# I Question about the physics term "work"

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1. Dec 27, 2017

### PeterO

I did look at the definition, and it confirms the answer is a scalar - therefore has no direction (and that's where the positive and negative comes in) Even if the "mathematical" calculation has a negative (or positive) sign on the numerical answer, the sign is ignored for a scalar is it not.

2. Dec 27, 2017

### A.T.

Look at the definition of a scalar.

3. Dec 27, 2017

### PeterO

OK I will re-word that.
Can you give an example of a scalar with anything other than magnitude?

4. Dec 27, 2017

### A.T.

Too vague. Do you mean a sign?

5. Dec 27, 2017

### Staff: Mentor

What if the vectors in the dot product are pointing in opposite directions?

6. Dec 27, 2017

### PeterO

The mathematics (arithmetic really) will have a negative sign, but when the answer is interpreted for to the situation (a scalar quantity) that sign is ignored.

Take the case of a mass moving in a circle. On opposite sides of the circle it is moving in opposite directions, but it does not have positive Kinetic energy on one side, and negative on the other. It just has kinetic energy; neither positive nor negative (because kinetic energy, like Work, is a scalar).

7. Dec 27, 2017

### A.T.

No, the sign is interpreted as the direction of energy transfer. When A is doing negative work on B, then energy is transferred from B to A.

8. Dec 27, 2017

### Staff: Mentor

The sign is not ignored. If A is doing negative work on B (i.e., with the sign included), it just means the B is doing positive work on A.

In thermodynamics, we encounter negative work all the time when we write $W=\int{PdV}$, where W is the work done by the system on the surroundings. If the gas is expanding, we have that the system is doing positive work on the surroundings, which, from the first law of thermodynamics, $\Delta U=Q-W$, means that the positive work that the system does is tending to decrease its internal energy U (and temperature). If the gas is being compressed by the surroundings, we have that the system is doing negative work on the surroundings, which, from the first law of thermodynamics, means that the negative work that the system does on the surroundings (i.e., the surroundings are doing positive work on the system) is tending to increase its internal energy (and temperature).

Just try doing thermo without the concept of work being allowed to be negative, and see how confused you will get.

9. Dec 27, 2017

### PeterO

I agree the negative sign that the arithmetic gives you can be used to interpret some details of the energy changes, but cannot agree that there is negative work. Work is a scalar quantity, and the idea of positive and negative are simply not relevant.

10. Dec 27, 2017

### Staff: Mentor

Then how do you account for our approach in thermodynamics, where it is extremely relevant (and automatically gives you the correct answer)?

11. Dec 27, 2017

### PeterO

The sign indicates what sort of change you will be having (increase or decrease), and the (unsigned) Work tells you by how much, but the work itself is still a scalar.
Imagining the work as positive and negative makes for simplicity in keeping track of energy in thermodynamics, but work itself is still a scalar.

12. Dec 27, 2017

### Staff: Mentor

Yes, work is a scalar, but you were saying that it can't have a negative sign. It obviously can, since the dot product of force and displacement can be negative. I guess we are just going to just have to agree to disagree.

13. Dec 27, 2017

### Staff: Mentor

So, when I have a body in motion, and I write $$\Delta (KE)=W$$the work is always positive and the change in kinetic energy is always positive, even when I am exerting a force to slow down the body?

Last edited: Dec 27, 2017
14. Dec 27, 2017

### A.T.

The sign of work is very relevant. Scalars can be negative or positive.

15. Dec 27, 2017

### weirdoguy

For example work done by kinetic friction is always negative.

16. Dec 27, 2017

### A.T.

Not really. The sign depends for which object the work is computed and in which frame of reference.

17. Dec 27, 2017

### Staff: Mentor

So when a body is being dragged forward by a moving boundary in the direction it is already moving, the work that the moving boundary is doing on the body is negative?

18. Dec 27, 2017

### jbriggs444

One can salvage the truth of this claim by considering the kinetic friction between two objects, A and B that are sliding past one another. The sum of the work done by A on B plus the work done by B on A will be negative.

19. Dec 27, 2017

### Mister T

A scalar can be positive or negative.

Perhaps you are confusing vector magnitudes with vector components. The magnitude of a vector is never negative, but the component of a vector can be either negative or positive. In the expression for kinetic energy, $\frac{1}{2}mv^2$, $v$ is a vector magnitude, the speed, and as such can never be negative. In an equation such as $v=v_o+at$, $v$ is actually a vector component and as such can either be negative or positive. This last bit is a source of confusion for students, which is why a few authors have taken to writing equations like that as $v_x=v_{ox}+a_xt$ to make that distinction clear.

20. Dec 28, 2017

### gleem

No. Only 2500J of work was done. Idealizing this situation, the person pushing the dresser is the source of the force and is fixed to the earth during the pushing. Consider the pushing to be a series of pushes in place, feet fixed arms acting as a spring for each push with steps taken by the person between pushes, ignoring the energy used to take the steps. The person does work on the dresser as we all agree. The reaction force of the dresser on the person does no work because during the pushing the person does not move. (ideally) because he is firmly connected to the immovable earth If the person is loosely coupled to the earth as when say standing on a slippery surface then when he pushes he will move backwards as the dresser moves forward. The energy needed to do the work is divided unevenly between the person and the dresser if the dresser moves more forward than the person moves backwards.

Regarding the sign of work I have always interpreted it when negative as the work done on an object in opposition to its motion as the brakes on an auto or friction on a sliding object. The force direction and displacement are in opposite directions.

21. Dec 28, 2017

### jbriggs444

If you are treating the arms as springs and ignoring the legs then the energy needed to do the work comes 100% from the arms.

22. Dec 28, 2017

### gleem

You can couple the legs and arms through the trunk if you want and consider the person as a spring. Real situations are difficult to model exactly by the point was to look at it simplistically to see the basic physics in play.