An important general consequence of Newton’s laws is the Work-Energy Theorem. This theorem enables us, in many situations, to give an explicit relation between the speed of a particle and its location in space. For example, we can use the work-energy theorem to explicitly describe this relation for a freely falling body. In this exposition we will show that this theorem can be applied to a much wider class of examples.
Suppose a force is acting on a particle which undergoes a very small displacement . The work done by the force on the particle is defined as
where is the angle between and . Note that it makes no difference whether we take as the interior or exterior angle between and since . The work can be positive or negative, depending on whether is between and or between and . Thus, if you are pushing a box uphill on an inclined plane, you are doing positive work on the box and gravity is doing negative work. If you are restraining the box as it slides downhill you are doing negative work and gravity is doing positive work. Note that if (force is perpendicular to displacement) the force does no work. Therefore, if a particle moves on a smooth surface, the normal force exerted by the surface does no work on the particle.
The definition (1) of work can be used even if the displacement is not very small, provided that the force does not vary during the displacement. If varies, the definition (1) would be ambiguous (what value of should we use?) and the only “natural” and useful definition of work is the one which follows.
Suppose that a particle is subjected to a displacement, not necessarily small, from an initial position to a final position . We also specify the path taken by the particle, which is not necessarily a straight line. Conceptually, we can subdivide the path into a sequence of very small displacements , each of which is straight (see Figure 2). Let be the force acting on the particle when it undergoes the displacement . The work done on the particle during this short step is , and the total work done on the particle as it goes from to is defined as
where “” means that we are interested in the limiting value of the sum as the length of the steps becomes shorter and shorter and the number of terms in the sum consequently becomes larger and larger.
The limit which we have defined is clearly a generalization of the notion of an integral and is generally represented by the symbol which is usually referred to as the “line integral of from to .” Thus we can write
The student may find it useful to remember (2) rather than (3), since the former is easily visualized. Depending on the nature of the force , the right side of (3) may or may not have the same value for all paths between a given pair of endpoints and . In the special case where has the same value at all points on the path we have (using the distributive property of the dot product)
Calculate the work done by gravity on a particle which moves from an initial position to a final position . It is assumed that and are close enough to the earth’s surface, and to each other, so that the gravitational force is constant; that is, .
We can write and a similar expression for . Using (4) and , , we find
Note that the work done by gravity depends only on the initial and final positions and does not depend on what path the particle traversed between those positions. This is true even when we take account of the variation of the magnitude and the direction of the gravitational force when the particle moves through large distances (the work done by gravity is the same for all paths between two given points but is not in general given by (5)). The signs with which and enter (5) are easily remembered by noting that gravity does positive work on a particle which moves downward (force is parallel to displacement) and negative work on a particle which moves upward.
Let us consider a particle of mass which at one instant of time is at a position and has velocity , and at a later instant is at a position and has velocity . Let be the total work done on the particle as it goes from to . The Work-Energy Theorem asserts that
The quantity is called the kinetic energy of the particle; thus the work-energy theorem can be stated in words as:
The work done on a particle during any period of time is equal to the change1 in its kinetic energy.
To prove this theorem we start with Newton’s second law and take the dot product of both sides with the instantaneous velocity vector , obtaining
Using we find Thus If we multiply both sides of (7) by a very short time interval , we obtain
Since where is the displacement which the particle undergoes during the time interval , the left side of (8) is the work done on the particle during the time interval . The right side of (8) is just the change in the quantity during the time interval . Thus, the work done on the particle during any short time interval is equal to the change in its kinetic energy during this time interval. Subdividing the period from to into many short time intervals, we see that the total work done on the particle during this period is equal to the final kinetic energy minus the initial kinetic energy, which proves (6).
Application of the work-energy theorem to a freely falling particle yields
This result follows immediately from the equation of constant acceleration where is the initial velocity, is the final velocity, is the acceleration, is the initial displacement, and is the final displacement. We can now show that (9) is also true in many situations where the particle is not freely falling. Consider, for example, a particle moving under the influence of gravity on a smooth surface of arbitrary shape. In this case, both the magnitude and direction of the particle’s acceleration will usually vary with time; thus the analysis of motion with constant acceleration is not applicable.
However, the work-energy theorem is always applicable, provided we take care to calculate the total work done on the particle by all the forces acting on it. We note that a smooth surface exerts no force parallel to the surface. In this case there are only two forces acting on the particle: the gravitational force and the normal force which is exerted by the surface. As we have already noted, any force perpendicular to the surface can do no work and hence does no work since if is a small displacement in the surface. The only force which does work is the gravitational force, and (9) is therefore true.
If we let the “final” state in (9) be an arbitrary point in the motion of the particle, we can omit the subscript “f” and write
Equation (10) makes it clear that the speed of the particle depends only on its altitude (and on the initial values and ), and not on the shape of the surface. If the particle starts from rest it can never reach an altitude greater than its initial altitude, since (10) would yield a negative value of if .
A particle (mass ) slides on a surface of a smooth inverted hemisphere (radius ), starting from rest at the top (an infinitesimal push starts the motion). At the instant when it has descended through an angle , what is its speed and what is the magnitude of the force exerted by the hemisphere on the particle? At what value of does the particle fly off the surface?
If we take our origin at the center of the hemisphere, then and (10) yields . To calculate the force exerted by the hemisphere on the particle, we must use Newton’s second law (). Two forces act on the particle:
There are no other forces acting on the particle.
The acceleration vector of the particle has a component directed radially inward and a component directed tangentially (downhill). We are interested only in the radial component of , which yields
Inserting the expression for which we obtained from the work-energy theorem, we find
The formula (13) for makes sense in two respects: when it yields and as increases decreases. (Warning. Some students will write (12) immediately, without writing (11). The person who does this is almost certainly thinking of as a third force on the particle and will eventually become confused. One should always start by putting all the forces on one side of the equal sign and on the other side.)
At what value of does the particle fly off? Many (or even most) students have difficulty in stating the criterion which determines the point where the particle leaves the surface. It is important to recognize that the surface can only push on the particle and cannot pull on it. Examining (13) we see that when , for , and for . A negative value of means that the surface is required to pull radially inward on the particle. Since the surface cannot do this, the particle will fly off when (). Note that if we were discussing a bead sliding on a smooth wire bent into the shape of an inverted semicircle, the wire could (and would) supply the necessary inward force when .
Oscillations are usually approached via a differential equation, but the work-energy theorem is sufficient for a complete analysis of the pendulum as we show in the next example.
A pendulum consists of a point mass attached to the ceiling by a massless string of length . The pendulum swings back and forth, remaining always in the same vertical plane. The angular amplitude of the oscillation is (i.e., when the pendulum is at one extreme of its motion, the angle between the string and the vertical is ).
The force exerted by the string is directed along the string and is perpendicular to the velocity of the point mass. Thus, in any small time interval the displacment of the point mass is perpendicular to the force exerted by the string, and the string therefore does no work. Only gravity does work on the mass so we can use (10). If we choose “0” as the instant when the string makes the maximum angle with the vertical (and therefore ) and “f” as the instant when the string makes angle with the vertical and has speed , then (10) yields
In writing (14) we took our origin at the ceiling and used the relation . Thus we find
Equation (15) gives the speed at any point in the pendulum’s motion. During a complete oscillation the pendulum passes through each point twice, once going to the right and once going to the left.
The radial component of yields where is the tension in the string. Solving for and using (14) we find This says that is largest when and smallest when . If is very small, then both cosines in the formula for are close to 1 and as expected.
We can use (15) to calculate the period of the pendulum. As the string’s angle with the vertical changes from to the mass travels a distance . The time required for the mass to travel this distance is
The time required for the pendulum to go from its low point to one extreme of its oscillation is found by integrating the right side of (16) with respect to from to . This time is equal to one-fourth of the period . Thus we find