# Definition of work done

1. Oct 27, 2014

### Deep_Thinker97

In my maths text book, it says that work done can be defined as Force x Distance moved in direction of force, AND change in kinetic energy. I feel both these definitions can be contradictory
Example:
A box moves at a constant velocity along a rough horizontal plane. It has a driving force of 5N and moves 3m. What is the work done against friction?
Well the frictional force is 5N since the box is at a constant velocity. Therefore, work done (against friction)=5N x 3m= 15Nm or 15J (this was the actual example in the book)
But, work done is also defined as the change in kinetic energy. This cannot be applied to this example as the box is travelling at a constant velocity so there is no change in kinetic energy.
I understand that there are some examples where work done does equal change in kinetic energy, but I don't understand what conditions must apply for this to be true (or not true)
How can the definition of work done be one thing under one circumstance and something different in another circumstance?
What is the actual, universal, definition of work done that is right in all circumstances?

2. Oct 27, 2014

### Staff: Mentor

The net work done by all forces acting on an object, equals the change in the object's KE.

In your example of the box sliding on a rough surface against friction, at constant velocity, the driving force does 15 J of work. The frictional force acting on the box does -15 J of work. The net work done on the box is 15 - 15 = 0 J.

3. Oct 27, 2014

### Staff: Mentor

That's correct and is the definition of work.

That's not quite correct. The net work on an object will equal the change in its kinetic energy. In your example, there is no net work done.

There are a few subtleties here, but that's the main idea.

4. Oct 27, 2014

### Andy Resnick

Just to add a bit: the "work-energy theorem" F⋅d = Δ(1/2 mv2) is a re-statement of F= ma (substituting one of the kinematics equations for 'a'). Thus, while it is a correct dynamical statement about forces, it is not a correct energy statement- and as you note, friction can't be simply plugged in. Neither can the normal force. Using the phrase "work done against friction" helps remind us of that: we couldn't say 'work done by friction'. Similarly, considering a person jumping vertically, so that their center of mass moves while still in contact with the ground, shows that the normal force can't do work.

Some people use the term 'pseudowork' when discussing friction and normal forces for this reason.

5. Oct 27, 2014

### Staff: Mentor

Exactly.

These are the precise 'subtleties' that I had in mind in my last post. :) (And I have written at length about pseudowork many times here.)