Work

We have seen that when a mass free-falls under the influence of gravity some of its kinetic energy is transformed into potential energy, or vice versa. Let us now investigate, in detail, how this transformation is effected. The mass falls because it is subject to a downwards gravitational force of magnitude $m g$. It stands to reason, therefore, that the transformation of kinetic into potential energy is a direct consequence of the action of this force.

This is, perhaps, an appropriate point at which to note that the concept of gravitational potential energy--although extremely useful--is, strictly speaking, fictitious. To be more exact, the potential energy of a body is not an intrinsic property of that body (unlike its kinetic energy). In fact, the gravitational potential energy of a given body is stored in the gravitational field which surrounds it. Thus, when the body rises, and its potential energy consequently increases by an amount ${\mit\Delta}U$; in reality, it is the energy of the gravitational field surrounding the body which increases by this amount. Of course, the increase in energy of the gravitational field is offset by a corresponding decrease in the body's kinetic energy. Thus, when we speak of a body's kinetic energy being transformed into potential energy, we are really talking about a flow of energy from the body to the surrounding gravitational field. This energy flow is mediated by the gravitational force exerted by the field on the body in question.

Incidentally, according to Einstein's general theory of relativity (1917), the gravitational field of a mass consists of the local distortion that mass induces in the fabric of space-time. Fortunately, however, we do not need to understand general relativity in order to talk about gravitational fields or gravitational potential energy. All we need to know is that a gravitational field stores energy without loss: i.e., if a given mass rises a certain distance, and, thereby, gives up a certain amount of energy to the surrounding gravitational field, then that field will return this energy to the mass--without loss--if the mass falls by the same distance. In physics, we term such a field a conservative field (see later).

Suppose that a mass $m$ falls a distance $h$. During this process, the energy of the gravitational field decreases by a certain amount (i.e., the fictitious potential energy of the mass decreases by a certain amount), and the body's kinetic energy increases by a corresponding amount. This transfer of energy, from the field to the mass, is, presumably, mediated by the gravitational force $-m g$ (the minus sign indicates that the force is directed downwards) acting on the mass. In fact, given that $U=m g h$, it follows from Eq. (127) that

$${\mit\Delta}K = f {\mit\Delta h}.$$ (131)

In other words, the amount of energy transferred to the mass (i.e., the increase in the mass's kinetic energy) is equal to the product of the force acting on the mass and the distance moved by the mass in the direction of that force.

In physics, we generally refer to the amount of energy transferred to a body, when a force acts upon it, as the amount of work $W$ performed by that force on the body in question. It follows from Eq. (131) that when a gravitational force $f$ acts on a body, causing it to displace a distance $x$ in the direction of that force, then the net work done on the body is

$$W = f x.$$ (132)

It turns out that this equation is quite general, and does not just apply to gravitational forces. If $W$ is positive then energy is transferred to the body, and its intrinsic energy consequently increases by an amount $W$. This situation occurs whenever a body moves in the same direction as the force acting upon it. Likewise, if $W$ is negative then energy is transferred from the body, and its intrinsic energy consequently decreases by an amount $\vert W\vert$. This situation occurs whenever a body moves in the opposite direction to the force acting upon it. Since an amount of work is equivalent to a transfer of energy, the mks unit of work is the same as the mks unit of energy: namely, the joule.

In deriving equation (132), we have made two assumptions which are not universally valid. Firstly, we have assumed that the motion of the body upon which the force acts is both 1-dimensional and parallel to the line of action of the force. Secondly, we have assumed that the force does not vary with position. Let us attempt to relax these two assumptions, so as to obtain an expression for the work $W$ done by a general force ${\bf f}$.

Let us start by relaxing the first assumption. Suppose, for the sake of argument, that we have a mass $m$ which moves under gravity in 2-dimensions. Let us adopt the coordinate system shown in Fig. 35, with $z$ representing vertical distance, and $x$ representing horizontal distance. The vector acceleration of the mass is simply ${\bf a}=(0,-g)$. Here, we are neglecting the redundant $y$-component, for the sake of simplicity. The physics of motion under gravity in more than 1-dimension is summarized by the three equations (64)-(66). Let us examine the last of these equations:

$$v^2 = v_0^{ 2} + 2 {\bf a}\!\cdot\!{\bf s}.$$ (133)

Here, $v_0$ is the speed at $t=0$, $v$ is the speed at $t=t$, and ${\bf s} = ({\mit\Delta}x$, ${\mit\Delta}z)$ is the net displacement of the mass during this time interval. Recalling the definition of a scalar product [i.e., ${\bf a}\!\cdot\!{\bf b}= (a_x b_x+ a_y b_y+a_z b_z)$], the above equation can be rearranged to give

$$\frac{1}{2} m v^2 - \frac{1}{2} m v_0^{ 2} = -m g {\mit\Delta}z.$$ (134)

Since the right-hand side of the above expression is manifestly the increase in the kinetic energy of the mass between times $t=0$ and $t=t$, the left-hand side must equal the decrease in the mass's potential energy during the same time interval. Hence, we arrive at the following expression for the gravitational potential energy of the mass:

$$U = m g z.$$ (135)

Of course, this expression is entirely equivalent to our previous expression for gravitational potential energy, Eq. (125). The above expression merely makes manifest a point which should have been obvious anyway: namely, that the gravitational potential energy of a mass only depends on its height above the ground, and is quite independent of its horizontal displacement.

Figure 35: Coordinate system for 2-dimensional motion under gravity

Let us now try to relate the flow of energy between the gravitational field and the mass to the action of the gravitational force, ${\bf f} = (0,-m g)$. Equation (134) can be rewritten

$${\mit\Delta} K = W = {\bf f}\!\cdot {\bf s}.$$ (136)

In other words, the work $W$ done by the force ${\bf f}$ is equal to the scalar product of ${\bf f}$ and the vector displacement ${\bf s}$ of the body upon which the force acts. It turns out that this result is quite general, and does not just apply to gravitational forces.

Figure 36 is a visualization of the definition (136). The work $W$ performed by a force ${\bf f}$ when the object upon which it acts is subject to a displacement ${\bf s}$ is

$$W = \vert{\bf f}\vert \vert{\bf s}\vert \cos\theta.$$ (137)

where $\theta$ is the angle subtended between the directions of ${\bf f}$ and ${\bf s}$. In other words, the work performed is the product of the magnitude of the force, $\vert{\bf f}\vert$, and the displacement of the object in the direction of that force, $\vert{\bf s}\vert \cos\theta$. It follows that any component of the displacement in a direction perpendicular to the force generates zero work. Moreover, if the displacement is entirely perpendicular to the direction of the force (i.e., if $\theta =90^\circ$) then no work is performed, irrespective of the nature of the force. As before, if the displacement has a component in the same direction as the force (i.e., if $\theta<90^\circ$) then positive work is performed Likewise, if the displacement has a component in the opposite direction to the force (i.e., if $\theta>90^\circ$) then negative work is performed.

Figure 36: Definition of work

Suppose, now, that an object is subject to a force ${\bf f}$ which varies with position. What is the total work done by the force when the object moves along some general trajectory in space between points $A$ and $B$ (say)? See Fig. 37. Well, one way in which we could approach this problem would be to approximate the trajectory as a series of $N$ straight-line segments, as shown in Fig. 38. Suppose that the vector displacement of the $i$th segment is ${\mit\Delta}{\bf r}_i$. Suppose, further, that $N$ is sufficiently large that the force ${\bf f}$ does not vary much along each segment. In fact, let the average force along the $i$th segment be ${\bf f}_i$. We shall assume that formula (136)--which is valid for constant forces and straight-line displacements--holds good for each segment. It follows that the net work done on the body, as it moves from point $A$ to point $B$, is approximately

$$W \simeq \sum_{i=1}^{N} {\bf f}_i\!\cdot\!{\mit\Delta}{\bf r}_i.$$ (138)

We can always improve the level of our approximation by increasing the number $N$ of the straight-line segments which we use to approximate the body's trajectory between points $A$ and $B$. In fact, if we take the limit $N\rightarrow\infty$ then the above expression becomes exact:

$$W = \lim_{N\rightarrow\infty} \sum_{i=1}^{N} {\bf f}_i\!\cdo... ...\Delta}{\bf r}_i = \int_A^B {\bf f}({\bf r})\!\cdot\!d{\bf r}.$$ (139)

Here, ${\bf r}$ measures vector displacement from the origin of our coordinate system, and the mathematical construct $\int_A^B {\bf f}({\bf r})\!\cdot\!d{\bf r}$ is termed a line-integral.

Figure 37: Possible trajectory of an object in a variable force-field

Figure 38: Approximation to the previous trajectory using straight-line segments

The meaning of Eq. (139) becomes a lot clearer if we restrict our attention to 1-dimensional motion. Suppose, therefore, that an object moves in 1-dimension, with displacement $x$, and is subject to a varying force $f(x)$ (directed along the $x$-axis). What is the work done by this force when the object moves from $x_A$ to $x_B$? Well, a straight-forward application of Eq. (139) [with ${\bf f} = (f,0,0)$ and $d{\bf r}=(dx,0,0)$] yields

$$W = \int_{x_A}^{x_B} f(x) dx.$$ (140)

In other words, the net work done by the force as the object moves from displacement $x_A$ to $x_B$ is simply the area under the $f(x)$ curve between these two points, as illustrated in Fig. 39.

Figure 39: Work performed by a 1-dimensional force

Let us, finally, round-off this discussion by re-deriving the so-called work-energy theorem, Eq. (136), in 1-dimension, allowing for a non-constant force. According to Newton's second law of motion,

$$f = m \frac{d^2x}{dt^2}.$$ (141)

Combining Eqs. (140) and (141), we obtain

$$W = \int_{x_A}^{x_B} m \frac{d^2 x}{dt^2} dx = \int_{t_A}^... ...}\!\left[\frac{m}{2}\!\left(\frac{dx}{dt}\right)^2\right] dt,$$ (142)

where $x(t_A)=x_A$ and $x(t_B)=x_B$. It follows that

$$W = \frac{1}{2} m v_B^{ 2} -\frac{1}{2} m v_A^{ 2} = {\mit\Delta}K,$$ (143)

where $v_A = (dx/dt)_{t_A}$ and $v_B = (dx/dt)_{t_B}$. Thus, the net work performed on a body by a non-uniform force, as it moves from point $A$ to point $B$, is equal to the net increase in that body's kinetic energy between these two points. This result is completely general (at least, for conservative force-fields--see later), and does not just apply to 1-dimensional motion.

Suppose, finally, that an object is subject to more than one force. How do we calculate the net work $W$ performed by all these forces as the object moves from point $A$ to point $B$? One approach would be to calculate the work done by each force, taken in isolation, and then to sum the results. In other words, defining

$$W_i = \int_A^B {\bf f}_i({\bf r})\!\cdot\!d{\bf r}$$ (144)

as the work done by the $i$th force, the net work is given by

$$W = \sum_i W_i.$$ (145)

An alternative approach would be to take the vector sum of all the forces to find the resultant force,

$${\bf f} = \sum_i {\bf f}_i,$$ (146)

and then to calculate the work done by the resultant force:

$$W = \int_A^B {\bf f}({\bf r})\!\cdot\!d{\bf r}.$$ (147)

It should, hopefully, be clear that these two approaches are entirely equivalent.