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Transcript
Note 08 Work and Kinetic Energy Sections Covered in the Text: Chapters 10 & 11, except 11.7. Chapter 12 read only. In this note we begin the study of mechanical energy and the ways in which mechanical energy can be transferred from one system to another. Energy is another word used in everyday speech that has a very specific and subtly different definition in physics. We shall see that certain problems in physics can be solved more easily by considering them from the point of view of mechanical energy than by applying Newton’s laws directly. We examine the concept of work and the relationship between work and mechanical energy. We begin with the idea of a system and its environment. A System and its Environment In physics it is useful to envoke the concept of a system and its environment. A system is usually an object under study, while an environment is what lies outside a system and what is therefore irrelevant to the problem. In any problem it is useful to identify the system clearly so as not to waste time with effects that have no bearing on the question being asked. For example, in Figure 8-1 is shown two views of an object falling toward the Earth. In (a) is shown the object itself and the force exerted on it: the force of gravity. In (b) is shown a wider perspective, including the Earth, the source of the force. The force the Earth exerts on the object is shown as well as the force the object exerts on the Earth (via the third law). For a certain class of problems the system is just the object itself, modelled as a particle. Even though the object is falling toward the Earth, the Earth plays no role in the problem other than to provide the force on the object. object object m But in other types of problems, and especially those pursued from the point of view of energy, the system must be taken as is shown in Figure 8-1b; the Earth must be included. This idea may not be clear at the moment, but we shall enlarge upon it presently in an effort to make it clearer. Work Done by the Agent of a Constant Force Definition of Work We all have a notion of the meaning of work as we use the word everyday. We say that the mastery of physics takes a lot of “work”. Perhaps you have had the experience of pushing on an object to try to make it move when it just would not move. Even if the object did not move you might claim to have done a lot of “work”, especially if the effort produced a good sweat. In physics, work is defined very precisely and in a way that is subtly different from its usage in everyday speech. If you push on an object and the object does not move, then according to its definition in physics, you do no work! The work done is non-zero only if the object on which the force is exerted moves through some displacement, as we shall see in the next section. Work Done by the Agent of a Constant Force We first define work in the simplest situation, when the force applied is constant—constant in the sense of being independent of time or position. Consider a particle subject to a constant force F moving through a displacement Δr along a straight line that makes an angle θ with Δr (Figure 8-2). m earth (a) Me (b) Figure 8-1. An effort to illustrate what is meant by a system and its environment. The Earth is part of the environment in (a), whereas in (b) it is part of the system. Figure 8-2. A constant force F is shown moving an object through some displacement. The agent of the force does an amount of work. 08-1 Note 08 The work done on an object by the agent of a force can be defined in these words:1 The work W done by an agent exerting a constant force of magnitude F on a particle is the product of the component Fcos θ of the force along the direction of the displacement of the point of application of the force and the magnitude ∆r of the displacement: W = FΔr cosθ . In Figure 8-3a, F is perpendicular to the displacement and therefore does no work (θ = π/2 and cosθ = 0). In (b) and (d) the work done by F is negative (cos θ is –ve since θ > π/2). Only in (c) is the work done by F positive. Let us consider an example. Example Problem 8-1 Calculating Work in a Simple Case …[8-1] Work is a scalar with dimensions M.L2.s–2 and units newton.meter (N.m). 1 newton.meter is given the name joule (abbreviated J) in honor of a 19th century physicist. Thus according to this definition if ∆r = 0, then W = 0, consistent with our earlier statement. But work defined by eq[8-1] depends on θ as well as on ∆r. Depending on the value of θ, work can be positive, negative or zero. To illustrate this, consider four cases of an object moving to the right (Figure 83). In each case a force of the same magnitude F is applied to the object, but in different directions. Is the work done by the agent of the force the same in all cases? A constant force of 1.50 N is applied at an angle of 60˚ above the horizontal to a 2.00 kg block in contact with a horizontal frictionless surface (Figure 8-4). The block moves a distance of 0.500 m along the surface. Calculate the work done by the agent of the force during this movement. 1.50 N 60˚ 2.00 kg Figure 8-4. A force applied to an object. Solution: Applying the definition of work done by the agent of a constant force, eq[8-1], we have W = FΔr cosθ = (1.50N)(0.500m)cos60 o = 0.375 J. It is useful to note that the work done is independent of the mass of the object. The most elegant definition of work uses the scalar dot product (discussed briefly in Note 02). In preparation we digress with a little vector analysis. Math Unit 6: Vector Analysis Figure 8-3. Four cases when a force of magnitude F is applied to an object in different directions. In each case the object is assumed to be moving to the right. 1 The work done is often referred to as the work done by the force. But the work is done by the source or agent of the force not the force itself. And the agent of the force is the environment. We shall assume this if the phrase “work done by the force” appears in these notes from now on. 08-2 Using the scalar (dot) product we can frame a more compact and general definition of work than is given by eq[8-1]. Scalar Product of Two Vectors As described in Note 02, the dot product of any two vectors A and B is the scalar quantity A • B ≡ AB cosθ , …[8-2] Note 08 where A and B are the magnitudes of the two vectors and θ is the smaller of the two angles between them (Figure 8-5). Solution: We shall use the definition of the vector dot product. First note that the magnitude of A is A = Ax2 + Ay2 + Az2 = 1+ 4 + 9 = 14 , B = 29 . and of B: € Figure 8-5. Two vectors. It is useful to express the vectors in terms of unit vectors. Recall that the concept of a unit vector was introduced in Note 02 for the purpose of indicating direction. We shall see that the following relationships between the dot products of the unit vectors will prove useful: ) ) ) ) ) ) i • i = j • j = k • k = 1, Therefore from eqs[8-2] and [8-3]: € A• B cos θ = AB = Ax Bx + Ay By + Az Bz = 0.992583 . AB Thus θ = ArcCos(0.992583) = 7˚. The smaller of the two angles between the two vectors is 7 ˚. ) ) ) ) ) ) i • j =i •k = j •k = 0. Using the vector dot product we can rewrite the definition eq[8-1] in the form: Thus any two vectors in 3D space can be expressed in component form as W = F • Δr . and ) ) ) A = Axi + Ay j + Azk and ) ) ) B = Bx i + By j + Bzk . Thus the scalar product of these vectors is A • B = Ax Bx + Ay By + Az Bz . …[8-3] We also have for the dot product of a vector with itself the result A • A = Ax2 + Ay2 + Az2 = A 2 . Let us consider an example. Example Problem 8-2 Finding the Angle Between Two Vectors Find the smallest angle between the vectors ) ) ) A = 1.00i + 2.00 j + 3.00k and ) ) ) B = 2.00i + 3.00 j + 4.00k . …[8-4] We repeat, eq[8-4] applies only if the force is constant . If the force is variable in the sense of depending on time or position then the work done must be found using calculus. This more general case we examine next. Work Done by the Agent of a Varying Force Work can be done by a force that varies with time, position or both. We begin by focussing our attention on a force that depends only on x. Consider a particle being moved along the x-axis from xi to xf by a force that depends in some way on x. In general, this force may be in any direction. Here we consider only the x-component of the force. The xcomponent might have the form shown in Figure 8-6a. To recycle the idea of a constant force and to utilize eq[8-4] we imagine the x-domain to be divided into a large number of equal intervals. Let us focus our attention on an arbitrary (nth) interval; call it ∆x. If the interval is narrow enough, then the x-component of force that applies over that interval is nearly a constant with some average value, say F x. The work done in that interval is the product: Wn ≈ Fx Δx . 08-3 Note 08 Now the area under the curve bounded by ∆x is just the width times the average height, or F x∆x. Thus the work done in the nth interval equals the area that interval occupies under the force curve (Figure 8-6a). We can find an approximate value for the total work done over the whole range by adding up the contributions of work done over all of these intervals. The result is: xf xf W = lim Wapprox = lim ∑ Fx Δx = ∫x Fx dx . Δx→ 0 Thus Δ x →0 i xi xf W = ∫x Fx dx , …[8-6a] i or, written in function form xf W = ∫x F(x)dx . …[8-6b] i According to calculus, this expression exactly equals the area under the force curve between the limits xi and xf (Figure 8-6b). For most purposes in a first course in physics, eq[86b] is sufficient for calculating the work done by a varying force. However, our treatment takes account of just one force. If a number of forces act on the object simultaneously then the expression for the force in eq[8-6] must be replaced by a sum of forces: xf Wnet = ∫x i (∑ F )dx . x And finally, the most general form of the work done allows for multiple forces acting in different directions (a situation likely to occur only in an advanced course in physics). This general expression involves the vector dot product: r2 Wnet = ∫r 1 Figure 8-6. How the work done by a force that varies with x is found using calculus. The total work done between the two positions xi and xf is equal to the area under the force curve between those positions. xf Wapprox = ∑ Fx Δx . …[8-5] xi This is only an approximation because the ∆xs are of finite size. To increase the accuracy we can take the limit of this expression as we make the width of the intervals vanishingly small, i.e., as ∆x → 0. The result is 08-4 (∑ F) • dr . …[8-7] A good example of a single varying force is the force exerted by a spring. This case we consider in the next section. Work Done by a Spring A classic example of a force that varies in a simple way with displacement is the force exerted by a spring on a mass connected to its end (Figure 8-7a). For convenience we assume that the mass rests on a horizontal frictionless surface. Robert Hooke (a contemporary of Isaac Newton) found experimentally that if the mass is displaced a small amount x from equilibrium (the position x = 0), then the spring exerts a force on the mass given by Fs (x) = –kx . …[8-8] k is a constant called the force constant of the spring and has units N.m–1. k is set by the construction of the Note 08 spring, being large for a “hard” spring and small for a “soft” spring.2 The negative sign in eq[8-8] indicates that the force is a restoring force—restoring in the sense that it tends to restore the mass to the system’s equilibrium position. If the mass is displaced to the right of the equilibrium position, then the force exerted by the spring is directed to the left. If the mass is displaced to the left of the equilibrium position, then the force exerted by the spring is directed to the right. Over small displacements, the variation of the force is linear, varying with x as sketched in Figure 8-7b. the area under the force curve between x = –xmax and x = 0. It is positive since the force and the displacement are in the same direction. This procedure can be applied to calculate the work done by the spring when the mass moves between any two positions, say xi and xf. The result is xf Ws = 1 ∫ (–kx)dx = 2 kx 2 i 1 2 – kx 2f . xi …[8-9] Clearly, in a movement of the mass from x = – xmax to x = +x max the work done by the spring is zero. In the segment between x = –xmax to x = 0, the spring does a positive amount of work. In the segment between x = 0 and x = +x max the spring does a negative amount of work. Let us consider a numerical example. Example Problem 8-4 Work Done by a Spring A mass is connected to the end of a spring in an arrangement as shown in Figure 8-7a. The spring constant is k = 0.020 N.m–1. The mass is released from a position x i = 0.25 m. How much work is done by the spring when the mass moves to the position xf = – 0.10 m? Solution: Using eq[8-9] we have 1 2 W = k(xi2 – x 2f ) Figure 8-7. In (a) is shown a mass m attached to the end of a spring. In (b) is shown a sketch of how the spring force varies with x. The variation is linear within the spring’s elastic limit. We can calculate the work done by the spring on the mass over any movement of the mass. Let us suppose that the mass is released from a position of x = –xmax. We can calculate the work done by the spring force as the mass moves from this position to the equilibrium position x = 0. According to eq[8-6] we have xf Ws = ∫ xi 0 Fs (x)dx = ∫ – xmax (–kx)dx = 1 2 kx2max . As can be seen from Figure 8-7b this result is equal to = 1 (0.020N.m–1 )[(0.25m) 2 – (–0.10m)2 ] 2 = 5.25 x 10–4 J. The spring does a positive amount of work. In physics, the work done on an object in some process is interpreted as the energy acquired by the object.3 This relationship is laid out in the so-called work-energy theorem. An object can possess energy by virtue of its motion. This kind of energy is called kinetic energy, as we shall see next. (In Note 09 we shall see that an object can also possess energy by virtue of its position relative to some reference 3 2 We emphasize for completeness that eq[8-8] applies only if the spring is not deformed or stretched beyond its elastic limit. And we are talking here about mechanical energy. The subject of the internal energy of an object will be discussed in the second half of this course. 08-5 Note 08 position.) difference between two terms of the form Kinetic Energy and the Work-Kinetic Energy Theorem In this section we introduce the idea of kinetic energy. Consider an object being moved in a straight line over a frictionless horizontal surface by a force or sum of forces ΣF that is constant (Figure 8-8). For convenience we assume that the force and displacement are in the same direction. In a certain elapsed time ∆t the object moves through a displacement ∆x. Since the force is constant the object accelerates continuously and so its speed increases from, say, vi to vf. Figure 8-8 An object is moved by the agent of a constant force and undergoes an acceleration and an increase in speed. We have seen that the work done on the object by the agent of the constant net force is given by xf Wnet = ∫x i ∑ Fdx . Substituting ΣF = ma and using the chain rule in calculus we have xf xf i i Wnet = ∫x madx = ∫x m xf dv dv dx dx = ∫x m dx i dt dx dt xf K= 1 2 mv2 . …[8-11] We can therefore write Wnet = K f – Ki = ΔK . …[8-12] K involves the square of the speed of the object and is therefore called kinetic energy. Kinetic energy is therefore the mechanical energy a body possesses by virtue of its motion. The unit of kinetic energy is the same as the unit of work, namely joule (J). Thus the workkinetic energy theorem states in so many words that the work done on an object (in a situation as is shown in Figure 8-8) equals the increase in the object’s kinetic energy. This result enables any number of physics problems to be solved in terms of mechanical energy rather than via the kinematic equations of motion. Sometimes the energy approach leads to a solution more quickly, other times the kinematic equations are faster. It depends on the problem. Let’s consider an example. Example Problem 8-4 Finding the Speed of an Object A block of mass 2.00 kg in a situation as shown in Figure 8-8 is initially at rest on a horizontal frictionless surface. A force of 2.00 N is applied to the block in the direction shown in the figure for 5.00 s. Calculate the final speed of the block. Solution: Applying the second law to the block, the block’s acceleration is a= F (2.00N) = = 1.00 m.s–2. m (2.00kg) The distance travelled in 5.00 s is = ∫x mvdv . Δx = ut + at 2 = 0 + (1.00m.s –2 )(5.00s)2 This amounts to a change in variable from x to v. Thus = 12.5m . 1 2 i 1 2 Wnet = mv 2f – 1 2 mvi2 . …[8-10] Since vf > vi, Wnet is positive. The work done equals the 08-6 1 2 Applying the work-energy theorem, the work done equals the increase in the block’s kinetic energy: Note 08 1 2 uct of the net applied force vector and the velocity vector. This result is useful in the solving of certain problems. FΔx = mv 2 or 1 2 (2.00N)(12.5m) = (2.00kg)v2 Example Problem 8-5 Average Power v = 5.00 m.s–1. Solving for v: It should be obvious this problem could have been solved just as well by kinematics alone. Power In physics, we are often interested in the rate at which energy changes, or is transferred from one system to another, as much as in the energy itself. The rate at which work is done or energy is transferred is called power. Two types of power are defined: average power and instantaneous power. Average Power If an amount of work ∆W is done in an elapsed time ∆t then the average power is defined as the work done divided by the elapsed time: P= ΔW . Δt The units of average power are joules per second (j.s ). 1 J.s –1 is given the special unit Watt (abbreviated W). Instantaneous Power The instantaneous power is defined as the limit of the average power as ∆t → 0: Δt →0 Δt → 0 ΔW dW = . Δt dt …[8-14] The unit of instantaneous power is also J.s–1 or Watt. Instantaneous power can also be expressed in terms of force and velocity. Substituting the definition of work, eq[8-7], into eq[8-14] we have P= dW dr =F• = F•v dt dt P= ΔW 2.00(J) = = 4.00 Watts. Δt 0.500(s) In the elapsed time given, energy is being transferred from the agent of the force to the object at the average rate of 4.00 joules per second. Example Problem 8-6 Instantaneous Power In the example shown in Figure 8-4 what is the instantaneous power at a clock time of 2.00 s? Solution: We intend to apply the expression given in eq[8-15b], namely P = Fv cos θ . To do this we need to find the velocity at the clock time of 2.00 s. Using kinematics we have v = 0 + at = Thus …[8-15b] (1.50N)cos60o (2.00s) m.s –1 (2.00kg) = 0.75 m.s–1. …[8-15a] or since this is a vector dot product, P = Fv cos θ , Solution: According to the definition of average power, eq[813], …[8-13] –1 P = lim P = lim The source of a force applied to an object does 2.00 J of work in an elapsed time of 0.500 s. Calculate the average power. P = (1.50N)(0.75m.s –1 )cos60o = 0.563 W. At the precise clock time of 2.00 s, energy is being transferred at the rate of 0.563 joules per second. where θ is the smallest angle between F and v . Thus the instantaneous power equals the vector dot prod- 08-7 Note 08 To Be Mastered • • • • • Definitions: system, environment Definitions: work done by a constant force, work done by a varying force Definitions: unit vector, vector dot product Definitions: work-kinetic energy theorem, kinetic energy Definitions: average power, instantaneous power Typical Quiz/Test/Exam Questions 1. (a) State Hooke’s Law. (b) Define the spring constant of an elastic spring. (c) Explain how a stretched spring can be used to measure force. 2. (a) Define what is meant by the work done on an object. (b) Give an example of the relationship between work and one type of mechanical energy. Illustrate your answer by means of a sketch showing all important quantities. 3. (a) Define what is meant by the work done by a constant force on an object. (b) Write down an expression for the work done by a force F(x) between xi and xf. 4. A block of mass 1.0 kg is in contact with a horizontal surface (see the figure). The coefficient of kinetic friction between the block and the surface is 0.6 N.kg–1. A constant force of 2.0 N is applied to the block in the direction shown. When the block moves a distance of 1.0 m along the surface how much work has been done by the agent of the force? 2.0 N 60˚ 1.0 kg 1.0 m 5. A block of mass 1.0 kg is in contact with an inclined plane (see the figure). The coefficient of kinetic friction between the block and the plane is 0.6 N.kg–1. A constant force of 2.0 N is applied to the block in the direction shown. When the block moves a distance of 1.0 m up the incline how much work has been done by the agent of the force? 08-8 Note 08 2.0 N 1.0 kg 1.0 m 30˚ 08-9