work comes from, and that is indeed quite subtle.

BERNIE649
The arm is like a coiled spring. As the spring extends it applies a force F to the body for a distance d and a time t. W= F x t. Impulse J = F x t. The wall does nothing.

#### Attachments

• Scan.jpg
9.4 KB · Views: 344
Gold Member
The arm is like a coiled spring. As the spring extends it applies a force F to the body for a distance d and a time t. W= F x t. Impulse J = F x t. The wall does nothing.
Well, I don't think you quite mean that, because an uncoiling spring that is not against a wall does not yield any net impulse on the spring's center of mass, the spring merely oscillates. That doesn't speak to the OP question. I like the spring analogy-- but it does not say the wall does nothing, it says the wall takes the kinetic energy of oscillation and turns it into kinetic energy of translation, that is, it provides the impulse but not the work. What we can say is that if an object's center of mass goes from stationary to moving, there is both an impulse and work (in a ratio whose scale is characterized by the velocity achieved), but they need not arise from the same source.

Last edited:
Aniket1
In this situation there is not just mechanical work being done but also chemical work (inside the body). When you apply the work energy theorem, you need to take into consideration the total work done and not just the mechanical work. So in this case:
Work done inside the body=Energy gained by you. The work done by the wall is indeed zero in this case..
A similar situation arises in an automobile. Work done by friction (of the road) is zero since there is no displacement in case of static friction. Here, again, we have chemical work by the combustion of fuel which imparts energy. Ofcourse, friction is necessary, but it does not provide any energy to the system. Energy comes from an external agent (fuel in this case)
In short, when your system is not purely mechanical, W=F.ds is not the only work done. You have to consider the work done by other agents.
One more similar example is climbing stairs. You gain potential energy mgh without doing any mechanical work :) Here again, work is done inside the body (chemical work)