73. Kinetic Energy and the WorkEnergy TheoremLearning Objectives
Work Transfers EnergyWhat happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Section 72(a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Section 72(d) is stored in the briefcaseEarth system and can be recovered at any time, as shown in Section 72(e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stoneEarth system and has the potential to do work. In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion. Net Work and the WorkEnergy TheoremWe know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion. Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, Figure 1(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an $F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $ vs. $d$ graph. In this case, $F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $ is constant. You can see that the area under the graph is $Fd\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $, or the work done. Figure 1(b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force $(F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta {)}_{i(\text{ave})}$. The work done is $(F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta {)}_{i(\text{ave})}{d}_{i}$ for each strip, and the total work done is the sum of the ${W}_{i}$. Thus the total work done is the total area under the curve, a useful property to which we shall refer later. Net work will be simpler to examine if we consider a onedimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 2. The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force ${\mathbf{F}}_{\text{app}}$ and the horizontal friction force $\mathbf{f}$. Thus, as expected, the net force is parallel to the displacement, so that $\theta =\mathrm{0\xba}$ and $\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =1$, and the net work is given by $${W}_{\text{net}}={F}_{\text{net}}d.$$The effect of the net force ${\mathbf{F}}_{\text{net}}$ is to accelerate the package from ${v}_{0}$ to $v$. The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 1.) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting ${F}_{\text{net}}=\text{ma}$ from Newton’s second law gives $${W}_{\text{net}}=\text{mad.}$$To get a relationship between net work and the speed given to a system by the net force acting on it, we take $d=x{x}_{0}$ and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance $d$ if the acceleration has the constant value $a$; namely, ${v}^{2}={{v}_{0}}^{2}+2\text{ad}$ (note that $a$ appears in the expression for the net work). Solving for acceleration gives $a=\frac{{v}^{2}{{v}_{0}}^{2}}{2d}$. When $a$ is substituted into the preceding expression for ${W}_{\text{net}}$, we obtain $${W}_{\text{net}}=m\left(\frac{{v}^{2}{{v}_{0}}^{2}}{2d}\right)d.$$The $d$ cancels, and we rearrange this to obtain $$W{}_{\text{net}}\text{}=\frac{1}{2}{\text{mv}}^{2}\frac{1}{2}{\text{mv}}_{0}^{2}\text{.}$$This expression is called the The WorkEnergy Theorem:The net work on a system equals the change in the quantity $\frac{1}{2}{\text{mv}}^{2}$. $${W}_{\text{net}}=\frac{1}{2}{\text{mv}}^{2}\frac{1}{2}{\text{mv}}_{0}^{2}$$The quantity $\frac{1}{2}{\text{mv}}^{2}$ in the workenergy theorem is defined to be the translational is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together. We are aware that it takes energy to get an object, like a car or the package in Figure 2, up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why highspeed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy. Example 1: Calculating the Kinetic Energy of a PackageSuppose a 30.0kg package on the roller belt conveyor system in Figure 2 is moving at 0.500 m/s. What is its kinetic energy? Strategy Because the mass $m$ and speed $v$ are given, the kinetic energy can be calculated from its definition as given in the equation $\text{KE}=\frac{1}{2}{\text{mv}}^{2}$. Solution The kinetic energy is given by $$\text{KE}=\frac{1}{2}{\text{mv}}^{2}\text{.}$$Entering known values gives $$\text{KE}=0\text{.}5(\text{30.0 kg})(\text{0.500 m/s}{)}^{2},$$which yields $$\text{KE}=\text{3.75 kg}\cdot {\mathrm{m}}^{2}{\text{/s}}^{2}=\text{3.75 J.}$$Discussion Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves. Example 2: Determining the Work to Accelerate a PackageSuppose that you push on the 30.0kg package in Figure 2 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N. (a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force. Strategy and Concept for (a) This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 2.) As expected, the net work is the net force times distance. Solution for (a) The net force is the push force minus friction, or ${F}_{\text{net}}\text{= 120 N \u2013 5}\text{.}\text{00 N = 115 N}$. Thus the net work is $$\begin{array}{lll}{W}_{\text{net}}& =& {F}_{\text{net}}d=\left(\text{115 N}\right)\left(\text{0.800 m}\right)\\ & =& \text{92.0 N}\cdot \mathrm{m}=\text{92.0 J.}\end{array}$$Discussion for (a) This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force. Strategy and Concept for (b) The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work. Solution for (b) The applied force does work. $$\begin{array}{lll}{W}_{\text{app}}& =& {F}_{\text{app}}d\phantom{\rule{0.25em}{0ex}}\text{cos}\left(\mathrm{0\xba}\right)={F}_{\text{app}}d\\ & =& \left(\text{120 N}\right)\left(\text{0.800 m}\right)\\ & =& \text{96.0 J}\end{array}$$The friction force and displacement are in opposite directions, so that $\theta =\text{180\xba}$, and the work done by friction is $$\begin{array}{lll}{W}_{\text{fr}}& =& {F}_{\text{fr}}d\phantom{\rule{0.25em}{0ex}}\text{cos}\left(\text{180\xba}\right)={F}_{\text{fr}}d\\ & =& \left(\text{5.00 N}\right)\left(\text{0.800 m}\right)\\ & =& \text{4.00 J.}\end{array}$$So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively, $$\begin{array}{lll}{W}_{\text{gr}}& =& \mathrm{0,}\\ {W}_{\mathrm{N}}& =& \mathrm{0,}\\ {W}_{\text{app}}& =& \text{96.0 J,}\\ {W}_{\text{fr}}& =& \text{4.00 J.}\end{array}$$The total work done as the sum of the work done by each force is then seen to be $${W}_{\text{total}}={W}_{\text{gr}}+{W}_{\mathrm{N}}+{W}_{\text{app}}+{W}_{\text{fr}}=\text{92}\text{.0 J}.$$Discussion for (b) The calculated total work ${W}_{\text{total}}$ as the sum of the work by each force agrees, as expected, with the work ${W}_{\text{net}}$ done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach. Example 3: Determining Speed from Work and EnergyFind the speed of the package in Figure 2 at the end of the push, using work and energy concepts. Strategy Here the workenergy theorem can be used, because we have just calculated the net work, ${W}_{\text{net}}$, and the initial kinetic energy, $\frac{1}{2}{m{v}_{0}}^{2}$. These calculations allow us to find the final kinetic energy, $\frac{1}{2}{\text{mv}}^{2}$, and thus the final speed $v$. Solution The workenergy theorem in equation form is $${W}_{\text{net}}=\frac{1}{2}{\text{mv}}^{2}\frac{1}{2}{m{v}_{0}}^{2}\text{.}$$Solving for $\frac{1}{2}{\text{mv}}^{2}$ gives $$\frac{1}{2}{\text{mv}}^{\text{2}}={W}_{\text{net}}+\frac{1}{2}{m{v}_{0}}^{2}\text{.}$$Thus, $$\frac{1}{2}{\text{mv}}^{2}=\text{92}\text{.}\mathrm{0\; J}+3\text{.}\text{75 J}=\text{95.}\text{75 J.}$$Solving for the final speed as requested and entering known values gives $$\begin{array}{lll}v& =& \sqrt{\frac{2\text{(95.75 J)}}{m}}=\sqrt{\frac{\text{191.5 kg}\cdot {\mathrm{m}}^{2}{\text{/s}}^{2}}{\text{30.0 kg}}}\\ & =& \text{2.53 m/s.}\end{array}$$Discussion Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package. Example 4: Work and Energy Can Reveal Distance, TooHow far does the package in Figure 2 coast after the push, assuming friction remains constant? Use work and energy considerations. Strategy We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing. Solution The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so $\theta =\text{180\xba}$. To reduce the kinetic energy of the package to zero, the work ${W}_{\text{fr}}$ by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus ${W}_{\text{fr}}=\text{95}\text{.}\text{75 J}$. Furthermore, ${W}_{\text{fr}}=fd\prime \phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \phantom{\rule{0.25em}{0ex}}\text{= \u2013}fd\prime $, where $d\prime $ is the distance it takes to stop. Thus, $$d\prime =\frac{{W}_{\text{fr}}}{f}=\frac{\text{95.75 J}}{\text{5.00 N}},$$and so $$d\prime =\text{19}\text{.2 m}.$$Discussion This is a reasonable distance for a package to coast on a relatively frictionfree conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy. Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone. Section Summary
Conceptual QuestionsExercise 1The person in Figure 3 does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it lose energy? Figure 3. Exercise 2Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement. Exercise 3When solving for speed in Example 3, we kept only the positive root. Why? Problems & ExercisesExercise 1Compare the kinetic energy of a 20,000kg truck moving at 110 km/h with that of an 80.0kg astronaut in orbit moving at 27,500 km/h. Show/Hide Solution Solution$1/\text{250}$ Exercise 2(a) How fast must a 3000kg elephant move to have the same kinetic energy as a 65.0kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates. Exercise 3Confirm the value given for the kinetic energy of an aircraft carrier in Section 77. You will need to look up the definition of a nautical mile (1 knot = 1 nautical mile/h). Show/Hide Solution Solution$1\text{.}1\times {\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{J}$ Exercise 4(a) Calculate the force needed to bring a 950kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a nonpanic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a). Exercise 5A car’s bumper is designed to withstand a 4.0km/h (1.1m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900kg car to rest from an initial speed of 1.1 m/s. Show/Hide Solution Solution$2\text{.}8\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ Exercise 6Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent’s face, if the glove and face compress 7.50 cm during a blow in which the 7.00kg arm and glove are brought to rest from an initial speed of 10.0 m/s. (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used and the knuckles and face would compress only 2.00 cm. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove? Exercise 7Using energy considerations, calculate the average force a 60.0kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him. Show/Hide Solution Solution102 N
