11.1.6.1: Illustrations

Illustration 1: Dot Products

We talk about work as the amount of force in the direction of an object's displacement, \(\Delta \mathbf{x}\), multiplied by the displacement, \(\Delta x\). No displacement, no work. Work is positive if \(\mathbf{F}\) and \(\Delta \mathbf{x}\) point in the same direction and negative if \(\mathbf{F}\) and \(\Delta \mathbf{x}\) point in opposite directions. This statement is fine if the force and displacement lie on the same axis (in other words lie on a line). What happens if they do not lie on the same axis? When working in two dimensions, the force and the displacement can point in any direction. So how much of a given force is in the direction of the displacement? (We could consider the equivalent description of the amount of the displacement in the direction of the force.)

To answer this question we must use the mathematical construction of the scalar or dot product. The dot product is defined as the scalar product of two vectors: \(\mathbf{A}\text{ dot }\mathbf{B}=\mathbf{A}\cdot\mathbf{B}=AB\cos(\theta )\), where \(\theta\) is the angle between the two vectors and \(A\) and \(B\) are the magnitudes of the vectors \(\mathbf{A}\) and \(\mathbf{B}\), respectively. Restart.

Drag the tip of either arrow (position is given in meters). The red arrow is \(\mathbf{A}\) and the green arrow is \(\mathbf{B}\). The magnitude of each arrow is shown and the dot product is calculated. When is the dot product zero? The dot product is zero when the vectors are perpendicular. For any two vectors, the magnitude of the dot product is a maximum when the vectors point in the same (or opposite) directions and a minimum when the vectors are perpendicular. Note also that the assignment of which vector is \(\mathbf{A}\) and which vector is \(\mathbf{B}\) does not matter.

So the dot product has the right properties to help us mathematically describe WORK. In general, for a constant force,

\[\text{WORK} =\mathbf{F}\cdot\Delta\mathbf{x} = F \Delta x \cos(\theta ),\nonumber\]

where \(\mathbf{F}\) is the constant force and \(\Delta\mathbf{x}\) is the displacement. \(F\) and \(\Delta x\) are the magnitude of the vectors, respectively. You may have heard or seen "\(\text{WORK} = Fd\)" which is not always correct. That statement ignores the vector properties of \(\mathbf{F}\) and \(\mathbf{x}\) and can lead you into thinking that the definition of \(\text{WORK}\) is \(\text{FORCE}\) times \(\text{DISTANCE}\), which it is not.

Illustration 2: Constant Forces

This Illustration shows a block on a table subjected to an applied force and a frictional force. The mass of the block, the force applied, and the coefficient of kinetic friction can be controlled by typing values for these variables into the text boxes and pressing the "set values and play" button. The coefficient of static friction is fixed at \(\mu_{s} = 0.4\). Restart.

Consider the animation with a \(100\text{-kg}\) block and vary the force applied from \(0\) to \(391\text{ N}\). What happens in the animation? The block does not move. What is the work done on the block due to the applied force? What is the work done on the block due to the force of static friction? What is the work done on the block due to the normal force? What is the work done on the block due to the force of gravity? The work done by each of these forces is zero. How do you know? First there is no displacement. If there is no displacement there can be no work. In addition, the normal force and the force of static friction can never do work. The normal force cannot do work because if there is a displacement it would always be perpendicular to the normal force, hence no work. The force of static friction can also never do work. When there is static friction there can never be a displacement. After all, static friction implies the block is static and thus not moving. Since there is no work done on the block, the block's kinetic energy cannot change.

Now consider the animation with a \(100\text{-kg}\) block and an applied force of \(446\text{ N}\). What happens in the animation? The block moves and, in fact, it accelerates. What is the work done on the block due to the applied force? What is the work done by the force of kinetic friction? What is the work done on the block due to the normal force? What is the work done on the block due to the force of gravity? The net work done by the sum of these forces is now not zero. How do you know? There is a change in kinetic energy. This can only happen when there is work done on the block. The force of gravity does not do any work on the block because the force is perpendicular to the displacement in this animation. The normal force, as said above, can never do work. The work done by the applied force will be positive.

The force of kinetic friction reduces the kinetic energy of the block by \(| F_{\text{k friction}}\:\Delta x |\) because the frictional force and the block's displacement are in opposite directions. Kinetic friction will always oppose motion, so it will always reduce kinetic energy. Note that we do not say the work done on the block by kinetic friction. This phrase is not correct. The work done by friction on the block is the energy the block loses, and this is not equal to \(- F_{\text{k friction}}\:\Delta x\). Some of the kinetic energy dissipated by friction, \(| F_{\text{k fricion}}\:\Delta x |\), is transferred to the table as thermal energy (the table heats up), while some of it remains with the block as thermal energy (the block heats up). Therefore, \(-F_{\text{k friction}}\:\Delta x\) is not the work done on the block by friction: \(- F_{\text{k friction}}\:\Delta x\) is the total work done by friction (done on the block and the table) and is the amount that the kinetic energy is reduced.

The change in kinetic energy of the block is the net force \(F_{\text{applied}}- F_{\text{k friction}}\) times the displacement (this is because the net force points in the direction of the displacement). Therefore, in the table, \(F_{\text{net}}\) times \(x = KE\) (because \(x\) is the displacement since the block starts at \(x = 0\text{ m}\) and the final \(KE\) is the change in \(KE\) since the block starts with no kinetic energy).

Illustration authored by Mario Belloni.
Script authored by Steve Mellema and Chuck Niederriter and modified by Mario Belloni.

Illustration 3: Force and Displacement

Two forces that we often use as examples when talking about work are the force of gravity and the elastic force of springs. We know from earlier chapters that the character of the two forces is different. The gravitational force on an object is always \(mg\), while the spring force is dependent on how much the spring is stretched or compressed from equilibrium. As a consequence, the form for the work done by each force will be different. Restart.

In general, for a constant force, \(\text{WORK} = \mathbf{F}\cdot\Delta\mathbf{x}= F\Delta x \cos(\theta )\), where \(\mathbf{F}\) is the constant force and \(\Delta\mathbf{x}\) is the displacement. \(F\) and \(\Delta x\) are the magnitude of the vectors, respectively.

The graph shows \(F\cos(\theta )\) vs. distance for a \(1\text{-kg}\) object near the surface of Earth (position is given in meters). By checking the box you make the graph represent the \(F\cos(\theta )\) vs. distance graph for a mass on a spring with \(k = 2\text{ N/m}\) (the equilibrium point of the spring is conveniently set at \(x = 0\text{ m}\)). You can enter in values for the starting and stopping points for the calculation of the work and then click the "evaluate area (integral)" button to calculate the work.

Begin by looking at the \(F\cos(\theta )\) vs. distance graph for gravity (\(F_{y}\) vs. \(y\)). Gravity is a constant force (near the surface of Earth). Therefore, the magnitude of work done by gravity will be \(| mg\Delta y |\). So, consider a ball at \(y = 0\text{ m}\) that drops to \(y = -2\text{ m}\). Is the work done by gravity positive or negative? Use the graph to calculate the work. It is indeed positive (a negative force in the \(y\) direction and a negative displacement in the \(y\) direction means \(\cos(\theta ) = 1\)). This is because the force is in the same direction as the displacement. What about lifting an object up from \(y = -2\text{ m}\) to \(y = 0\text{ m}\)? The work done is negative since the force is in the opposite direction from the displacement \([\cos(\theta ) = -1]\). We can use \(| F\Delta y |\) because the force does not vary over the displacement. But what if the force does vary, as in the case of a spring?

Check the check box to see the graph representing a spring force. Enter in values for the starting and stopping points for the calculation of the work and then click the "evaluate area (integral)" button. Enter in \(x = 0\text{ m}\) for the starting point and \(x = 4\text{ m}\) for the ending point, representing the stretching of a spring. Is the magnitude of the work done \(| F\Delta x |\)? Why or why not? The magnitude of the work is not \(| F\Delta x |\). In the case of the spring, the magnitude of the work is \(0.5\ast k x^{2}\), which is the area under the force function (it is also the integral of \(F dx\)). Note also that the work is negative: The force and the displacement are in the opposite direction \([\cos(\theta ) = -1]\).

Enter in \(x = 4\text{ m}\) for the starting point and \(x = 0\text{ m}\) for the ending point. What happens to the sign of the work done by the spring now?

Illustration 4: Springs

The fact that the spring force varies with position means that while we can determine the force, we cannot determine the velocity of an object attached to a stretched spring using kinematic equations for constant acceleration. Why? The force is not constant (it varies with position) and therefore the acceleration is not constant. What can we do? We can use the work-energy theorem.

The spring-ball system shown in the animation can be stretched by click-dragging the \(1\text{-kg}\) dark blue ball (position is given in meters and time is given in seconds). The black arrow attached to the ball shows the net, i.e., total, force on the ball. The pale blue ball on the left is the free-body diagram for the dark blue ball. The red and green arrows attached to the pale blue ball show the spring and gravitational forces, respectively. The acceleration due to gravity is \(9.8\text{ m/s}^{2}\) in this animation. Restart.

Hooke's law states that the force that the spring exerts is \(F = -k x\), where \(k\) is the spring constant and \(x\) is measured from the equilibrium position of the spring. In this Illustration the initial position of the spring and the spring constant can be changed by using the text boxes.

So how do we determine the work done by the spring? We need to calculate the integral of \(F\cos(\theta )\Delta x\), where \(F\) and \(\Delta x\) are the magnitude of the force vector and the displacement vector, respectively. We must calculate the integral because the force is not a constant.

Consider \(k = 2\text{ N/m}\) and \(y = 5\text{ m}\) and run the animation. Initially the spring is compressed and the net force points down and the infinitesimal displacement points down; therefore, \(\cos(\theta ) = 1\). We determine that the work done by the spring is initially positive, yielding an increasing kinetic energy. After passing through equilibrium (\(y = 0.1\text{ m}\) for this \(k\)), however, the net force is now upward, while the infinitesimal displacement still points down. Thus \(\cos(\theta ) = -1\), and the work is negative. Therefore, the kinetic energy decreases until the spring is maximally extended and \(v = 0\text{ m/s}\). The process then reverses with a positive amount of work until the mass again passes through equilibrium and the spring does a negative amount of work until the mass is at rest at its starting position \(y = 5\text{ m}\). The process repeats indefinitely if there are no resistive forces.

Illustration 5: Circular Motion

A \(1\text{-kg}\) black ball is constrained to move in a circle as shown in the animation (position is given in meters, time is given in seconds, and energy on the bar graph is given in joules). Restart. In the no external force animation the wire is horizontal on a frictionless tabletop and the force of the wire is the only force that acts. In the gravity animation the wire is vertical and the ball is subjected to gravity (downward as usual) as well as the force of the wire. You may set the initial velocity and then choose either configuration. The blue arrow represents the net force acting on the mass, while the bar graph displays its kinetic energy in Joules.

For no external force, select various initial velocities and then set v and play: no external force. In what direction does the net force point? Here the only force acting is the force of the wire pulling on the ball to make it go in a circle. The direction of this force is always toward the center of the circle (a centripetal force). With this force, does the black ball's speed change? No. The ball's velocity changes, but its speed does not. The work-energy theorem tells us that since there is no work done by the force of the wire (its force is perpendicular to the ball's displacement) there can be no change in the ball's kinetic energy. In general, any centripetal force cannot ever do work.

For gravity, select various initial velocities and then set v and play: gravity. In what direction does the net force point now? Well, this is a bit more complicated. There is a force toward the center of the circle as there was before (again due to the wire), but now there is also the force of gravity downward. Therefore, the net force does not point toward the center of the circle any more. With this force, does the black ball's speed change? Yes. While the part of the force due to the wire cannot do any work, the part of the force that is due to gravity can and does do work.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.