We consider the case of a ball thrown vertically into the air. We know from Newton's Second Law that the change in the "motion" of the ball, where we use Galileo's description of the "motion" as the velocity, is given by a = F/m = (mg)/m = g, pointed in the vertically down direction. Therefore, the equation for motion for the connection between velocity and position is
If we divide both sides of the equation by 2 and multiply by the mass of the ball, we get
We can make some comments about this expression. First, note that the expression above works perfectly well even if the motion is not constrained to one dimension! Imagine that we threw the ball so that there was some initial horizontal velocity in addition to the original vertical velocity. Now the initial velocity squared would be v02 = v0x2 + v0y2. Since there is no acceleration in the horizontal direction for the free-fall trajectory of the ball, the final velocity for any particular vertical position y, will be v2 = v0x2 + vy2 so that
| v2 - v02 | = [vx2 + vy2] - [v0x2 + v0y2] |
| = vy2 - v0y2 | |
| v2 - v02 | = -Fgrav.(y - y0) |
Thus, the change in the velocity squared in any number of dimensions is just equal to the change in the motion caused by the gravitational force acting along the vertical direction. Physicists have generalized this notion with the concept of the work done by a force where the work, W, is defined as
where s is the path of the object and the dot indicates that we are making a dot product. The dot product of two vectors is the product of the magnitudes of the vectors times the cosine of the angle between them.
Thus, the work is a way of expressing the change in the motion of an object through the action of a force through a displacement. If you think about it, this notion of work will appear counterintuitive since it says that if the object does not undergo a displacement which has some component parallel to the force or vice versa, then the work done (and hence the change in the motion) attributable to that force is zero. It's not as counter-intuitive as it sounds however. Consider the example of an object of mass m sliding without friction across a horizontal tabletop:
In this case, the force of gravity does no work on the object. We could say that's because the normal force is also doing work and that that work opposes that done by gravity, but the definition of work is based on Newton's Second Law and its connection to the equations for motion. When we want the work done by gravity, we talk only of the action of gravity. Since there is no displacement along the vertical direction, gravity does nothing to change the motion of the object, hence gravity does no work in this case. Neither does the normal force. In fact, one important point to take away from this consideration is that the normal force never does any work on an object since, by definition, the normal force goes to zero if there's any component of the motion that is parallel to the direction of the normal, i.e. to move along the normal direction is to lose contact with the surface that creates the normal force in the first place.
Since we make use of our original equation, W = F · s = ½ mv2 - ½ mv02 so often, we give the quantity ½ mv2 a name. We call it the kinetic energy, usually denoted by the letter K, and the expression W = Kf - K0 is called the work-energy theorem.
Although the work-energy theorem definition may seem like a lot of formality, the fact that both W and K are scalars and not vectors makes an enormous difference in our ability to solve problems, as you'll see with the examples below.
Solution: Even though we could do this problem with the equations for motion, it's instructive to do it with the work-energy approach. Begin by noting that gravity is the only force doing work on the block, so the change in kinetic energy, by the work energy theorem is
| Wgrav. | = Kf - K0 |
| mg · s | = ½ mv2 - ½ mv02 |
| g*sin(q)*(h/sin(q)) | = ½ v2 - 0 ==> |
| v | = sqrt[2gh] |
As stated, we could have gotten this result by the equations for motion since the acceleration is constant for the entire trip. However, this result is valid even if the acceleration is not constant. Consider the case below.
Solution: Nothing changes for the result! The work-energy theorem says that the work done by gravity cares only about the change in the vertical position from initial to final position of the mass. Since these positions are the same (at least along the vertical) as before, the result stays the same for the final velocity. This problem would be impossible to do by equations for motion (the acceleration along the curvy incline is not constant) and practically impossible by Newton's Second Law (we'd have to know the exact shape of the incline and integrate along small pieces of the path to find the speed after every short piece).