Physics
Work Energy Theorem
The Work-Energy Theorem states that the work done on an object is equal to the change in its kinetic energy. In other words, the net work done on an object is equal to the change in its kinetic energy. This theorem provides a useful way to analyze the motion of objects and the energy transfers involved.
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12 Key excerpts on "Work Energy Theorem"
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2018(Publication Date)
- Wiley(Publisher)
6.2 The Work–Energy Theorem and Kinetic Energy 149 The work–energy theorem may be derived for any direction of the force relative to the displace- ment, not just the situation in Interactive Figure 6.5. In fact, the force may even vary from point to point along a path that is curved rather than straight, and the theorem remains valid. According to the work–energy theorem, a moving object has kinetic energy, because work was done to accelerate the object from rest to a speed υ f . † Conversely, an object with kinetic energy can perform work, if it is allowed to push or pull on another object. Example 4 illustrates the work–energy theorem and considers a single force that does work to change the kinetic energy of a space probe. † Strictly speaking, the work–energy theorem, as given by Equation 6.3, applies only to a single particle, which occupies a mathematical point in space. A macroscopic object, however, is a collection or system of particles and is spread out over a region of space. Therefore, when a force is applied to a macroscopic object, the point of application of the force may be anywhere on the object. To take into account this and other factors, a discussion of work and energy is required that is beyond the scope of this text. The interested reader may refer to A. B. Arons, The Physics Teacher, October 1989, p. 506. Analyzing Multiple-Concept Problems EXAMPLE 4 The Physics of an Ion Propulsion Drive The space probe Deep Space 1 was launched October 24, 1998, and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only small amounts of fuel. Suppose the probe, which has a mass of 474 kg, is traveling at an initial speed of 275 m/s. No forces act on it except the 5.60 × 10 ‒2 -N thrust of its engine. This external force F → is directed parallel to the displacement s → , which has a magnitude of 2.42 × 10 9 m (see Figure 6.6).
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
Kinetic energy, like work, is a scalar quantity. These are not surprising observations, because work and kinetic energy are closely related, as is clear from the following statement of the work–energy theorem. THEWORK– ENERGYTHEOREM WhenanetexternalforcedoesworkWonanobject,thekineticenergyofthe objectchangesfromitsinitialvalueofKE 0 toafinalvalueofKE f ,thedifference betweenthetwovaluesbeingequaltothework: W= KE f − KE 0 = 1 _ 2 m υ f 2 − 1 _ 2 m υ 0 2 (6.3) Work done by net ext. force *For extra emphasis, the final speed is now represented by the symbol υ f , rather than υ. INTERACTIVE FIGURE 6.5 A constant net external force Σ → Facts over a displacement → s and does work on the plane. As a result of the work done, the plane’s kinetic energy changes. s v 0 v f Final kinetic energy = m f 2 _ 1 2 1 2 Initial kinetic energy = m 0 2 _ ΣF ΣF υ υ 6.2 The Work–Energy Theorem and Kinetic Energy 163 The work–energy theorem may be derived for any direction of the force relative to the displacement, not just the situation in InteractiveFigure6.5. In fact, the force may even vary from point to point along a path that is curved rather than straight, and the theorem remains valid. According to the work–energy theorem, a moving object has kinetic energy, because work was done to accelerate the object from rest to a speed υ f . † Conversely, an object with kinetic energy can perform work, if it is allowed to push or pull on another object. Example 4 illustrates the work–energy theorem and considers a single force that does work to change the kinetic energy of a space probe. Description Symbol Value Comment ExplicitData Mass m 474 kg Initial speed υ 0 275 m/s Magnitude of force F 5.60 × 10 −2 N Magnitude of displacement s 2.42 × 10 9 m ImplicitData Angle between force → Fand displacement → s θ 0° The force is parallel to the displacement.
eBook - PDF- John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
In physics, when a net force performs work on an object, there is always a result from the effort. The result is a change in the kinetic energy of the object. As we will now see, the relation- ship that relates work to the change in kinetic energy is known as the work–energy theorem. This theorem is obtained by bringing together three basic concepts that we’ve already learned about. First, we’ll apply Newton’s second law of motion, SF 5 ma, which relates the net force SF to the acceleration a of an object. Then, we’ll determine the work done by the net force when the object moves through a certain distance. Finally, we’ll use Equation 2.9, one of the equations of kinematics, to relate the distance and acceleration to the initial and final speeds of the object. The result of this approach will be the work–energy theorem. To gain some insight into the idea of kinetic energy and the work–energy theorem, look at Figure 6.5, where a constant net external force S F B acts on an airplane of mass m. This net force is the vector sum of all the external forces acting on the plane, and, for simplicity, it is assumed to have the same direction as the displacement s B . According to Newton’s second law, the net force produces an acceleration a, given by a 5 SF/m. Check Your Understanding (The answers are given at the end of the book.) 1. Two forces F B 1 and F B 2 are acting on the box shown in the drawing, causing the box to move across the floor. The two force vectors are drawn to scale. Which one of the following statements is correct? (a) F B 2 does more work than F B 1 does. (b) F B 1 does more work than F B 2 does. (c) Both forces do the same amount of work. (d) Neither force does any work. 2. A box is being moved with a velocity v B by a force P B (in the same direction as v B ) along a level horizontal floor. The normal force is F B N , the kinetic frictional force is f B k , and the weight is m g B .
eBook - PDF- Richard C. Hill, Kirstie Plantenberg(Authors)
- 2013(Publication Date)
- SDC Publications(Publisher)
However, the work-energy method is not very useful for determining a body's acceleration or the direction of its velocity. The main idea with an energy approach is to analyze how a body's energy changes as it moves. The amount and form of a body's energy will change due to the body either doing work or due to work being done on the body. Conceptual Dynamics Kinetics: Chapter 7 – Particle Work and Energy 7 - 3 7.1) WORK Work is a concept that has meaning in our everyday lives. Everyone has done “work.” We “work” for a living, we “work”out and we “work” hard. This is not exactly what we will be talking about here. If you are doing physical “work” like pushing or lifting something, then you are doing the type of work that we will learn how to calculate in this chapter. In the study of dynamics, work has a very specific mathematical definition that is given by Equation 7.1-1. This equation gives the most general equation for work; however, work in its most basic form is simply a force applied over a distance (U = Fd ). So if you push a box across the floor from the living room to the kitchen, then you are doing work. However, if the box is very heavy and you push and push and sweat and sweat and the box does not move, then you have done no work. Even though you are exhausted, the box did not move and, therefore, the force you applied to the box did not do any work. Work: 2 1 1 2 U d r r F r (7.1-1) U 1-2 = work done by F from r 1 to r 2 (work is a scalar) F = force vector r = position vector Work is the amount of energy transferred by a force acting through a distance. One definition of work is “The amount of energy transferred by a force acting through a distance.” What does that mean in the context of dynamics? If a force is applied to a particle and the force causes that particle to move through a distance, the force has done work. This also means that the force has transferred some energy to (or from) the particle.
eBook - PDF- Michael Tammaro(Author)
- 2019(Publication Date)
- Wiley(Publisher)
The higher an object’s speed, the higher its kinetic energy. Kinetic energy, like work, is a scalar quantity—it has no direction. Its SI unit, like work, is joules (J). Since the speed v and mass m are positive numerically, kinetic energy is an inherently positive quantity. The Work–Energy Theorem Although Equation 6.2.1 was derived for an object moving in a straight line, where the net force was the frictional force, it is true in general. In the general case, it is known as the work–energy theorem: The Work-Energy Theorem The net work done on an object equals the change in the object’s kinetic energy: W K K K net f i = Δ = − (6.2.3) In Equation 6.2.3, W net is the work done by the net force, K i is the object’s initial kinetic energy, and K f is the object’s final kinetic energy. In other words, while the work W net is being done on the object, its kinetic energy changes from K i to K f . So far we have considered only constant forces, but Equation 6.2.3 is valid for forces that vary, too. If an object’s speed changes from v i to v f , then the change in its kinetic energy can be written explicitly as Δ = − K mv mv 1 2 f 2 1 2 i 2 (6.2.4) Given the forces acting on an object and its displacement, the net work can be written as W F r ( cos ) , net net θ = Δ where F net is the magnitude of the net force (assumed to be constant), r Δ is the magnitude of the displacement, and θ is the angle between the displacement and I N T E R A C T I V E F E A T U R E 150 | Chapter 6 the net force. Substituting this relationship for W net in Equation 6.2.3 and the relationship for ΔK in Equation 6.2.4, the work–energy theorem becomes θ ( ) Δ = − F r mv mv cos net 1 2 f 2 1 2 i 2 (6.2.5) Equation 6.2.5 makes intuitive sense. If the net force is zero, then there is no accelera- tion, and the kinetic energy does not change. But more generally, the kinetic energy does not change when the net force does no work. Is it possible for a nonzero net force to do no work? Yes.
eBook - PDFThe Engineering Dynamics Course Companion, Part 1
ParticlesKinematics and Kinetics
- Edward Diehl(Author)
- 2022(Publication Date)
- Springer(Publisher)
where KE 1 D Kinetic Energy at state 1 PE 1 D Potential Energy at state 1 U 1!2 D External Work acting on system between states 1 and 2 KE 2 D Kinetic Energy at state 2 PE 2 D Potential Energy at state 2 The Work-Energy equation might seem familiar to students who have taken or are taking Fluid Mechanics or Thermodynamics. You can think of this as the solid mechanics form of energy accounting. It’s good to use the following grouping to organize how you think of energy accounting: KE 1 C PE 1 „ ƒ‚ … Initial Energy C U 1!2 „ƒ‚… Outside Work Happens D KE 2 C PE 2 „ ƒ‚ … Final Energy : Note that this equation can be used in multiple locations in between, not just initial and final. We can break problems apart into stages, and this can be a very useful solution strategy in many problems. 8.2. WORK (U) 131 F F ∆x ∆y Figure 8.2: Newtdog works by applying forces in the direction of motion (©E. Diehl). Energy methods are powerful tools to solve many kinds of engineering problems, often as an alternative to other solution techniques that become excessively complicated. We’ll see that the kinetic energy change is closely related to N2L since it’s derived from it. We’ll start the derivation/explanation with the definition of work. 8.2 WORK (U) What is the definition of “work”? You might answer with an equation, but perhaps a layman’s answer like “effort to move stuff ” or “getting stuff done” is more descriptive. As a technical definition we might phrase this as “work is force through a distance.” There is emphasis on the word “through” because this is a necessary part of the concept. Work requires both force and movement, and only when the force and movement align is work done. In Figure 8.2, Newtdog is pushing the loaded wheel barrow up the hill. He’s specifically pushing the handles in the direction of motion. The force he’s exerting on the wheel barrow multiplied by the distance it moves in the direction of the force is the work he is applying.
eBook - PDF- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
In particular, you will see how the work-energy theorem is useful in relating the speeds of a particle, at different points along its trajectory, to the forces acting on it, even when the trajectory is otherwise too complicated to deal with. Thus, some aspects of motion can be addressed with fewer equations and without vector decompositions. Chapter 7 | Work and Kinetic Energy 327 7.1 | Work Learning Objectives By the end of this section, you will be able to: • Represent the work done by any force • Evaluate the work done for various forces In physics, work represents a type of energy. Work is done when a force acts on something that undergoes a displacement from one position to another. Forces can vary as a function of position, and displacements can be along various paths between two points. We first define the increment of work dW done by a force F → acting through an infinitesimal displacement d r → as the dot product of these two vectors: (7.1) dW = F → · d r → = | F → | | d r → | cos θ. Then, we can add up the contributions for infinitesimal displacements, along a path between two positions, to get the total work. Work Done by a Force The work done by a force is the integral of the force with respect to displacement along the path of the displacement: (7.2) W AB = ∫ path AB F → · d r → . The vectors involved in the definition of the work done by a force acting on a particle are illustrated in Figure 7.2. Figure 7.2 Vectors used to define work. The force acting on a particle and its infinitesimal displacement are shown at one point along the path between A and B. The infinitesimal work is the dot product of these two vectors; the total work is the integral of the dot product along the path. We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle between them, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes and angles.
eBook - PDF- Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
F B CHAPTER 11 CHAPTER 11 ENERGY 1: WORK AND KINETIC ENERGY W e have seen how Newton’s laws are useful in un- derstanding and analyzing a wide variety of problems in mechanics. In this and the following two chapters we consider a different approach based on one of the truly fundamental and universal concepts in physics: energy. There are many kinds of energy. In this chapter we consider one particular form — kinetic energy, the energy associated with a body because of its motion. We also introduce the concept of work, which is re- lated to kinetic energy through the work – energy theorem. This theorem, derived from Newton’s laws, pro- vides new and different insight into the behavior of mechanical systems. In Chapter 12 we introduce a sec- ond kind of energy — potential energy — and begin developing a conservation law for energy. In Chapter 13 we discuss energy in a more comprehensive way and generalize the law of conservation of energy, which is one of the most useful laws of physics. application moves through some distance, and one way to define the energy of a system is a measure of its capacity to do work. In the case of the wheelchair rider, he does work because he exerts a force as the wheelchair moves forward through some distance. For him to do work, he must ex- pend some of his supply of energy — that is, the chemical energy stored in his muscle fibers — which can be replen- ished from his body’s store of energy through resting and which ultimately comes from the food he eats. The energy stored in a system may take many forms: for example, chemical, electrical, gravitational, or mechanical. In this chapter we study the relationship between work and one particular type of energy — the energy of motion of a body, which we call kinetic energy. 11- 2 WORK DONE BY A CONSTANT FORCE Figure 11-2a shows a block of mass m being lifted through a vertical distance h by a winch that is turned by a motor.
eBook - PDFApplied Mathematics
Made Simple
- Patrick Murphy(Author)
- 2014(Publication Date)
- Butterworth-Heinemann(Publisher)
The simplest mechanical example is the typical hydroelectric scheme where a lake of water is dammed and the water flow then regulated to feed through tunnels to drive machinery in order to generate electricity. This is a straightforward conversion of the water's potential energy into kinetic energy. But our task here is to summarize these results into a form for applying energy principles to solve our problems. For simplicity we restrict ourselves to two methods of solution. Method 1: In any motion of a body where the force of gravity is the only force on the body which does any work, the sum of the potential energy and the kinetic energy is constant. We express this in the form F p + E k = constant Method 2: If the body moves under the action of a system of constant forces, let this system have a resultant of magnitude R. Since this acts on the body it follows from Newton's Second Law that R = ma, where a is the acceleration of the body. If the body moves a distance s under the action of the resultant R, then the work done by all the forces in the system is the same as the work done by the resultant R of the system, and the work done is Rs. From the equation v 2 = u 2 + las multiplied by m we have mas — imv 2 — mu 2 Rs = mas = imv 2 — mu 2 We see therefore that the change in kinetic energy of the body is equal to the total work done by all the forces acting on the body. We shall now apply these two methods to some problems. Example: A bullet of mass 15 g is fired into a fixed block of wood with a horizontal velocity of 400 m s 1 . If the bullet comes to rest after penetrating a distance of 80 mm, calculate the resistance of the block assuming the resistance to be constant. SOLUTION: The only two forces on the bullet are its weight and the resistance F of the block. The work done by the weight is zero since the displacement of the bullet is at right-angles to the line of action of mg.
No longer available |Learn more- Irving Granet, Maurice Bluestein(Authors)
- 2014(Publication Date)
- CRC Press(Publisher)
As stated earlier, work is a transitory effect and is neither a property of a system nor stored in a system. There is one process, however, that does permit the evalua-tion of the work done, because the path is uniquely defined. This process is frictionless and quasi-static, and we shall find it useful in subsequent discussions. To describe this process, we first define equilibrium state in the manner given by Hatsopoulos and Keenan (1961): A state is an equilibrium state if no finite rate of change can occur without a finite change, temporary or permanent, in the state of the environment. The term permanent change of state refers to one that is not canceled out before completion of the process. Therefore, the frictionless quasi-static process can be identified as a succession of equilibrium states. Involved in this definition is the concept of a process carried out infinitely slowly, so that it is in equilibrium at all times. The utility of the frictionless, quasi-static process lies in our ability to evaluate the work terms involved in it, because its path is uniquely defined. Energy or work done per unit time is called power or the rate of energy change. Energy, work, and power units as well as their conversions are detailed in Table 1.8. 2.4 Internal Energy To this point, we have considered the energy in a system that arises from the work done on the system. However, it was noted in Chapter 1 and earlier in this chapter that a body possesses energy by virtue of the motion of the molecules of the body. In addition, it pos-sesses energy due to the internal attractive and repulsive forces between particles. These forces become the mechanism for energy storage whenever particles become separated, such as when a liquid evaporates or the body is subjected to a deformation by an external energy source. Also, energy may be stored in the rotation and vibration of the molecules.
eBook - PDF- Joaquim A. Batlle, Ana Barjau Condomines(Authors)
- 2022(Publication Date)
- Cambridge University Press(Publisher)
5.21 338 Work–Energy Theorem mechanical energy of the system E RGal T RGal þ U RGal ð Þ, a scalar variable depending only on the system’s state. In a system where only the conservative forces do a nonzero work, the mechanical energy remains constant: there is a conservation of mechanical energy: Δ T RGal þ U RGal ð Þ 2 1 ¼ 0 ) E RGal ¼ T RGal þ U RGal ¼ constant: (5.38) In those systems, if T RGal increases, U RGal decreases, and the other way round. Taking into account that interactions through drivers, constraint forces and friction phenomena are not conservative, the WET may be written as: ΔE RGal syst ð Þ 2 1 ¼ W drivers RGal 2 1 þ W RGal F const from moving obst: i 2 1 þ W RGal F kin:fric: from fixed obst: i þ W RGal F kin:fric: from moving obst: i 2 1 þ W drivers 2 1 þ W kin:fric: 2 1 : (5.39) From all the terms in the right-hand side of Eq. (5.39), the only ones with a definite sign (always < 0) are the third and the last one (kinetic frictions with fixed obstacles and internal kinetic frictions, respectively), the other terms may increase or decrease the mechanical energy. Potential Energy Increment Associated with a Uniform Gravitational Field As already mentioned, the weight of a system is a conservative force (relative to the rigid body R generating the gravitational attraction) under the assumption of constant gravitational field. For a system with mass m and center of mass G, the potential energy increment between two G locations with a height difference h is (Fig. 5.22): ΔU weight R i 2 1 ¼ W weight R i 2 1 ¼ mgh ð Þ ¼ mgh: (5.40) The gravitational potential energy increases when G goes up (relative to the initial location), and decreases when it goes down. The height difference which may have either sign) has to be expressed as a function of the system coordinates. R P 1 h P 2 G g mg Fig. 5.22 339 5.9 Potential Energy ► Example 5.11 An elevator goes up with variable speed relative to the ground (E).
eBook - PDF- David Agmon, Paul Gluck;;;(Authors)
- 2009(Publication Date)
- WSPC(Publisher)
(b) In keeping with the remarks of part a we write the work done by friction as heat energy, so that in the course of coming to a halt the kinetic energy of the body as a whole is completely converted into heat. Energy conservation gives mV 2 /2 = | W f | = jumgL from which L = V 2 /2^g. This example leads us to the following generalization of (6.10). When a body is acted upon by both external conservative and non-conservative forces F c and F nc , doing work W c ^ { and W nc ,ext, respectively, the total work must equal the change in kinetic energy Recall now from (6.7) that A£ p = -W c ^ xt . Combining these two equations we obtain W nc , exl =AE p +AE k (6.12) Note that A£ p is the sum total of the changes of all forms of potential energy (gravitational, elastic, etc.). In words (6.12) says that the work done by non-conservative forces equals the sum of the changes of the potential and kinetic energies. If part of that work appears as heat it may be transferred to the right-hand side and identified as heat, as we did in the present example. This theorem is also valid for a many-body system, in which case A£ p and A£ k refer to the sum total of changes for all the bodies in the system. Example 18 The work of friction and heat. A small block of mass m rests on top of one side of a double slope of angle a and is released from rest from a height H, as shown. It is given that H tan a> jU s > jU k . The block moves to and from alternate slopes until it comes to rest, passing the joint smoothly without loss of speed, (a) Where will it come to rest? (b) What is the total distance covered before it comes to a halt? Chapter 6 Work and Energy 197 Solution (a) Since tan a> jU s > jU k , the only possible point at which the block can be in equilibrium is the lowest point P. (b) One could use energy or dynamical considerations to find the distance 5 covered till the first stop, repeat for all subsequent stops, and sum all the contributions.
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