# Work Conundrum

1. Jul 8, 2013

### kopite1892

Please could someone help me understand why there is no "work" when carrying a heavy object

I was trying to workout the work & power equasions for the strongmen when they carry heavy objects (yoke, farmers carry etc)

But because they carry the object perpendicular to the force work = 0.........

Is there a exception to this or another formular to use?

For example
If a 80kg man carries 200kg for 100m in 120 seconds can we work out joules and watts?

Thanks

2. Jul 9, 2013

### SteamKing

Staff Emeritus
Who said that there is no work done when carrying a heavy object?

3. Jul 9, 2013

### ModusPwnd

This is a common example that is used when the concept of work is first taught. Of course the teacher and text would probably say it more carefully, there is no work done on the object by the man carrying it.

kopite1892, Your question might be better thought of as a question in biology. There is no work done by the man on the object. We intuitively think he is doing work on the object because we associate the definition of work in physics with the notion of work outside of science. But "work" in science has a specific definition and per this definition, no work is done on the object by the person because no component of force is parallel to the distance traveled (you have your person traveling at a constant speed). But inside the man's body his cells are performing work, just not on the object. Muscle cells contract and relax and when they do so they perform work which we feel.

This explains it better than I can,
http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html#nwk

Last edited: Jul 9, 2013
4. Jul 9, 2013

### mayble

Haha, that really is confusing. My take on the question is, how much work does it take for a person to walk and how much work does it take for a heavier person to walk? (Heavier person is the same as someone carrying a heavy object, in terms of the point forces relevant in this problem.)

I don't know anything about the mechanics of human bodies, but I'm pretty sure there's not going to be a cut and dry equation. If I had to approximate the problem, I would think about how walking works: the back foot exerts a diagonal force downwards and backwards. These forces are counteracted by normal force and static friction, respectively. The work to walk one step would be based on the static friction force and the distance of one step: W = (F_sf)x(D_step). Multiply that by the number of steps required to transverse the space in question.

5. Jul 9, 2013

### Lsos

To move a heavy object from place to place, no work needs to be done. The only reason that a human gets tired when doing so is because a human is incredibly inefficient in doing this particular action. All of the energy expended goes into struggling to support himself under the weight. Put the same weight on a wheeled cart, and you'll see that the amount of work needed is dramatically reduced, possibly by many orders of magnitude.

Another example is that of a helicopter. It might require 1000s of horsepower to remain hovering in the air, but all the work is wasted into simply moving air around...none of it actually goes into the helicopter. Replace the helicopter with a table and you'll see that you can accomplish the same thing with 0 horsepower.

6. Jul 10, 2013

### mayble

Lsos -- I'm sorry, I don't follow either paragraph.

Say the heavy object starts at point A. Assume it starts at rest and does not spontaneously disappear from point A and reappear at point B. To get to point B, some force needs to be applied to the object. And if a force is applied over a displaced distance... isn't that the definition of work? It shouldn't matter if it's a person, a box, or a wheel borough -- a wheel changes the amount of work needed, not the need for work at all. Is my understanding of mechanical work completely off?

How can a table accomplish hovering with 0 horsepower?

7. Jul 10, 2013

### jbriggs444

To get the object started on its trip from point A to point B you are correct, you need some initial force. However, once the object is moving, no net force is required for it to continue moving to point B. When the object approaches point B, a final force is required to bring the object to a stop.

The work done by the starting force and the stopping force can be arbitrarily tiny and arbitrarily brief. So the work done by the two can be negligible. In addition, it turns out that the work done by the stopping force is equal and opposite to the work done by the starting force. So the net work done is zero.

All of the forces involved are static. There is no motion. No distance covered. Even though the forces are non-zero, all the relevant distances are zero and no work is done by those forces.

The tabletop does no work on an object resting on its center. The table legs do no work on the tabletop. The floor does no work on the table legs. If the table is on good wheels, it can take negligible work to move table and object from one side of the room to the other.

8. Jul 10, 2013

### PhanthomJay

A car accelerating from rest requires a net external force, which is provided by the static friction force between the road and driving tires. The static friction force however does no work, and yet there is a KE change of the cars center of mass. So wE cannot say that no power is delivered by the car (internally) through its engine. The concept of Work does not sometimes "work" well when objects with linkages or deformations are involved.

9. Jul 10, 2013

### Lsos

The force accelerating the car DOES do work on the car: the same amount of work that the car gains in KE. However, in order to stop the car we need to apply the same exact force x distance as we did to get it moving, but in the opposite direction. The end result is that there is no work on the car. This stopping energy usually gets wasted as heat, but that needn't be the case. Hybrids do a good job of getting it back, for example. Nevertheless, there are inefficiencies. ALL of the gas you burn when driving around is due to these inefficiencies. In theory, and even in practice, we can engineer almost all of these inefficiencies away. It's just usually not worth it.

We can easily make a machine that goes from point A to point B with almost zero energy expenditure. It just gets difficult depending how far away the points are from each other, and whether they move around.

10. Jul 10, 2013

### Staff: Mentor

Not really, since the point of application of the static friction--the contact patch of the tires--does not move (instantaneously). The work done by the static friction is zero. (The ground is not a source of energy. The engine is!)

You can certainly calculate F*Δx, where Δx is the displacement of the center of mass and F is the friction force from the road. And that will certainly equal the change in KE (ignoring other forces, such as air resistance). I suspect that's what you're thinking of.