Work and Kinetic Energy

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 which 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 . 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 iswhere

x = (*v* + *v*_{0}) / 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* - *v*_{0})
/ Dt * (*v* + *
v*_{0}) / 2 * Dt .

Simplifying the preceding mess,
w = 1/2 * m * *v*^{2} - 1/2 *
m * *v*_{0}^{2} .

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
*v _{0}* 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*

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**v*^{2}. 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**v*^{2}. Remember that *
v*^{2} 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 theroem.

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 at left), 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/s^{2}. The person
lifting the box applies a force of 20kg*9.8m/s^{2} =
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/s^{2})*1.5m. So the net work done in lifting
the box was zero.

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 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,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, m|*F*| = m_{k} *
|*N*| , so m_{k} = 70N / 161N = 0.435
.

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.

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