Question about the physics term "work"

[PeterO]

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

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.

 

[Chestermiller]

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.

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.

 

[Chestermiller]

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.

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?

 

[A.T.]

Work is a scalar quantity, and the idea of positive and negative are simply not relevant.

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

 

[weirdoguy]

For example work done by kinetic friction is always negative.

 

[A.T.]

For example work done by kinetic friction is always negative.

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

 

[Chestermiller]

For example work done by kinetic friction is always negative.

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?

 

[jbriggs444]

For example work done by kinetic friction is always negative.

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.

 

[Mister T]

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.

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.

 

[gleem]

So I used 5000 joules of energy?

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.

 

[jbriggs444]

arms acting as a spring

ignoring the energy used to take the steps

The energy needed to do the work is divided unevenly between the person and the dresser

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.

 

[gleem]

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.

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.