# Having trouble understanding/accepting the definition of work. Thought exercise

1. Aug 7, 2012

One concept in physics that has never set well with me is the way work and energy are defined.

According to all the physics sources I've looked at, work is defined as:

$$W = \vec{F} \cdot \vec{d}$$

(for a constant force over a distance)

However, intuitively the notion of taking the dot product of F with the displacement doesn't seem quite right to me. I am well familiar with the properties and interpretation of the dot product, but I still feel like something is missing.

Consider the following thought exercise:

Imagine we have two separate objects in space, object A and object B. Object A is at rest and object B is moving at a constant velocity through space. In other words, there is no net force acting on either object.

Suppose that we now apply a force $F_1$ to object A and a separate force $F_2$ to object B for an equal amount of time and that $F_1 = F_2$ and that both forces point in the same direction as the velocity of object B does. Object A will cover a distance $d_A$ and object B will cover a distance $d_B$ in this time interval.

Because A started at rest it will cover a smaller distance over the time interval than object B will cover. Therefore according to the definition of work the work done on object B over the time interval will be greater than the work done on object A, even though equal force was applied to both in equal time.

This in turn implies that the amount of kinetic energy applied to A to cause the resulting change in momentum would not be sufficient to apply the same change in momentum to object B, even though we already know that the same amount of force over time was applied to achieve it. This seems like a contradiction to me.

It seems to me that a more intuitively accurate formula for work might be something like this:

$$W = \vec{F} \cdot \frac{\vec{d}}{ \lVert \vec{d} \rVert}\, t$$

(where t is the amount of time over which the force is applied to the object)

Notice that this alternate formula gives the same value for both objects A and B, whereas the standard formula does not.

It seems like you can't really cheat by just changing frames of reference, because in this example we're trying to calculate energy totals for our current frame of reference as the observer. The standard formula seems to imply that more kinetic energy must be put into changing the momentum of already moving objects than of objects at rest, which seems to violate the basic laws of physics.

Furthermore, I can't help but notice that the energy laws, such as conservation of energy are very similar to the laws of conservation of momentum. Perhaps they are in fact one and the same when proper analysis is applied. Perhaps change in momentum in a direction (as given by my alternate formulation of work above) is in fact the real unit of mechanical work. Granted, one would need to reformulate other equations in physics that depended on it because of the change, but maybe it could in fact work.

Energy is basically supposed to measure a systems "ability" to enact a change in momentum right (assuming we're considering only mechanical energy), and isn't it true that if a particle collides with a particle at rest then it transfers momentum into the other, and by so doing enacts a change in momentum? Thus doesn't this further support the argument that energy may in fact be a measure of momentum in disguise (in the purely mechanical cases at least).

Can you disprove my reasoning and clarify what work actually is and why the standard definition is what it is? Or perhaps, could my alternate definition be more correct somehow?

All the sources I've found on work and energy don't really discuss why it's defined to be the way it is; they instead just repeatedly reference other definitions and terms without really linking any of those terms to any real reason. How were these things originally thought up and why are the formulas what they are?

2. Aug 7, 2012

### Muphrid

Re: Having trouble understanding/accepting the definition of work. Thought exercise..

It seems to me you want to look at impulse instead of work. You're talking about a situation where a force acts between two objects for a constant time, and in that case, the impulse--the change in momentum imparted--is relatively simple to calculate:

$$\Delta p = F \Delta t$$
Saying that the two objects have a different amount of work done on them gets a, "So what?" response from me. You want something to be the same between the two of them, impulse is it. Your situation of a force that goes on and off based on time is problematic to analyze because these two particles would have potential energy based on their positions that just goes poof when the force is turned off.

From a Newtonian mechanics perspective, energy is just passed around between objects, either stored as potential energy due to outside forces or converted into kinetic energy. You can say that energy is conserved because there is a symmetry of the laws of physics with respect to time. Any symmetry leads to a corresponding conserved quantity, and time symmetry leads to energy.

To understand why time and energy are related in this way, I suggest you read up on the quantity called action. The action has the units of Energy x Time, Distance x Momentum, or Angle x Angular Momentum, and the symmetries of the laws of physics with respect to shifts in time, translations, and rotations (time, distance, and angle) generate the symmetries that conserve energy, momentum, and angular momentum. You could say that action is the most fundamental quantity, and we just had to give names for "action / distance (that is, momentum)" and so on because the conservation laws made them useful. Most higher level physics doesn't deal directly with Newton's laws anymore, only the equivalent equations of motion based on an action principle (even for classical physics).

The work done on a particle by a force is just the amount of energy converted from potential energy of the particle to kinetic energy.

3. Aug 7, 2012

### tiny-tim

Force is rate of increase of momentum.

Force is not rate of increase of energy.

4. Aug 8, 2012

### Staff: Mentor

Re: Having trouble understanding/accepting the definition of work. Thought exercise..

You can test for yourself what impact speed has on work using a bike and a hill. Ride a bike partway up a hill for a certain amount of time. Then do it again for the same amount of time, but faster. See if you are more tired the second time.

Last edited: Aug 8, 2012
5. Aug 9, 2012

### Lsos

Re: Having trouble understanding/accepting the definition of work. Thought exercise..

Same force does not mean same energy. You can see this in everyday examples.

Accelerating a car from 60-120 is at least 3x harder than accelerating from 0-60, and not just because of air resistance, but because W=Fd. Even with the same force, you will still cover 3x the distance due to the higher initial speed. Indeed you will also burn 3x more fuel. What actually happens in a real car is that the force is NOT the same at higher speeds, because you have to be in a higher gear. You had to trade some of the force in into distance, and so your force (and acceleration) will drop....but the engine ends up working just as hard.

Likewise, stopping a car from 120 will heat up the brakes 4x more than stopping from 60. Interestingly, you also need to jump from the 4th floor to hit the ground 2x as fast as jumping from the 1st floor (ground floor being 0).

But you were talking about objects in space. F=wd still applies in space, but it's a bit more complicated because you are probably using rockets, and a rocket takes it's mass with it. So, while it might be just as easy to accelerate a rocket moving at 10 as it is to accelerate an identical rocket moving at 20, remember that the faster moving rocket had to have something bringing it up to its speed in the first place. After all, they both started on earth from 0. which means that when standing on the launchpad, the faster rocket still needs to be much larger (4x, although probably more, I don't understand the rocket equation enough) than the slower one.

You then might be tempted to ignore the fact that both rockets started at 0, and just look at their starting speeds as being "0". But then you'd be looking at two different frames of reference, and work is indeed different depending on which frame of reference you're looking at. That's essentially the problem with rockets, they need to bring their frame of reference with them wherever they go, and that gets heavy.

6. Aug 9, 2012