Work and kinetic energy

The work of a curious fellow

All work and no play...

Up to now we have been using Newton's laws of motion to predict the future of dynamical systems. Sir Isaac had some other ideas about dynamics that have proven to be useful. In this section of the program we will explore a few of them.

Let's begin with the notion of "work". Work is defined as force multiplied by displacement. Remember in the general case, force and displacement are vectors. Work is a scalar quantity so the multiplication operation we are talking about here is the dot product as discussed back in Vector Arithmetic. If the fellow in the picture at the right was actually holding up his end of that bit of pipe while the machine held up the other, neither of them would be doing any work since the force (upward) is perpendicular to the motion (horizontal).

Consider a free particle of mass, m, at rest in our reference frame. Next suppose that we apply a force in the x direction to the particle and that the magnitude of that force is a function of x, f=f(x). Since our particle was at rest before we applied this force to it, all the other forces on the particle, if any, balance each other out and the applied force is the net or "resultant" force. Now let's consider a tiny time interval Dt short enough so that the force is nearly constant during the interval. The average acceleration of the particle over the interval dt is a = (v - v0) / Dt, where v0 is the initial velocity and v is the final velocity. The distance x traveled by the particle during time interval Dt is the average velocity times the time interval, like this, x = (v + v0) / 2 * Dt.

The work, w, done on the particle is the force f times the distance x. The force f though, from Newton's second law is m*a so the work is w = m * a * x or, w = m * (v - v0) / Dt * (v + v0) / 2 * Dt. Simplifying the preceding mess, w = 1/2 * m * v2 - 1/2 * m * v02.

no working here
football having been 

worked on devilish hard Now here is another of those insights that seem sort of magical. In the expression above for work, w is the difference between two terms that have identical form. The only difference in the terms is that v appears in one and v0 appears in the other. So it looks like the work done on a particle, which is free to move, during a time interval Dt, is equal to the change in the quantity 1/2*m*v2 over that same interval. If we choose our reference frame so that the initial velocity is zero, the work done on a free particle shows up as velocity and the amount of work is equal to the quantity 1/2*m*v2. This quantity evidently has some close relationship to work.

Suppose now we apply to the particle set in motion in the preceding paragraph, a force in the opposite direction. During the application of this force, the particle will experience an opposite acceleration and slow down eventually to a stop. In this case the work, w, will be negative since the final velocity is less than the initial velocity. The magnitude however will be exactly the same as was the case when the work went to speed up the particle. Again equal to the change in the quantity 1/2*m*v2. The negative sign on w simply means that in this case the particle was doing work, rather than having work done on it. The moving particle apparently has an ability to do work that the particle at rest did not have.

The ability to do work is defined as "energy" and the ability of a particle to do work by virtue of its motion is defined as "kinetic energy". The kinetic energy of a particle of mass m moving at velocity v is in fact 1/2*m*v2. Remember that v2 is the scalar product v·v, so kinetic energy is a scalar even if we are working in more than one dimension. We will use the symbol ke to represent kinetic energy. What our calculations have shown is that the work done on a particle by the resultant or net force is equal to its change in kinetic energy. This result is known as the work-energy theorem. If something was at rest in our reference frame and is now in motion, you may be sure that some work was done on it.

Our uncovering of the work-energy theorem by working with Newton's laws is an example of what sometimes happens in physics. Newton's laws apply only in restricted circumstances out of the realm of quantum mechanics, very small systems, and relativistic mechanics, very fast systems. The work-energy theorem extends to these other realms also, so it is more fundamental or closer to Nature's truth than the laws we used to derive it.

Since work is force, whose units are Newtons, times displacement, whose units are meters, the units on work are Newton meters. The Newton meter is given the name Joule, in honor of Mr. Joule I believe (shown here), so we speak of work in terms of Joules. Kinetic energy must have the same units as work since the change in kinetic energy is equal to the work of a resultant force.

I have been making a point that our particle was unconstrained (free) so that it could respond to the applied force. Let's look now at a box of rocks of mass m=20kg sitting on the floor. What is the work involved in lifting it to a height of 1.5 meter. Assume that the box is lifted with a force of m*g where g is the acceleration due to gravity, 9.8 m/s2. The person lifting the box applies a force of 20kg*9.8m/s2 = 196N. The force is applied over a distance of 1.5 meters so the work of the lifter is 294J. If this were all there were to it, we would expect the change in kinetic energy to be 294 Joules. But the box arrives at the 1.5 meter altitude with zero velocity.

What we overlooked in the analysis so far was the interaction of the box with the Earth. It is true that the lifter did 294J of work on the box. At the same time, the Earth did -294J of work on the box, or the box did 294J of work on the Earth if you prefer that view of things. The work of the Earth on the box was 20kg*(-9.8m/s2)*1.5m. So the net work done in lifting the box was zero.

Mr. Joule
Pulling a load Next let's tie a rope to the 20kg box of rocks and drag it a distance of 2 meters along the floor. We will begin our observation of the box as soon as it is moving and drag it over the 2 meter mark at a constant velocity. The angle that the rope makes with the floor as we are dragging the box is 30 degrees. The force applied to the rope is 70N. What is the work done on the box?

The constant velocity condition in the problem tells us that there is no change in kinetic energy of the box so we know that the net work done on the box is zero. It appears that the professor has asked a trivial question. Sometimes we do that you know. Perhaps the question should have been what was the work done by the person pulling the rope. The force, f, on the rope was at a 30-degree angle but the displacement, r, of the end of the rope was horizontal so the work was f·r = |f|*|r|*cos(30) = 70N*2m*.866 = 121.24J .

So the rope puller did work but it did not show up as a change in kinetic energy of the box and since the elevation of the box never changed, the Earth did not undo the work of the rope puller. If that makes you curious about where the work went then you are thinking the right way about work and energy. Something must have done negative work on the box to balance the positive work done through the rope. Remember our old friend, friction? The force of friction which is always directed opposite to the displacement would always contribute negative work. The condition that the velocity was constant tells us that the friction force exactly balanced the force from the rope, otherwise there would have been some acceleration. So the combined work of the rope puller and the friction added up to zero.

What then must have been the coefficient of kinetic friction, mk, between the box and the floor? The normal force between the box and the floor is the weight of the box minus any lift provided by the rope. The weight is 20kg*9.8m/s2 = 196N. The rope pulls up on the box with a force 70N*sin(30) = 35N, so the net normal force is 196N-35N = 161N. Remember the magnitude of the frictional force is given by the equation |F| = mk * |N| , so mk = 70*cos(30)N / 161N = 0.3765. There is a fundamental difference between the lifted and dragged box situation, which we will explore in the next lesson.

For now let's look at another example of work, the work done by a spring. A spring is an object which exerts a force proportional to its displacement from some neutral or equilibrium position. The direction of that force is towards the neutral position. The Work by Spring display illustrates the effect.

A spring that has been stretched or compressed has had work done on it. The work is stored in the inter-atomic bonds that hold the spring together and determine its unstressed shape. When the spring is released, it exerts a force on any object attached to it, as the inter-atomic bonds in the spring material try to return to their lower energy configuration. This force times the distance the attached object is moved is the work done by the spring on the object. As we know, that work shows up as kinetic energy of the object. When the spring reaches its unstressed length, the object will have some kinetic energy which will carry it beyond the spring's preferred length, stressing the inter-atomic bonds for another half cycle of the object's motion.

In the next section we will continue the story of energy, taking a closer look at the spring and block system.

Are there any questions?

work by spring
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